Answer:
0.5 week
Step-by-step explanation:
The time spent by a pound of meat in this system is given by the average number of pounds of meat on inventory (2500 pounds) divided by the average weekly meat consumption (5000 pounds per week).
The time, in weeks, is:
[tex]t=\frac{2500\ pounds}{5000\ pounds/week} \\t= 0.5\ week[/tex]
The average time spent by a pound of meat in production is 0.5 week.
If the filling equipment is functioning properly what is the probability that the volume of oil in a randomly selected barrel will be 55.4 gallons or more? Show your work?
Answer:
[tex]P(X>55.4)=P(\frac{X-\mu}{\sigma}>\frac{55.4-\mu}{\sigma})=P(Z>\frac{55.4-55}{0.5})=P(Z>0.8)[/tex]
[tex]P(Z>0.8)=1-P(Z\leq 0.8)[/tex]
[tex]P(Z>0.8)=1-0.788=0.212[/tex]
Step-by-step explanation:
Assuming this previous info : "Trucks carry barrels of crude oil from a port in Texas to a distributer in New Mexico on long trailers. Filling equipment is used to fill the barrels with the oil. When functioning properly, the actual volume of oil loaded into each barrel by the filling equipment at the port is approximately normally distributed with a mean of 55 gallons and a standard deviation of 0.5 gallons. If the mean is greater than 55.4 gallons, the filling mechanism is overfilling."
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".
Solution to the problem
Let X the random variable that represent the amount filling of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(55,0.5)[/tex]
Where [tex]\mu=55[/tex] and [tex]\sigma=0.5[/tex]
We are interested on this probability
[tex]P(X>55.4)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X>55.4)=P(\frac{X-\mu}{\sigma}>\frac{55.4-\mu}{\sigma})=P(Z>\frac{55.4-55}{0.5})=P(Z>0.8)[/tex]
And we can find this probability using the complement rule:
[tex]P(Z>0.8)=1-P(Z\leq 0.8)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(Z>0.8)=1-0.788=0.212[/tex]
Final answer:
The probability of a randomly selected barrel containing 55.4 gallons or more is calculated using normal distribution and z-scores. Since the value of 55.4 gallons is significantly higher than the mean with a large positive z-score, the probability is very small and close to 0.
Explanation:
The question involves finding the probability that the volume of oil in a randomly selected barrel will be 55.4 gallons or more, given that the normal operating conditions of a tank are 45 gallons with a 3-gallon standard deviation. This is a problem of probability involving normal distribution, where we are looking for the area under the normal curve to the right of 55.4 gallons.
To find this probability, we first calculate the z-score, which is the number of standard deviations away from the mean the value of 55.4 gallons is. The z-score formula is given by Z = (X - μ) / σ, where X is the value of interest (55.4 gallons), μ is the mean (45 gallons), and σ is the standard deviation (3 gallons). Plugging in the numbers, we get Z = (55.4 - 45) / 3 = 3.467.
Next, we use z-score tables or a calculator to find the probability corresponding to the calculated z-score. The probability represents the area to the left of the z-score under the normal curve. Since we are interested in the probability of the volume being more than 55.4 gallons, we need to subtract this value from 1 to find the area to the right of the z-score.
If using a standard normal distribution table, which gives the area to the left, we find the value corresponding to Z = 3.467. If such a precise value is not available in the table (which is likely because 3.467 is quite high for typical z-score tables), we would take the value for the closest z-score provided or use software or a calculator with statistical functions. Most standard normal tables do not extend beyond a z-score of 3.49, which is associated with a probability very close to 1. Therefore, the probability of the barrel having 55.4 gallons or more is very small and could be approximated as P(Z > 3.467) ≈ 0.
