Answer:
(1) The probability that the time to deliver a pizza is at least 32 minutes is 0.70.
(2a) The percentage of results more than 45 is 79.67%.
(2b) The percentage of results less than 85 is 91.77%.
(2c) The percentage of results are between 75 and 90 is 15.58%.
(2d) The percentage of results outside the healthy range 20 to 100 is 2.64%.
Step-by-step explanation:
(1)
Let Y = the time taken to deliver a pizza.
The random variable Y follows a Uniform distribution, U (20, 60).
The probability distribution function of a Uniform distribution is:
[tex]f(x)=\left \{ {{\frac{1}{b-a};\ x\in [a, b] } \atop {0};\ otherwise} \right.[/tex]
Compute the probability that the time to deliver a pizza is at least 32 minutes as follows:
[tex]P(Y\geq 32)=\int\limits^{60}_{32} {\frac{1}{b-a} } \, dx \\=\frac{1}{60-20} \int\limits^{60}_{32} {1 } \, dx\\=\frac{1}{40}\times[x]^{60}_{32}\\=\frac{1}{40}\times[60-32]\\=0.70[/tex]
Thus, the probability that the time to deliver a pizza is at least 32 minutes is 0.70.
(2)
Let X = results of a certain blood test.
It is provided that the random variable X follows a Normal distribution with parameters [tex]\mu = 60[/tex] and [tex]s = 18[/tex].
The probabilities of a Normal distribution are computed by converting the raw scores to z-scores.
The z-scores follows a Standard normal distribution, N (0, 1).
(a)
Compute the probability that the results are more than 45 as follows:
[tex]P(X>45)=P(\frac{X-\mu}{\sigma}> \frac{45-60}{18})=P(Z>-0.833)=P(Z<0.833)=0.7967[/tex]
The percentage of results more than 45 is: [tex]0.7967\times100=79.67\%[/tex]
Thus, the percentage of results more than 45 is 79.67%.
(b)
Compute the probability that the results are less than 85 as follows:
[tex]P(X<85)=P(\frac{X-\mu}{\sigma}< \frac{85-60}{18})=P(Z<1.389)=0.9177[/tex]
The percentage of results less than 85 is: [tex]0.9177\times100=91.77\%[/tex]
Thus, the percentage of results less than 85 is 91.77%.
(c)
Compute the probability that the results are between 75 and 90 as follows:
[tex]P(75<X<90)=P(\frac{75-60}{18}<\frac{X-\mu}{\sigma}< \frac{90-60}{18})\\=P(0.833<Z<1.67)\\=P(Z<1.67)-P(Z<0.833)\\=0.9525-0.7967\\=0.1558[/tex]
The percentage of results are between 75 and 90 is: [tex]0.1558\times100=15.58\%[/tex]
Thus, the percentage of results are between 75 and 90 is 15.58%.
(d)
Compute the probability that the results are between 20 and 100 as follows:
[tex]P(20<X<100)=P(\frac{20-60}{18}<\frac{X-\mu}{\sigma}< \frac{100-60}{18})\\=P(-2.22<Z<2.22)\\=P(Z<2.22)-P(Z<-2.22)\\=0.9868-0.0132\\=0.9736[/tex]
Then the probability that the results outside the range 20 to 100 is: [tex]1-0.9736=0.0264[/tex].
The percentage of results outside the range 20 to 100 is: [tex]0.0264\times100=2.64\%[/tex]
Thus, the percentage of results outside the healthy range 20 to 100 is 2.64%.
Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist. (Enter your answer using interval notation.) (t − 5)y' + (ln t)y = 6t, y(1) = 6
Answer:
0 < t < 5 is the required interval for the differential equation (t - 5)y' + (ln t)y = 6t to have a solution.
Step-by-step explanation:
Given the differential equation
(t - 5)y' + (ln t)y = 6t
and the condition y(1) = 6
We can rewrite the differential equation by dividing it by (t - 5) as
y' + [(ln t)/(t - 5)]y = 6t/(t - 5)
(ln t)/(t - 5) is continuous on the interval (0, 5) and (5, +infinity).
6t/(t - 5) is continuous on (-infinity, 5) and (5, +infinity)
We see that for these expressions, we have continuity at the intervals (0, 5) and (5, +infinity).
But the initial condition is y = 6, when t = 1.
The solution to differential equation is certain to exist at (0, 5)
Which implies that
0 < t < 5
is the required interval.
Determine whether the series is convergent or divergent. 1 2 3 4 1 8 3 16 1 32 3 64 convergent divergent Correct:
The given series is an alternating harmonic series. It is convergent because the sequence of absolute values decreases to zero.
Explanation:The given series is: 1/2 , 3/4 , 1/8 , 3/16 , 1/32 , 3/64 . . . As you can see, the numerator alternates between 1 and 3, while the denominator doubles each time. This type of series is known as an alternating harmonic series.
To determine if an alternating series converges or diverges, we can apply the Alternating Series Test. This test says that an alternating series converges if the sequence of absolute values decreases to zero. In our case, the absolute value of the terms based solely on the denominator is decreasing to zero, hence, the series converges.
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Because of safety considerations, in May 2003 the Federal Aviation Administration (FAA) changed its guidelines for how small commuter airlines must estimate passenger weights. Under the old rule, airlines used 180 lb as a typical passenger weight (including carry-on luggage) in warm months and 185 lb as a typical weight in cold months. The Alaska Journal of Commerce (May 25, 2003) reported that Frontier Airlines conducted a study to estimate average passenger plus carry-on weights. They found an average summer weight of 183 lb and a winter average of 190 lb. Suppose that each of these estimates was based on a random sample of 100 passengers and that the sample standard deviations were 20 lb for the summer weights and 23 lb for the winter weights. A. Construct and interpret a 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers.B. Construct and interpret a 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers.C. The new FAA recommendations are 190 lb for sum- mer and 195 lb for winter. Comment on these recommen- dations in light of the confidence interval estimates from Parts (a) and (b).
Answer:
a) [tex]183-1.984\frac{20}{\sqrt{100}}=179.032[/tex]
[tex]183+1.984\frac{20}{\sqrt{100}}=186.968[/tex]
So on this case the 95% confidence interval would be given by (179.032;186.968)
b) [tex]190-1.984\frac{23}{\sqrt{100}}=185.437[/tex]
[tex]190+1.984\frac{23}{\sqrt{100}}=194.563[/tex]
So on this case the 95% confidence interval would be given by (185.437;194.563)
c) For Summer the confidence interval was (179.032;186.968) and as we can see our upper limit is <190 so then we can conclude that they are below the specification of 190 at 5% of significance
For Winter the confidence interval was (185.437;194.563) and again the upper limit is <190 so then we can conclude that they are below the specification of 195 at 5% of significance
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n represent the sample size
Part a : Summer
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=100-1=99[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,99)".And we see that [tex]t_{\alpha/2}=1.984[/tex]
Now we have everything in order to replace into formula (1):
[tex]183-1.984\frac{20}{\sqrt{100}}=179.032[/tex]
[tex]183+1.984\frac{20}{\sqrt{100}}=186.968[/tex]
So on this case the 95% confidence interval would be given by (179.032;186.968)
Part b: Winter
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=100-1=99[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,99)".And we see that [tex]t_{\alpha/2}=1.984[/tex]
Now we have everything in order to replace into formula (1):
[tex]190-1.984\frac{23}{\sqrt{100}}=185.437[/tex]
[tex]190+1.984\frac{23}{\sqrt{100}}=194.563[/tex]
So on this case the 95% confidence interval would be given by (185.437;194.563)
Part c
For Summer the confidence interval was (179.032;186.968) and as we can see our upper limit is <190 so then we can conclude that they are below the specification of 190 at 5% of significance
For Winter the confidence interval was (185.437;194.563) and again the upper limit is <190 so then we can conclude that they are below the specification of 195 at 5% of significance
Final answer:
Using statistical methods, we have constructed 95% confidence intervals for the mean summer and winter weights of Frontier Airlines passengers, finding that the new FAA recommendations align with our calculated intervals but are positioned conservatively towards the upper limits.