All human blood can be "ABO-typed" as one of O,A, B, or AB, but the distribution of thetypes varies a bit among groups of people. Here is the distributionof blood types for a randomly chosen person in the United States.Blood type O A B ABU.S. probability 0.46 0.41 0.12 0.01Choose a married couple at random. It is reasonable to assume thatthe blood types of husband and wife are independent and follow thisdistribution.a. What is the probability that the wife has type A and the husbandhas type B?b. What is the probability that one of the couple has type A blood andthe other has type B?
Answer:
(a) 0.0492
(b) 0.0984
Step-by-step explanation:
The probability distribution of blood type in the US is:
[tex]\begin{array}{cc}Type&Probability\\O&0.46\\A&0.41\\B&0.12\\AB&0.01\end{array}[/tex]
(a) The probability that the wife has type A and the husband has type B is:
[tex]P(w=A\ and\ h=B) =0.41*0.12\\P(w=A\ and\ h=B) =0.0492[/tex]
(b) The probability that one of the couple has type A blood and the other has type B is given by the probability that the wife has type A and the husband has type B added to the probability that the wife has type B and the husband has type A.
[tex]P= P(w=A\ and\ h=B) + P(w=B\ and\ h=A)\\P=0.41*0.12+0.12*0.41=0.0984[/tex]
Final answer:
The probability that the wife has type A blood and the husband has type B blood is 4.92%. The probability that one of the couple has type A blood and the other has type B blood is 9.84%.
Explanation:
The student is being asked to calculate probabilities related to the blood types of a randomly chosen married couple in the United States using the given distribution of ABO blood types. The probabilities for each blood type are: type O at 0.46, type A at 0.41, type B at 0.12, and type AB at 0.01. To solve these questions, we use the principle that the probability of independent events occurring together is the product of their respective probabilities.
Probability wife has type A and husband has type B: This is a straightforward calculation using the probabilities of each event happening independently. Since the blood type of the wife and the husband are independent events, you multiply their individual probabilities:P(wife has type A and husband has type B) = P(wife has type A) × P(husband has type B) = 0.41 × 0.12 = 0.0492, or 4.92%.Probability one has type A and the other has type B: Here we need to calculate the probability for two scenarios - Scenario 1: Wife has type A and husband has type B, and Scenario 2: Wife has type B and husband has type A. We add the probabilities of these two independent scenarios together.P(one has type A and the other has type B) = P(wife has type A and husband has type B) + P(wife has type B and husband has type A) = 0.0492 + 0.0492 = 0.0984, or 9.84%.A disease has a constant force of mortality, µ. Historically 10% of all people with the disease die within 20 years. A more virulent strain of the disease is encountered with a constant force of mortality 2µ. What is the probability of an individual who has the new strain of the disease dying within 20 years?
Answer:
Step-by-step explanation:
For the disease having force of mortality µ.
we apply exponential distribution law
probability of dying within 20 years
p( x <20) = [tex]1-e^{-\mu}[/tex] = .1
[tex]e^{-\mu}[/tex] = .9
For the disease having force of mortality 2µ.
we apply exponential distribution law
probability of dying within 20 years
p( x <20) = 1 - [tex]e^{-2\mu}[/tex]
= 1 - ( .9)²
= 1 - .81
= .19
or 19%
The tensile strength of a metal part is normally distributed with mean 40 pounds and standard deviation
5 pounds. A part is considered defective if the tensile strength is less than 35 pounds.
(a) what is the probability that a part is defective?
(b) If a testing sample consists of 5 parts, what is the expected number of parts defective in
such a sample? Assume that each part is independent of the others.
Answer:
a) There is a 15.87% probability that a part is defective.
b) The expected number of parts defective in such a sample is 0.7935.
Step-by-step explanation:
To solve this question, we use concepts of the normal probability distribution and the binomial probability distribution.
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
(a) what is the probability that a part is defective?
The tensile strength of a metal part is normally distributed with mean 40 pounds and standard deviation
5 pounds. A part is considered defective if the tensile strength is less than 35 pounds.
Here we have [tex]\mu = 40, \sigma = 5[/tex]
This probability is the pvalue of Z when X = 35.