Explanation:
To answer the student's question about constructing and interpreting 95% confidence intervals for the mean summer and winter weights of Frontier Airlines passengers, and then commenting on the new FAA recommendations, we'll follow statistical methods.
Part A: Summer Weight
For the summer weight, the mean is 183 lb, the standard deviation is 20 lb, and the sample size (n) is 100 passengers. The formula for a 95% confidence interval is: mean ± (z*standard error), where the standard error is (standard deviation/√n). Using z=1.96 for 95% confidence, we calculate the interval as 183 ± (1.96*(20/√100)), which simplifies to 183 ± 3.92. Thus, the 95% confidence interval for summer is ±183 lb ± 3.92 lb.
Part B: Winter Weight
Similar calculations for the winter weight (mean = 190 lb, standard deviation = 23 lb, n = 100) yield a 95% confidence interval of 190 ± (1.96*(23/√100)), or 190 ± 4.51. Therefore, the 95% confidence interval for winter is ±190 lb ± 4.51 lb.
Part C: Comment on FAA Recommendations
The new FAA recommendations are 190 lb for summer and 195 lb for winter. These fall within our confidence intervals, but lean towards the upper limit, suggesting a conservative approach by the FAA, possibly to ensure safety by accommodating potential underestimations of passenger weights.
A student has to take twelve hours of classes a week. Due to her extracurricular activities, she must take at least three hours of classes on Monday, at least two on Tuesday, and at least one on Friday. In how many ways can she do this
The student has 210 ways to schedule the remaining six hours of classes across the week, after meeting specific daily minimum class hour requirements.
The student wants to schedule twelve hours of classes across a week while meeting minimum class hour requirements on specific days.
Monday requires at least three hours, Tuesday at least two, and Friday at least one hour of classes.
Considering the constraints, we must calculate the number of ways she can distribute her remaining class hours across the week.
Steps to Calculate the Distribution:
Determine the total hours already allocated due to constraints: 3 (Monday) + 2 (Tuesday) + 1 (Friday) = 6 hours.Subtract the allocated hours from the total required hours to find the remaining hours to be distributed: 12 total hours - 6 allocated hours = 6 hours remaining.Assuming the remaining six hours can be distributed among the five weekdays, the problem becomes a partitioning problem, where we need to find the number of ways to partition six hours into five parts (days), where a part could be zero, indicating no class on that day.Use the formula for partitions of n into k parts: P(n+k-1,k) which is equivalent to C(n+k-1, k-1) where C represents combination.Calculate the number of combinations: C(6+5-1,5-1) = C(10,4).Evaluate the combination: C(10,4) = 10! / (4!(10-4)!) = 210 ways.Therefore, the student has 210 ways to schedule the remaining six hours of classes across the week.
Explain the difference between the interquartile range and the range. Which is more sensitive to extreme values? Explain your thinking.
Answer:
Interquartile range: Difference between the third and the first quartile.
Range: Difference between the highest and the lowest value of a dataset.
Since the range is not in the middle of a set(like the quartiles), it is more sensitive to extreme values.
Step-by-step explanation:
The interquartile range is the difference between the third quartile and the first quartile.
The first quartile(Q1) separates the lower 25% from the upper 75% of a set. So 25% of the values in a data set lie at or below the first quartile, and 75% of the values in a data set lie at or above the first quartile.
The third quartile(Q3) separates the lower 75% from the upper 25% of a set. So 75% of the values in a data set lie at or below the third quartile, and 25% of the values in a data set lie at or the third quartile.
The range is the difference between the highest and the lowest value of a dataset.
Lets see two different sets of cardinality 8
Set 1: 5,5,6,7,8,9,13,13
Range = 13-5 = 8
Q1 = Element at the position 0.25*8 = 2 = 5
Q3 = Element at the position 0.75*8 = 6 = 9
IQR = Q3 - Q1 = 9 - 5 = 4
Now, replacing the first and the last element by extreme values
Set 1: 0,5,6,7,8,9,13,20
Range = 20-0 = 20
Q1 = Element at the position 0.25*8 = 2 = 5
Q3 = Element at the position 0.75*8 = 6 = 9
IQR = Q3 - Q1 = 9 - 5 = 4
The range changes and the IQR does not. Since extreme values are not in the middle of the distribution(in which the quartiles are), the range is more sensitive to extreme values, as this example showed.
Final answer:
The interquartile range measures the spread of the middle 50% of data and is less sensitive to extreme values compared to the range, which measures the spread of all data points by looking at the difference between the maximum and minimum values.
Explanation:
Difference Between Interquartile Range and Range in Data Analysis:
The difference between the interquartile range (IQR) and the range is that while the range measures the spread of all data points by subtracting the smallest value from the largest, the interquartile range measures the spread of the middle 50% of data, between the 25th and 75th percentiles (Q1 and Q3 respectively). To calculate the range in a dataset, if the highest score is 100 and the lowest is 70, the range is 100 - 70 = 30. On the other hand, if Q1 is 72.25 and Q3 is 86.75, the interquartile range would be 86.75 - 72.25 = 14.50.
The range is more sensitive to extreme values or outliers because it takes into account the most extreme scores in the dataset. In contrast, the IQR is less sensitive to extreme values since it focuses on the central portion of the data set and ignores the extremes. This is why IQR is often used to identify potential outliers, which are suspected if they lie more than 1.5 times the IQR below Q1 or above Q3.
Therefore, the IQR provides a more robust measure of data variability, especially when dealing with skewed distributions or datasets with outliers.
100 5 90 10 80 15 70 next term in the sequence?
Answer:
20
Step-by-step explanation:
Find the probability for the experiment of tossing a coin three times. Use the sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
1. The probability of getting exactly one tail
2. The probability of getting a head on the first toss
3. The probability of getting at least one head
4. The probability of getting at least two heads
Answer:
1) 0.375
2) 0.5
3) 0.875
4) 0.5
Step-by-step explanation:
We are given the following in the question:
Sample space, S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
[tex]\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}[/tex]
1. The probability of getting exactly one tail
P(Exactly one tail)
Favorable outcomes ={HHT, HTH, THH}
[tex]\text{P(Exactly one tail)} = \dfrac{3}{8} = 0.375[/tex]
2. The probability of getting a head on the first toss
P(head on the first toss)
Favorable outcomes ={HHH, HHT, HTH, HTT}
[tex]\text{P(head on the first toss)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5[/tex]
3. The probability of getting at least one head
P(at least one head)
Favorable outcomes ={HHH, HHT, HTH, HTT, THH, THT, TTH}
[tex]\text{P(at least one head)} = \dfrac{7}{8} = 0.875[/tex]
4. The probability of getting at least two heads
P(Exactly one tail)
Favorable outcomes ={HHH, HHT, HTH,THH}
[tex]\text{P(Exactly one tail)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5[/tex]
Final answer:
The question involves calculating probabilities based on the outcomes of tossing a coin three times. The probabilities for getting exactly one tail, a head on the first toss, at least one head, and at least two heads are found to be 3/8, 1/2, 7/8, and 1/2 respectively, by counting the favorable outcomes within the complete sample space.
Explanation:
The question asks for the probability of various outcomes when tossing a coin three times, given the sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. Let's solve each part step-by-step.
Probability of getting exactly one tail: There are three outcomes (HTT, THT, TTH) out of eight that satisfy this condition, so the probability is 3/8.
Probability of getting a head on the first toss: Four outcomes (HHH, HHT, HTH, HTT) satisfy this, so the probability is 4/8 or 1/2.
Probability of getting at least one head: Every outcome except TTT includes at least one head, so the probability is 7/8.
Probability of getting at least two heads: There are four outcomes (HHH, HHT, HTH, THH) that satisfy this condition, leading to a probability of 4/8 or 1/2.
Each probability is based on the fundamental principle of counting favorable outcomes over the total number of outcomes.
Lewis earned $1800 from his summer job at the grocery store. This is $250 more than twice what his friend Tara earned. Right and solve an equation to find out how much Tara earned from her summer job
Answer: Tara earned $775 from her summer job.
Step-by-step explanation:
Let x represent the amount of money that Tara earned from her summer job.
Lewis earned $1800 from his summer job at the grocery store. This is $250 more than twice what his friend Tara earned. The equation would be
2x + 250 = 1800
Subtracting 250 from the left hand side and the right hand side of the equation, it becomes
2x + 250 - 250 = 1800 - 250
2x = 1550
Dividing the left hand side and the right hand side of the equation by 2, it becomes
2x/2 = 1550/2
x = 775
A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of babies born in New York. The mean weight was grams with a standard deviation of grams. Assume that birth weight data are approximately bell-shaped. Estimate the number of newborns who weighed between grams and grams. Round to the nearest whole number.The number of newborns who weighed between grams and grams is .
Answer:
a) The the approximate number of babies between 2363 and 3234 are:
n = 0.683*621= 423.95 and that's approximately 424 babies
b) The approximate number of babies between 1492 and 4976 are:
n = 0.955*621= 592.74 and that's approximately 593 babies
Step-by-step explanation:
Assuming this problem:"A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 621 babies born in New York. The mean weight was 3234 grams with a standard deviation of 871 grams. Assume that birth weight data are approximately bell-shaped.
Estimate the number of newborns who weighed between 2363 grams and 4105 grams. Round to the nearest whole number.
The number of newborns who weighed between 1492 grams and 4976 grams is . "
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Part a
Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(3234,871)[/tex]
Where [tex]\mu=3234[/tex] and [tex]\sigma=871[/tex]
We select a sample size of n = 621. And we want to find this probability:
[tex] P(2363 < X < 4105) [/tex]
[tex]P(2363<X<4105)=P(\frac{2363-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{4105-\mu}{\sigma})=P(\frac{2363-3234}{871}<Z<\frac{4105-3234}{871})=P(-1<z<1)[/tex]
And we can find this probability taking this difference:
[tex]P(-1<z<1)=P(z<1)-P(z<-1)=0.841-0.159=0.683 [/tex]
And the the approximate number of babies between 2363 and 3234 are:
n = 0.683*621= 423.95 and that's approximately 424 babies
Part b
[tex] P(1492 < X < 4976) [/tex]
[tex]P(1492<X<4976)=P(\frac{1492-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{4976-\mu}{\sigma})=P(\frac{1492-3234}{871}<Z<\frac{4976-3234}{871})=P(-1<z<1)[/tex]
And we can find this probability taking this difference:
[tex]P(-1<z<1)=P(z<1)-P(z<-1)=0.977-0.0228=0.955 [/tex]
And the the approximate number of babies between 1492 and 4976 are:
n = 0.955*621= 592.74 and that's approximately 593 babies
Answer:
-2
1
3125
Step-by-step explanation:
What is solution to130 more than or equal to-29 - z
Answer:
The answer to your question is z ≥ -159
Step-by-step explanation:
Process
1.- Write the inequality
130 ≥ - 29 - z
2.- Add 29 to both sides of the inequality
130 + 29 ≥ - 29 + 29 - z
3.- Simplify
159 ≥ - z
4.- Divide by -1 and change the inequality
159/-1 ≤ -z/-1
5.- Simplify
-159 ≤ z
A computer chess game and a human chess champion are evenly matched. They play ten games. Find probabilities for the following events. a. They each win five games. b. The computer wins seven games. c. The human chess champion wins at least seven games.
To calculate the probabilities, we can use the binomial distribution formula. For each event, substitute the appropriate values into the formula and calculate the probabilities using the combination symbol and the probabilities of winning and losing a game. For part c, sum the probabilities of winning at least 7 games.
Explanation:To find the probabilities for the given events, we can use the binomial distribution formula. Let p be the probability of winning a game for both the computer and the human, and q be the probability of losing a game. Since they are evenly matched, p = q = 0.5.
a. The probability that they each win five games is P(X = 5), where X follows a binomial distribution with n = 10 (number of games) and p = 0.5. We can calculate this probability using the formula P(X = 5) = C(10, 5) * (0.5)^5 * (0.5)^5.
b. The probability that the computer wins seven games is P(X = 7), where X follows a binomial distribution with n = 10 and p = 0.5. We can calculate this probability using the formula P(X = 7) = C(10, 7) * (0.5)^7 * (0.5)^3.
c. The probability that the human chess champion wins at least seven games can be calculated by summing the probabilities of winning exactly 7, 8, 9, and 10 games: P(X >= 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10).
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An obstetrician knew that there were more live births during the week than on weekends. She wanted to determine whether the mean number of births was the same for each of the five days of the week. She randomly selected eight dates for each of the five days of the week and obtained the following data:a. Write the null and alternative hypotheses.b. State the requirements that must be satisfied to use the one-way ANOVA procedurec. On which day or dates are there more births?