So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{35 - 40}{5}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.1587.
There is a 15.87% probability that a part is defective.
(b) If a testing sample consists of 5 parts, what is the expected number of parts defective in such a sample? Assume that each part is independent of the others.
This is the expected value of a binomial distribution when [tex]n = 5, p = 0.1587[/tex].
So
[tex]E(X) = np = 5*0.1587 = 0.7935[/tex]
The expected number of parts defective in such a sample is 0.7935.
What is the probability that more than twelve loads occur during a 4-year period? (Round your answer to three decimal places.)
Answer:
Given that an article suggests
that a Poisson process can be used to represent the occurrence of
structural loads over time. Suppose the mean time between occurrences of
loads is 0.4 year. a). How many loads can be expected to occur during a 4-year period? b). What is the probability that more than 11 loads occur during a
4-year period? c). How long must a time period be so that the probability of no loads
occurring during that period is at most 0.3?Part A:The number of loads that can be expected to occur during a 4-year period is given by:Part B:The expected value of the number of loads to occur during the 4-year period is 10 loads.This means that the mean is 10.The probability of a poisson distribution is given by where: k = 0, 1, 2, . . ., 11 and λ = 10.The probability that more than 11 loads occur during a
4-year period is given by:1 - [P(k = 0) + P(k = 1) + P(k = 2) + . . . + P(k = 11)]= 1 - [0.000045 + 0.000454 + 0.002270 + 0.007567 + 0.018917 + 0.037833 + 0.063055 + 0.090079 + 0.112599 + 0.125110+ 0.125110 + 0.113736]= 1 - 0.571665 = 0.428335 Therefore, the probability that more than eleven loads occur during a 4-year period is 0.4283Part C:The time period that must be so that the probability of no loads occurring during that period is at most 0.3 is obtained from the equation:Therefore, the time period that must be so that the probability of no loads
occurring during that period is at most 0.3 is given by: 3.3 years
Step-by-step explanation:
In square inches, the area of the square is 4x^2 - 2x - 6 and the area of the triangle 2x^2 + 4x - 5. What polynomial represents the area of the shaded region?
Answer: the polynomial representing the shaded region is
2x² - 6x - 1
Step-by-step explanation:
The area of the square is expressed as 4x² - 2x - 6
The area of the triangle is expressed as 2x² + 4x - 5. The area of the shaded region would be the area of the square - the area of the area of the triangle.
The polynomial that represents the area of the shaded region would be
4x² - 2x - 6 - (2x² + 4x - 5)
Collecting like terms, it becomes
= 4x² - 2x² - 2x - 4x - 6 + 5
= 2x² - 6x - 1
Final answer:
The area of the shaded region is represented by the polynomial [tex]2x^2 - 6x - 1[/tex] square inches, which is the difference between the area of the square and the area of the triangle.
Explanation:
The student asked for the polynomial that represents the area of the shaded region given the area of the square and the area of the triangle within it. To find this area, we must subtract the area of the triangle from the area of the square. The given area of the square is [tex]4x^2 - 2x - 6[/tex] square inches, and the area of the triangle is [tex]2x^2 + 4x - 5[/tex] square inches. Therefore, the polynomial representing the area of the shaded region is found by the following subtraction:
Area of the square - Area of the triangle = [tex](4x^2 - 2x - 6) - (2x^2 + 4x - 5) = 2x^2 - 6x - 1[/tex]
This simplifies to 2x^2 (from the square's area) minus 6x (the change in linear dimensions caused by the triangle) minus 1 (the constant term from completing the square). Thus, the polynomial representing the shaded region is [tex]2x^2 - 6x - 1[/tex] square inches.
The weights of bags of peas are normally distributed with a mean of 13.50 ounces and a standard deviation of 1.06 ounces. Bags in the upper 4% are too heavy and must be repackaged. What is the most that a bag can weigh and not need to be repackaged?