Answer:
The complete question is stated below:
An obstetrician knew that there were more live births during the week than on weekends. She wanted to determine whether the mean number of births was the same for each of the five days of the week. She randomly selected eight dates for each of the five days of the week and obtained the following data:
Monday: Tuesday: Wednesday: Thursday: Friday:
10,456 11,621 11,084 11,171 11,545
10,023 11,944 11,570 11,745 12,321
10,691 11,045 11,346 12,023 11,749
10,283 12,927 11,875 12,433 12,192
10,265 12,577 12,193 12,132 12,422
11,189 11,753 11,593 11,903 11,627
11,198 12,509 11,216 11,233 11,624
11,465 13,521 11,818 12,543 12,543
a. Write the null and alternative hypotheses.
b. State the requirements that must be satisfied to use the one-way ANOVA procedure.
c On which day or dates are there more births?
Answer:
a. Null Hypothesis: There is no statistically significant difference between the mean number of babies born alive on Mondays, Tuesdays, Wednesdays, Thursdays and Fridays.
Alternative Hypothesis: The mean number of babies born alive on each weekday from Monday to Friday are not the same.
b) 1. The dependent variable should be measured at ratio or interval level
2. The independent variable should contain at least two groups that are categorical and independent
3. There must independence of observations between and within the various groups
4. No significant outliers should exist
5. for each group of the independent variable, the dependent variable should be approximately normally distributed.
6. the variances should be homogeneous
c. There are more births on Tuesdays and Fridays
Step-by-step explanation:
a. A Null Hypothesis is one that states that no statistical significance exist between the two variables in the hypothesis. It is the null hypothesis that the researcher tries to disprove. While the alternative hypothesis is simply the opposite of the null hypothesis, it hypothesizes that there is indeed a statistically significant relationship between the variables in question.
b. 1. The dependent variable should be measured at ratio or interval level; among the various levels of measurements such as ordinal, nominal, ratio or interval, the dependent variable must be either on the interval or ratio level.
2. The independent variable should contain at least two groups that are categorical and independent; the independent variable, in this case, the number of live births, should be two groups or more, and they should be categorical.
3. There must independence of observations between and within the various groups; the values observed within each group should be independent of the other. For example, no one pregnant woman should be involved in more than one group.
4. No significant outliers should exist; outliers are single data points within the measured variable that do not follow the usual point pattern of the other values, either values that are too high or too low. they reduce the accuracy of the one-way ANOVA.
5. for each group of the independent variable, the dependent variable should be approximately normally distributed; violations of normality causes the result to be invalid. It can only hold little variations to produce valid results.
6. the variances should be homogeneous; the variances (distribution or spread around the mean) of the two or more test groups must be considered equal.
c. Tuesday with a mean livebirth of 12,237 births and Friday with an average livebirth of 12,002 births have the most number of births.
Rapid HIV tests allow for quick diagnosis without expensive laboratory equipment. However, their efficacy has been called into question. In a population of 1517 tested individuals in Uganda, 4 had HIV but tested negative (false negatives), 166 had HIV and tested positive, 129 did not have HIV but tested positive (false positives), and 1218 did not have HIV and tested negative. A randomly slected person from this population tests negative for HIV. What is the probability that this person has HIV?
Answer:
The probability that a person has HIV given that the test negative is 0.0033.
Step-by-step explanation:
Denote the events as follows:
X = a person in Uganda had HIV
Y = a person in Uganda was tested positive for HIV.
The information provided is:
[tex]P(Y^{c}\cap X)=\frac{4}{1517}\\[/tex]
[tex]P(Y\cap X)=\frac{166}{1517}\\[/tex]
[tex]P(Y\cap X^{c})=\frac{129}{1517}\\[/tex]
[tex]P(Y^{c}\cap X^{c})=\frac{1218}{1517}\\[/tex]
The probability that a person has HIV given that he/she was tested negative is:
[tex]P(X|Y^{c})=\frac{P(Y^{c}\cap X)}{P(Y^{c})}[/tex]
Compute the probability of a person not having HIV as follows:
[tex]P(Y^{c})=P(Y^{c}\cap X)+P(Y^{c}\cap X^{c})=\frac{4}{1517}+\frac{1218}{1517} =\frac{4+1218}{1517} =\frac{1222}{1517}[/tex]
Compute the value of [tex]P(X|Y^{c})[/tex] as follows:
[tex]P(X|Y^{c})=\frac{P(Y^{c}\cap X)}{P(Y^{c})}=\frac{4}{1517}\times\frac{1517}{1222}=0.0033[/tex]
Thus, the probability that a person has HIV given that he/she was tested negative is 0.0033.
Final answer:
The probability that a randomly selected person from this population who tests negative actually has HIV is approximately 0.327%. This accounts for the possibility of false negatives. However, confirmatory testing is necessary to accurately diagnose HIV.
Explanation:
To calculate the probability that a randomly selected person from the population who tested negative for HIV actually has the virus, we need to focus on the false negatives and true negatives provided in the figures from Uganda.
Out of 1517 tested individuals, there were 4 false negatives and 1218 true negatives.
Thus, for someone who tests negative, to find out the probability that they actually have HIV (false negative), we use the formula: Probability = Number of false negatives / (Number of false negatives + Number of true negatives).
In this case, we have Probability = 4 / (4 + 1218) = 4 / 1222. When we calculate this probability, we get approximately 0.00327, or 0.327%.
the Dimensions of Prokaryotic Cells and their Constituents Escherichia coli cells are about 2 mm (microns) long and 0.8 mm in diameter. (Section 1.5) a. How many E. coli cells laid end to end would fit across the diameter of a pinhead? (Assume a pinhead diameter of 0.5 mm.)
Answer:
Try Photomath !
Step-by-step explanation:
A statistics professor finds that when he schedules an office hour for student help, an average of 3.3 students arrive. Find the probability that in a randomly selected office hour, the number of student arrivals is 6.
Answer:
[tex] P(X=6)[/tex]
If we use the probability mass function we got:
[tex] P(X=6) = \frac{e^{-3.3} 3.3^6}{6!}= 0.0662[/tex]
Step-by-step explanation:
Previous concepts
The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:
Solution to the problem
Let X the random variable that represent the number of students arrive at the office hour. We know that [tex]X \sim Poisson(\lambda=3.3)[/tex]
The probability mass function for the random variable is given by:
[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]
And f(x)=0 for other case.
For this distribution the expected value is the same parameter [tex]\lambda[/tex]
[tex]E(X)=\mu =\lambda=3.3[/tex]
And we want this probability:
[tex] P(X=6)[/tex]
If we use the probability mass function we got:
[tex] P(X=6) = \frac{e^{-3.3} 3.3^6}{6!}= 0.0662[/tex]
Answer:
0.0662
Step-by-step explanation:
The given problem indicates the Poisson experiment.
The pdf of Poisson experiment is as follow
P(X=x)=μ^x(e^-μ)/x!
Here, the average student visits in office hour=3.3=μ.
We have to find the probability that the number of student arrivals is 6 in a randomly selected office hour.
So, x=6 and μ=3.3
The probability that the number of student arrivals is 6 in a randomly selected office hour
P(X=6)=3.3^6(e^-3.3)/6!
P(X=6)=1291.468(.0369)/720
P(X=6)=47.6552/720
P(X=6)=0.0662.
Thus, the probability that the number of student arrivals is 6 in a randomly selected office hour is 6.62 or 0.0662.
Find the probability for the experiment of drawing a card at random from a standard deck of 52 playing cards.
1. The card is a face card.
2. The card is not a face card.
3. The card is a red face card.
4. The card is a 6 or lower. (Aces are low.)
5. The card is a 9 or lower. (Aces are low.)
Answer:
1. 12/52 or 0.2308
2. 10/13 or 0.7692
3. 3/26 or 0.1154
4. 3/26 or 0.1154
5. 9/52 or 0.1731
Step-by-step explanation:
1.
The face card are king,queen and jack.
There are 12 face card in a deck of 52 cards i.e. 4 king, 4 queen and 4 jack.
P(FC)=12/52 or 0.2308
So, the probability that the card is a face card is 0.2308.
2.
The card is not a face card.
P(FC')=1-P(FC)=1-(12/52)=(52-12)/52=40/52=10/13
P(FC')=10/13 or 0.7692
So, the probability that the card is not a face card is 0.7692.
3.
The card is a red face card
There are 6 red face card in a deck of 52 cards i.e. 2 king, 2 queen and 2 jack.
P(RFC)=6/52
P(RFC)=3/26 or 0.1154
So, the probability that the card is a red face card is 0.1154.
4.
The card is a 6 or lower.
There are 6 cards in a deck of 52 cards that are 6 or lower i.e. 6,5,4,3,2 and ace.
P(6 or lower)=6/52
P(6 or lower)=3/26 or 0.1154
So, the probability that the card is a 6 or lower is 0.1154.
5.
The card is a 9 or lower.
There are 9 cards in a deck of 52 cards that are 9 or lower i.e. 9, 8, 7, 6, 5, 4, 3, 2 and ace.
P(9 or lower)=9/52
P(9 or lower)=9/52 or 0.1731
So, the probability that the card is a 9 or lower is 0.1731.
In a pack or deck of 52 playing cards, they are divided into 4 suits of 13 cards each i.e. spades , hearts , diamonds and clubs. King, Queen and Jack are face cards. So, there are 12 face cards in the deck of 52 playing cards.
In a Deck of 52 playing cards,
Number of face card = 12
number of non face card = 40
Number of red face card = 6
Number of Ace cards = 4
So, Probability of the card is face card is , = 12/52 =3/13
Probability of the card is not face card ,= 40/52 = 10/13
Probability of the card is a red face card is,=6/52 = 3/26
The probability of card is a 6 or lower is, = 24/52 = 6/13
The probability of card is a 9 or lower is, = 36/52 = 9/13.
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Check my work Check My Work button is now disabled3Item 7Item 7 10 points An article describes a study in which a new type of ointment was applied to forearms of volunteers to study the rates of absorption into the skin. Eight locations on the forearm were designated for ointment application. The new ointment was applied to four locations, and a control was applied to the other four. How many different choices were there for the four locations to apply the new ointment?
Answer:
70 combinations
Step-by-step explanation:
Given:
- The total number of designated location n = 8
- The number of location the ointment is to be applied r = 4
Find:
How many different choices were there for the four locations to apply the new ointment?
Solution:
- For this question we will apply combinations to the given problem. We have total of 8 available places out of which we have to "choose" 4 locations to apply the new ointment. We will use combinations " C " operator.
- Out of 8 "choose" 4 = 8_C _4 = 70 possible combinations.
- Since there is requirement of "order" so its only limited to the selection process and not an arrangement.
The atmospheric pressures at the top and the bottom of a mountain are read by a barometer to be 93.8 and 100.5 kPa. If the average density of air is 1.25 kg/m3 , what is the height of the mountain
Answer:
546.94 meters is the height of the mountain.
Step-by-step explanation:
Pressure at the top of mountain = [tex]P_1=93.8 kPa[/tex]
Pressure at the bottom of the mountain = [tex]P_2=100.5 kPa[/tex]
Pressure difference =[tex]P_2-P_1=100.5kPa-93.8kPa=6.7 kPa[/tex]
6.7 kPa = 6.7 × 1000 Pa (1 kPa= 1000 pa)
Density of the air = d = [tex]1.25 kg/m^3[/tex]
Acceleration due to gravity = g = [tex]9.8 m/s^2[/tex]
Height of the mountain = h
[tex]P=h\times d\times g[/tex]
[tex]h=\frac{P}{d\times g}=\frac{6.7\times 1000 Pa}{1.25 kg/m^3\times 9.8 m/s^2}[/tex]
[tex]h=546.94 m[/tex]
546.94 meters is the height of the mountain.
A person draws a five-card hand from a standard 52-card deck. If the order of the cards matters (i.e. ABCDE is a distinct hand from ABCED), use the Subtraction Principle to count how many hands contain at least one hearts card.
Answer:
242,784,360 hands.
Step-by-step explanation:
The number of hands that contain at least one hearts card is given by the total possible number of hands subtracted by the number of hands with no hearts card.
The total possible number of hands is:
[tex]N_T=\frac{52!}{(52-5)!}=\frac{52!}{(47)!} =52*51*50*49*48\\N_T=311,875,200[/tex]
Since there are 13 hearts cards in a deck, the number of possible hands with no hearts card is:
[tex]N_{H=0} =\frac{52-13!}{(52-13-5)!}=\frac{39!}{(34)!} =39*38*37*36*35\\N_{H=0}=69,090,840[/tex]
The number of hands with at least one hearts card is:
[tex]N_{H>0} = N_T-N{H=0}\\N_{H>0}=311,875,200-69,090,840\\N_{H>0}=242,784,360[/tex]
242,784,360 hands contain at least one hearts card.
A group of students estimated the length of one minute without reference to a watch or clock, and the times (seconds) are listed below. Use a 0.10 significance level to test the claim that these times are from a population with a mean equal to 60 seconds. Does it appear that students are reasonably good at estimating one minute? 75 88 51 73 49 31 69 74 72 59 72 81 99 101 73 What are the null and alternative hypotheses? A. Upper H 0: muequals60 seconds Upper H 1: munot equals60 seconds B. Upper H 0: munot equals60 seconds Upper H 1: muequals60 seconds C. Upper H 0: muequals60 seconds Upper H 1: muless than60 seconds D. Upper H 0: muequals60 seconds Upper H 1: mugreater than60 seconds Determine the test statistic. nothing (Round to two decimal places as needed.) Determine the P-value. nothing (Round to three decimal places as needed.) State the final conclusion that addresses the original claim. ▼ Fail to reject Reject Upper H 0. There is ▼ sufficient not sufficient evidence to conclude that the original claim that the mean of the population of estimates is 60 seconds ▼ is is not correct. It ▼ appears does not appear that, as a group, the students are reasonably good at estimating one minute.
Answer:
It does not appear that, as a group, the students are reasonably good at estimating one minute.
Step-by-step explanation:
We are given the following data in the question:
75, 88, 51, 73, 49, 31, 69, 74, 72, 59, 72, 81, 99, 101, 73
Formula:
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{1067}{15} = 71.13[/tex]
Sum of squares of differences = 4739.733
[tex]S.D = \sqrt{\frac{4739.733}{14}} = 18.39[/tex]
Population mean, μ = 60 minutes
Sample mean, [tex]\bar{x}[/tex] = 71.13 minutes
Sample size, n = 15
Alpha, α = 0.10
Sample standard deviation, s = 18.39 minutes
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 60\text{ minutes}\\H_A: \mu \neq 60\text{ minutes}[/tex]
We use Two-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{71.13 - 60}{\frac{18.39}{\sqrt{15}} } = 2.34[/tex]
Calculating the p-value from the table, we have,
P-value = 0.034354
Since the p-value is lower than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.