Answer:
The most that a bag can weigh and not need to be repackaged is 15.355 ounces.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 13.50, \sigma = 1.06[/tex]
Bags in the upper 4% are too heavy and must be repackaged. What is the most that a bag can weigh and not need to be repackaged?
Bags in the upper 4% have a pvalue of 1-0.04 = 0.96. So the most that a bag can weight and not need to be repackaged is the value of X when Z has a pvalue of 0.9599. So this is X when [tex]Z = 1.75[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.75 = \frac{X - 13.50}{1.06}[/tex]
[tex]X - 13.50 = 1.75*1.06[/tex]
[tex]X = 15.355[/tex]
The most that a bag can weigh and not need to be repackaged is 15.355 ounces.
Of the 44 students in a class, 38 are taking the class because it is a major requirement, and the other 6 are taking it as an elective. If two students are selected at random from this class, what is the probability that the first student is taking the class as an elective and the second is taking it because it is a major requirement?
Answer:
The probability that of the 2 students selected 1st is taking the class for an elective and 2nd for major is 0.121.
Step-by-step explanation:
The number of students taking the class because it is a major requirement is, n (M) = 38.
The number of students taking the class because it as an elective is, n (E) = 6.
The total number of students is, N = 44.
Two students are selected at random.
Assume that the selection is without replacement.
Compute the probability that the 1st student selected is taking the class as an elective and the 2nd is taking it because it is a major requirement as follows:
P (1st Elective ∩ 2nd Major) = P (1st Elective) × P (2nd Major)
[tex]=\frac{6}{44}\times \frac{38}{43}\\ =0.120507\\\approx0.121[/tex]
Thus, the probability that of the 2 students selected 1st is taking the class for an elective and 2nd for major is 0.121.
According to a recent poll 53% of Americans would vote for the incumbent president. If a random sample of 100 people results in 40% who would vote for the incumbent, test whether the claim that the actual percentage is different from 53% is supported or not supported.
(1) State the null hypothesis.
(2) State the alternative hypothesis.
(3) What is the test statistic used for the test (z or t)?
(4) State the significance or alpha (α) level?
Answer:
1) Null hypothesis:[tex]p=0.53[/tex]
2)Alternative hypothesis:[tex]p \neq 0.53[/tex]
3) [tex]z=\frac{0.4 -0.53}{\sqrt{\frac{0.53(1-0.53)}{100}}}=-2.605[/tex]
4) We assume that [tex]\alpha=0.05[/tex]
[tex]p_v =2*P(z<-2.605)=0.0092[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who would vote for the incumbent is different from 0.53.
Step-by-step explanation:
Data given and notation
n=100 represent the random sample taken
[tex]\hat p=0.4[/tex] estimated proportion of people who would vote for the incumbent
[tex]p_o=0.53[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level (assumed)
Confidence=95% or 0.95 (Assumed)
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the true proportion is 0.53 or not.:
1) Null hypothesis:[tex]p=0.53[/tex]
2)Alternative hypothesis:[tex]p \neq 0.53[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.4 -0.53}{\sqrt{\frac{0.53(1-0.53)}{100}}}=-2.605[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level assumed [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
[tex]p_v =2*P(z<-2.605)=0.0092[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who would vote for the incumbent is different from 0.53 .
A sales representative makes presentations about a product in three homes per day. In each home, there may be a sale (denote by S) or there may be no sale (denote by F).
Determine the sample space for the experiment.
Answer:
Sample Space = {SSS,SSF,SFF,SFS,FSS,FSF,FFS,FFF}
Step-by-step explanation:
Given a sale representative makes presentations about a product in three homes per day.
In each home, Sale is denoted by S and No Sale is denoted by F.