Thus, we conclude that it does not appear that, as a group, the students are reasonably good at estimating one minute.
Option A is the correct set of hypotheses for testing if students can estimate one minute. The test statistic and p-value are needed to make a decision, typically using a t-test. The conclusion depends on the p-value relative to the significance level.
The null and alternative hypotheses you are working with for testing whether students are good at estimating one minute are:
H0: \\(\mu = 60\\) seconds - This states that the true population mean estimation time is 60 seconds.Ha: \\(\mu \not= 60\\) seconds - This hypothesis asserts that the true population mean estimation time is not 60 seconds.To determine if students are good at estimating one minute, we have option A as correct, which is H0: \\(\mu = 60\\) and Ha: \\(\mu \not= 60\\).
The test statistic and p-value should be calculated using appropriate statistical methods (typically a t-test assuming we do not know the population standard deviation, which would require the mean, standard deviation, and sample size). After calculating the test statistic and p-value, you will compare the p-value to the significance level (alpha). If the p-value is less than alpha, you reject H0; otherwise, you fail to reject H0.
Based on the decision to reject or fail to reject H0, you can make a conclusion regarding the original claim about whether students are reasonably good at estimating one minute.
In 1998 the World Health Organization reported the findings of a major study on the quality of blood pressure monitoring around the world. In its report it stated that for Canada the results for diastolic blood pressure had a mean of 78 mmHg and a standard deviation of 11 mmHg. Assuming that diastolic blood pressure measurements are Normally distributed, the DBP reading that represents the 80th percentile of the distribution is _____.
Answer:
87.26 mm Hg is the reading that represents the 80th percentile of the distribution.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 78 mm Hg
Standard Deviation, σ = 11 mm Hg
We are given that the distribution of diastolic blood pressure is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
We have to find the value of x such that the probability is 0.80
P(X < x)
[tex]P( X < x) = P( z < \displaystyle\frac{x - 78}{11})=0.8[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(z < 0.842) = 0.8[/tex]
[tex]\displaystyle\dfrac{x - 78}{11} = 0.842\\\\x = 87.262 \approx 87.26[/tex]
87.26 mm Hg is the reading that represents the 80th percentile of the distribution.
To find the DBP reading representing the 80th percentile, we can use the z-score formula. The DBP reading is 86.8 mmHg.
Explanation:To find the diastolic blood pressure (DBP) reading that represents the 80th percentile of the distribution, we can use the z-score formula. The z-score is calculated by subtracting the mean from the desired value and dividing it by the standard deviation. Then, we can use the z-score table or calculator to find the corresponding percentile. In this case, since we want the 80th percentile, we need to find the z-score that corresponds to an area of 0.8 to the left of it.
The formula for the z-score is: z = (x - mean) / standard deviation
Using the given values for Canada (mean = 78 mmHg, standard deviation = 11 mmHg), we can substitute them into the formula and solve for x:
0.8 = (x - 78) / 11
Multiplying both sides by 11, we get:
8.8 = x - 78
Adding 78 to both sides, we get:
x = 86.8 mmHg
Therefore, the DBP reading that represents the 80th percentile of the distribution is 86.8 mmHg.
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At time, t, in seconds, your velocity, v, in meters/second, is given by v(t)=6+(t^2), 0 (=)t(=)6. Use (delta)t= 2 to estimate the distance traveled during this time.
Find the upper and lower estimates, and then average the two. find the upper estimate of distance traveled, find the lower estimate of distance traveled, and find the average.
Answer:
Total distance is 8 meters.
Step-by-step explanation:
First, let's take the equation and analyse it:
v(t) = 6 + t²
The delta t = 2
To find the acceleration, we need to differentiate v with respect to t like this:
dv/dt = d/dt (6 + t²)
= 2t
At t = 2, the change in velocity (dv/dt) = 2×2 = 4 m/s². This is the acceleration,a
The final velocity is given by v = 6 + (2)²
= 10 m/s
The distance is given by the formula:
s = ut + 1/2at²
initial velocity, u = 0
Therefore, the equation becomes s = 1/2at²
= 1/2× 4×4
= 8 m Ans
The upper estimate of the distance traveled is 150 meters, the lower estimate is 12 meters, and the average of these estimates is 81 meters.
Explanation:This math problem involves the concept of approximating the distance traveled using velocity and small intervals of time. The distance traveled, in approximation, can be calculated by taking the velocity at a certain time, and multiplying it by the change in time, which is given as 2 seconds in this case.
To find the upper estimate, we calculate using the endpoint of the interval, t = 6. To find the lower estimate, we calculate using the beginning of the interval, t = 0.
For upper estimate: v(6) x 2 = [6 + (6^2)] x 2 = 150 meters
For lower estimate: v(0) x 2 = [6 + (0^2)] x 2 = 12 meters
The average of the two estimates is then found by adding them together and dividing by 2, which results in 81 meters.
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In a town there are on average 2.3 children in a family and a randomly chosen child has on average 1.6 siblings. Determine the variance of the number of children in a randomly chosen family.
The variance of the number of children in a randomly chosen family is approximately 9.4148.
To determine the variance of the number of children in a randomly chosen family, we can use the properties of the Poisson distribution, assuming that the number of children in a family follows a Poisson distribution with a mean of 2.3.
The Poisson distribution has a variance equal to its mean, so if the mean number of children per family is 2.3, then the variance is also 2.3.
However, we can also calculate the variance directly from the given information.
Let ( X ) be the random variable representing the number of children in a family.
Given:
- The mean number of children in a family, [tex]\( \mu = 2.3 \)[/tex]
- Each child has, on average, 1.6 siblings.
To find the variance, we use the formula:
\[ \text{Variance} = \text{E}(X^2) - [\text{E}(X)]^2 \]
[tex]First, we find \( \text{E}(X^2) \), the expected value of \( X^2 \):\[ \text{E}(X^2) = \text{Variance} + [\text{E}(X)]^2 \]\[ \text{E}(X^2) = 1.6^2 \times 2.3 + (2.3)^2 \]\[ \text{E}(X^2) = 4.096 \times 2.3 + 5.29 \]\[ \text{E}(X^2) = 9.4148 + 5.29 \]\[ \text{E}(X^2) = 14.7048 \][/tex]
Then, using the formula for variance:
[tex]\[ \text{Variance} = \text{E}(X^2) - [\text{E}(X)]^2 \]\[ \text{Variance} = 14.7048 - (2.3)^2 \]\[ \text{Variance} = 14.7048 - 5.29 \]\[ \text{Variance} = 9.4148 \][/tex]
So, the variance of the number of children in a randomly chosen family is approximately 9.4148.