Now the cases for required sample space will be :
Sale in first home, Sale in second home and Sale in third home.Sale in first home, Sale in second home and No Sale in third home.Sale in first home, No Sale in second home and No Sale in third home.Sale in first home, No Sale in second home and Sale in third home.No Sale in first home, Sale in second home and Sale in third home.No Sale in first home, Sale in second home and No Sale in third home.No Sale in first home, No Sale in second home and Sale in third home.No Sale in first home, No Sale in second home and No Sale in third home.So our Required sample space for the experiment
= {SSS,SSF,SFF,SFS,FSS,FSF,FFS,FFF}.
A levee was designed to protect against floods with an annual exceedance probability of 0.02. A larger flood would cause the levee to fail. What is the risk that the levee will NEVER fail in the next 20 years?
Answer:
66.76% probability that the levee will NEVER fail in the next 20 years.
Step-by-step explanation:
For each year, there are only two possible outcomes. Either a levee fails during the year, or no levees fail. In each year, the probabilities of levees failing are independent from each other. So we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
A levee was designed to protect against floods with an annual exceedance probability of 0.02. This means that [tex]p = 0.02[/tex]
What is the risk that the levee will NEVER fail in the next 20 years?
This is [tex]P(X = 0)[/tex] when [tex]n = 20[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{20,0}*(0.02)^{0}*.(0.98)^{20} = 0.6676[/tex]
66.76% probability that the levee will NEVER fail in the next 20 years.
The risk that a levee, designed to withstand floods with an annual exceedance probability of 0.02, will never fail in the next 20 years is approximately 67%, assuming each year's flood risk is independent and identical.
Explanation:The question asks about a levee's failure probability over the next 20 years. If a levee is designed to protect against floods with an annual exceedance probability of 0.02, this means that there's a 2% chance each year that the levee will fail due to a flood.
To determine the risk that the levee will never fail in 20 years, we must first understand the probability that it won't fail in any given year, which is 1 - 0.02 = 0.98 (or 98%). The probability that the levee will not fail in 20 consecutive years is therefore 0.98^(20), or approximately 0.67 (or 67%), assuming each year's flood risk is independent and identical.
This calculation, while seemingly straightforward, doesn't account for variables like changes in climate, improvements in the levee's design and maintenance, or other unforeseen factors that might affect the levee's performance.
Learn more about Probability here:https://brainly.com/question/32117953
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The monthly salary of all adults in a certain US town is known to be right skewed with a mean of $3700. If a random sample of size 40 adults were selected from that town, what would be the shape of the distribution of the sample mean?
Answer:
The shape of the distribution of the sample mean would be bell-shaped(normally distributed).
Step-by-step explanation:
The Central Limit Theorem answers this question.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size, of at least 30, can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].
In this problem, we have that
The population distribution is right-skewed.
Sampling distribution of the sample mean with size 40.
The Central Limit Theorem estabilishes that the sampling distribution of the sample mean will be normally distributed, even if the distribution of the population is not.
So the shape of the distribution of the sample mean would be bell-shaped(normally distributed).
Tom practiced piano 1 and 1/3 hours on Monday. If he spent half as much time practicing on Tuesday, how long did he practice on Tuesday? Please show/tell me how you got the answer!!
Answer: 2/3 hours
Step-by-step explanation: 1 1/3 will equal 4/3 as you have a whole number equaling 3/1 so then you divide by two and get 2/3
Paul consumes only books and DVDs. At his current consumption bundle, his marginal utility from DVDs is 1212 and from books is 44. Each DVD costs $66, and each book costs $33. Is he maximizing his utility? Explain. Let MU Subscript Upper BMUB be the marginal utility of books, MU Subscript Upper DMUD be the marginal utility from DVDs, Upper P Subscript Upper BPB be the price of books, Upper P Subscript Upper DPD be the price of DVDs, and MRS be the marginal rate of substitution. Paul is
Answer: MUb / Pb ∠ MUd / Pd
Step-by-step explanation:
First let us define the parameters given;
we have that :
Marginal utility of from book MUb = 44
Marginal utility of from DVD MUd = 1212
Cost of DVD Pd = $ 66
Cost of Book Pb = $ 33
Using,
⇒ MUb / Pb = 44 / 33 = 1.33
⇒ MUd / Pd = 1212 / 66 = 18.36
From this we can see that;
MUb / Pb ∠ MUd / Pd
The question asked is if he is maximizing his utility?