Question:
In a town there are on average 2.3 children in a family and a randomly chosen child has on average 1.6 siblings. Determine the variance of the number of children in a randomly chosen family.
The probability that an international flight leaving the United States is delayed in departing (event D) is .29. The probability that an international flight leaving the United States is a transpacific flight (event P) is .59. The probability that an international flight leaving the U.S. is a transpacific flight and is delayed in departing is .11. (a) What is the probability that an international flight leaving the United States is delayed in departing given that the flight is a transpacific flight
Answer:
[tex] P(D|P)= \frac{0.11}{0.59}=0.186[/tex]
Step-by-step explanation:
For this case we have defined the following events:
D= "An international flight leaving the United States is delayed in departing"
P="An international flight leaving the United States is a transpacific flight "
And we have defined the probabilities:
[tex] P(D)= 0.29 , P(P) = 0.59[/tex]
And for the event: "an international flight leaving the U.S. is a transpacific flight and is delayed in departing" [tex] D \cap P [/tex] we know the probability:
[tex] P(D \cap P) =0.11[/tex]
We want to find this probability:
What is the probability that an international flight leaving the United States is delayed in departing given that the flight is a transpacific flight
So we want this probability:
[tex] P(D|P)[/tex]
And we can use the conditional formula from the Bayes theorem given two events A and B:
[tex] P(A|B) = \frac{P(A \cap B)}{P(B)}[/tex]
And if we use this formula for our case we have:
[tex] P(D|P)= \frac{P(D \cap P)}{P(P)}[/tex]
And if we replace the values we got:
[tex] P(D|P)= \frac{0.11}{0.59}=0.186[/tex]
Find the distance from (3, −4, 8) to each of the following. (a) the xy-plane √7 (b) the yz-plane (0,−4,8) (c) the xz-plane (3,0,8) (d) the x-axis (3,0,0) (e) the y-axis (0,−4,0) (f) the z-axis (0,0,8)
Answer:
a) 8 units
b) 3 units
c) 4 units
d) [tex]4\sqrt{5}\text{ units}[/tex]
e) [tex]\sqrt{73}\text{ units}[/tex]
f) 5 units
Step-by-step explanation:
We are given the following:
Point (3, −4, 8)
We have to find the distance of the point from the following:
Distance formula:
[tex](x_1,y_1,z_1),(x_2,y_2,z_2)\\\\d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}[/tex]
(a) the xy-plane
We have to find the distance from (3, −4, 8) to (3, −4, 0)
[tex]d = \sqrt{(3-3)^2 + (-4+4)^2 + (0-8)^2} = 8\text{ units}[/tex]
(b) the yz-plane
We have to find the distance from (3, −4, 8) to (0, −4, 8)
[tex]d = \sqrt{(0-3)^2 + (-4+4)^2 + (8-8)^2} = 3\text{ units}[/tex]
(c) the xz-plane
We have to find the distance from (3, −4, 8) to (3, 0, 8)
[tex]d = \sqrt{(3-3)^2 + (0+4)^2 + (8-8)^2} = 4\text{ units}[/tex]
(d) the x-axis
We have to find the distance from (3, −4, 8) to (3, 0, 0)
[tex]d = \sqrt{(3-3)^2 + (0+4)^2 + (0-8)^2} = \sqrt{80} = 4\sqrt{5}\text{ units}[/tex]
(e) the y-axis (0,−4,0)
We have to find the distance from (3, −4, 8) to (0, -4, 0)
[tex]d = \sqrt{(0-3)^2 + (-4+4)^2 + (0-8)^2} = \sqrt{73}\text{ units}[/tex]
(f) the z-axis (0,0,8)
We have to find the distance from (3, −4, 8) to (0, 0, 8)
[tex]d = \sqrt{(0-3)^2 + (0+4)^2 + (8-8)^2} = \sqrt{25} = 5\text{ units}[/tex]
Suppose that 30% of all students who have to buy a text for a particular course want a new copy (the successes!), whereas the other 70% want a used copy. Consider randomly selecting 15 purchasers.
(a) What are the mean value and standard deviation of the number who want a new copy of the book?(b) What is the probability that the number who want new copies is more than two standard deviations away from the mean value?
Answer:
a. Mean = 4.5, Standard Deviation = 1.775
b. 0.0152
Step-by-step explanation:
Given
n = 15 purchasers
p = Success = 30%
p = 0.3
q =Failure = 70%
q = 0.7
a.
Mean = np
Mean = 15 * 0.3
Mean = 4.5
Standard Deviation = √Variance
Variance = npq
Variance = 15 * 0.3 * 0.7
Variance = 3.15
Standard Deviation = √3.15
Standard Deviation = 1.774823934929884
Standard Deviation = 1.775 ---------- Approximated
b.
The probability that the number who want new copies is more than two standard deviations away from the mean value
Standard Deviation = 1.775
Mean = 4.5
2 Standard Deviation and Mean = 2 * 1.775 + 4.5
= 3.55 + 4.5
= 8.05
P(X>8.05) = P(9) + P(10) +........+ P(15)
Using the binomial distribution
(p + q) ^ n where p = 0.3, q = 0.7 , n = 15
Expanding (p+q)^n where n = 15 and r > 8
We have
15C9 p^9 q^6 + 15C10 p^10 q^5 + 15C11 p^11 q⁴ + 15C12 p^12 q³ + 15C13 p^13 q² + 15C14 p^14 q + p^15
= 5005 (0.3)^9 (0.7)^6 + 3003 (0.3)^10 (0.7)^5 + 1365 (0.3)^11 (0.7)⁴ + 455 (0.3)^12 (0.7)³ + 105 (0.3)^14 (0.7)² + 15 (0.3)^14 (0.7) + 0.3^15
=0.015234
A Web ad can be designed from four different colors, three font types, five font sizes, three images, and five text phrases. A specific design is randomly generated by the Web server when you visit the site. Let A denote the event that the design color is red and let B denote the event that the font size is not the smallest one. Use the addition rules to calculate the following probabilities.
P(AuB)
p(AuB')
P(A'uB')
Answer:
P(A∪B)=17/20 or 0.85
P(A∪B')=2/5 or 0.4
P(A'∪B')=4/5 or 0.8
Step-by-step explanation:
There are four font colors so each color had equal chance and thus,
P(A)=1/4
There are 5 font sizes and so not the smallest fonts are 4.Thus,
P(B)=4/5
P(A∪B)=P(A)+P(B)-P(A∩B)
The design is generated randomly so event A and event B are independent.
P(A∩B)=P(A)*P(B)
P(A∩B)=1/4(4/5)=1/5
P(A∪B)=P(A)+P(B)-P(A∩B)
P(A∪B)=1/4+4/5-1/5=1/4+3/5
P(A∪B)=17/20 or 0.85
P(A∪B')=P(A)+P(B')-P(A∩B')
P(B')=1-P(B)=1-4/5=1/5
P(A∩B')=P(A)*P(B')=1/4*1/5=1/20
P(A∪B')=P(A)+P(B')-P(A∩B')
P(A∪B')=1/4+1/5-1/20=9/20-1/20=8/20
P(A∪B')=2/5 or 0.4
P(A'∪B')=P(A∩B)'
P(A'∪B')=1-P(A∩B)
P(A'∪B')=1-1/5=4/5
P(A'∪B')=4/5 or 0.8
The probability of event A or B occurring can be calculated using the addition rule. The probability of A occurring is 1/4, the probability of B occurring is 4/5, and the probability of both A and B occurring is 1/5. Using the addition rule, the probability of A or B occurring is 13/20.
Explanation:To calculate the probability of events A or B occurring, we can use the addition rule. The addition rule states that the probability of A or B occurring is equal to the sum of their individual probabilities minus the probability that both A and B occur together.
First, let's calculate P(A): There are four possible colors, and one of them is red. So the probability of A, P(A), is 1/4. Next, let's calculate P(B): There are five possible font sizes, and the smallest size is not included in B. So the probability of B, P(B), is 4/5. Now, let's calculate P(A ∪ B): We need to account for the possibility that both A and B occur together. Since A and B are independent events, we can multiply their individual probabilities: P(A ∩ B) = P(A) * P(B) = (1/4) * (4/5) = 1/5. Finally, we can use the addition rule to calculate P(A ∪ B): P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 1/4 + 4/5 - 1/5 = 13/20. Learn more about Probability here:
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What will be the cost of gasoline for a 4,700-mile automobile trip if the car gets 41 miles per gallon, and the average price of gas is $2.79 per gallon?
$ 320
Step-by-step explanation:41 miles ............ 1 gallon
4700 miles .......x gallon
x = 4700×1/41 = 114.63 gallons
114.63 gallons×$2.79 ≈ $ 320
To find the cost of gasoline for a 4,700-mile automobile trip, divide the total distance by the car's miles per gallon, then multiply by the price of gas per gallon, finally we get a total cost of $318.65.
Explanation:To find the cost of gasoline for a 4,700-mile automobile trip, we need to divide the total distance by the car's miles per gallon to get the number of gallons of gas needed.
Then, we multiply the number of gallons by the price of gas per gallon to get the total cost.
In this case, the car gets 41 miles per gallon, so the number of gallons needed is 4700 miles / 41 miles per gallon = 114.63 gallons.
Multiplying this by the average price of gas, $2.79 per gallon, we get a total cost of $318.65.
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Three identical fair coins are thrown simultaneously until all three show the same face. What is the probability that they are thrown more than three times?
Answer:
The probability that the coins are thrown more than three times to show the same face is 0.3164.
Step-by-step explanation:
The problem is related to Geometric distribution.
The Geometric distribution defines the probability distribution of X failures before the first success.
The probability distribution function is:
[tex]P(X=k)=(1-p)^{k}p;\ k = 0, 1, 2, ...[/tex]
First compute the probability that in the [tex]i^{th}[/tex] throw all the three coins will show the same face.
P (All the 3 coins shows the same face) = P (All the three coins shows Heads) + P (All the three coins shows Tails)
[tex]=(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2} )+(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2} )\\=\frac{1}{8}+\frac{1}{8}\\ =\frac{2}{8} \\=\frac{1}{4}[/tex]
Now compute the probability that it takes more than 3 throws for the coins to show the same face.
P (X > 3) = 1 - P (X ≤ 3)
[tex]=1-[P(X=1)+P(X=2)+P(X=3)]\\=1-[[(1-\frac{1}{4} )^{0}\times\frac{1}{4}]+[(1-\frac{1}{4} )^{1}\times\frac{1}{4}]+ [(1-\frac{1}{4} )^{2}\times\frac{1}{4}]+[(1-\frac{1}{4} )^{3}\times\frac{1}{4}]]\\=1-[0.2500+0.1875+0.1406+0.1055]\\=1-0.6836\\=0.3164[/tex]
Thus, the probability that it takes more than 3 throws for the coins to show the same face is 0.3164.
The probability that it takes more than 3 throws for the coins to show the same face is 0.3164.
What is probability?Probability means possibility. It deals with the occurrence of a random event. The value of probability can only be from 0 to 1. Its basic meaning is something is likely to happen. It is the ratio of the favorable event to the total number of events.
Given
Three identical fair coins are thrown simultaneously until all three show the same face.
Then the total event will be
[tex]\rm Total \ event = 2^3 = 8[/tex]
All the possibilities.
[tex]\rm (HHH),( HHT ),(HTH),( THH),(HTT),(THT),(TTH),(TTT)[/tex]
For getting the same face, then the probability will be
[tex]\rm P(same\ face) = \dfrac{2}{8} = \dfrac{1}{4}[/tex]
Now compute the probability that it takes more than 3 throws for the coins to show the same face.
[tex]\rm P (X > 3) = 1 - P (X ≤ 3)\\\\P (X > 3) = 1 -[P(x=1) +P(x=2) + P(x=3)]\\\\P (X > 3) =1 -\{[(1-\dfrac{1}{4})^0 * \dfrac{1}{4}] +[(1-\dfrac{1}{4})^1 * \dfrac{1}{4}] +[(1-\dfrac{1}{4})^2 * \dfrac{1}{4}] +[(1-\dfrac{1}{4})^3 * \dfrac{1}{4}] \}\\\\P (X > 3) =1 - 0.6836\\\\P (X > 3) =0.3164[/tex]
Thus, the probability that it takes more than 3 throws for the coins to show the same face is 0.3164.
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If you bet $1 that a sum of 5 will turn up, what should the house pay (plus returning your $1 bet) if a sum of 5 turns up in order for the game to be fair?
Here is the complete question.
1). What are the odds for rolling a sum of 5 in a single roll of two fair dice?
2). If you bet $1 that a sum of 5 will turn up, what should the house pay (plus returning your $1 bet) if a sum of 5 turns up in order for the game to be fair?
Answer:
1). [tex]\frac{1}{8}[/tex]
2). $8.
Step-by-step explanation:
1).
The sample space (S) for rolling two fair dice is given as the following parameters illustrated below:
[tex]\left[\begin{array}{cccccc}(1,2)&(1,2)&(1,3)&(1,4)&(1,5)&(1,6)\\(2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6)\\(3,1)&(3,2)&(3,3)&(3,4)&(3,5)&(3,6)\end{array}\right] \left[\begin{array}{cccccc}(4,1)&(4,2)&(4,3)&(4,4)&(4,5)&(4,6)\\(5,1)&(5,2)&(5,3)&(5,4)&(5,5)&(5,6)\\(6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6)\end{array}\right][/tex]
Let F represent the event that the sum of both dice turns up is 5.
F = [(1,4),(2,3),(3,2),(4,1)}
∴ The probability for an event F is illustrated below as:
[tex]\frac{P(F)}{P(F')}[/tex] [tex]=\frac{\frac{4}{36} }{1-\frac{4}{36} }[/tex]
[tex]=\frac{\frac{4}{36} }{\frac{32}{36} }[/tex]
[tex]={\frac{4}{36}}*\frac{36}{32}[/tex]
= [tex]\frac{1}{8}[/tex]
2). From above, we can see that the probability for rolling a sum of 5 are 1 to 8. Therefore, if you roll a sum of 5, in order for the game to be fair, the house is required to pay $8.