From the law of equi-marginal utility, a consumer will maximize utility only when the utility derived from every unit of spending on goods and services is the same. That is to say that for this to happen,
MUb / Pb = MUd /Pd.
But from the question solved, this is not true, so from the answer gotten The consumer is not maximizing his utility because;
MUb / Pb ∠ MUd / Pd
cheers i hope this helps.
Trevor decides to start saving money for a new car. He knows he can invest money into an account which will earn 6.8% APR, compounded weekly, and would like to have saved $12,000 after 4 years. How much money will he need to invest into the account now so that he has $12,000 after 4 years.
Answer:
PV=$9,143.88
Step-by-step explanation:
Compound Interest
When a principal amount (also called present value PV) is saved at some rate of interest r for some time t, the future value FV that includes the original investment plus the interests is computed as
[tex]FV=PV\left(1+r\right)^t[/tex]
If the investment is compounded other than annually, then r and t must be scaled to the proper time units.
If we already know the future value, then the present value is computed by solving the above equation
[tex]PV=FV\left(1+r\right)^{-t}[/tex]
The Annual Percentage Rate (APR) is 6.8% compounded weekly, so the value of r is (assuming 52 weeks per year)
[tex]\displaystyle r=\frac{6.8}{100\times 52}=0.00131[/tex]
And the time is computed in weeks
t=4*52=208
The present value of the investment Trevor needs to save now is
[tex]PV=12,000\left(1+0.00131\right)^{-208}[/tex]
[tex]\boxed{PV=\$9,143.88}[/tex]
Trevor will need to invest $9,143.88 into the account to have $12,000 after 4 years
Eighty percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive vehicles pass or fail independently of one another, calculate the following probabilities. (Enter your answers to three decimal places.)(a)P(all of the next three vehicles inspected pass)(b)P(at least one of the next three inspected fails)(c)P(exactly one of the next three inspected passes)(d)P(at most one of the next three vehicles inspected passes)(e) Given that at least one of the next three vehicles passes inspection, what is the probability that all three pass (a conditional probability)? (Round your answer to three decimal places.)
Answer:
(a) P(all of the next three vehicles inspected pass) = 0.512 .
(b) P(at least one of the next three inspected fails) = 0.488 .
(c) P(exactly one of the next three inspected passes) = 0.096 .
(d) P(at most one of the next three vehicles inspected passes) = 0.104 .
(e) Probability that all three pass given that at least one of the next three vehicles passes inspection = 0.516 .
Step-by-step explanation:
We are given the Probability of all vehicles examined at a certain emissions inspection station pass the inspection to be 80%.
So, Probability that the next vehicle examined fails the inspection is 20%.
Also, it is given that successive vehicles pass or fail independently of one another.
(a) P(all of the next three vehicles inspected pass) = Probability that first vehicle, second vehicle and third vehicle also pass the inspection
= 0.8 * 0.8 * 0.8 = 0.512
(b) P(at least one of the next three inspected fails) =
1 - P(none of the next three inspected fails) = 1 - P(all next three passes)
= 1 - (0.8 * 0.8 * 0.8) = 1 - 0.512 = 0.488 .
(c) P(exactly one of the next three inspected passes) is given by ;
First vehicle pass the inspection, second and third vehicle doesn't passSecond vehicle pass the inspection, first and third vehicle doesn't passThird vehicle pass the inspection, first and second vehicle doesn't passHence, P(exactly one of the next three inspected passes) = Add all above cases ;
(0.8 * 0.2 * 0.2) + (0.2 * 0.8 * 0.2) + (0.2 * 0.2 * 0.8) = 0.096 .
(d) P(at most one of the next three vehicles inspected passes) = P(that none
of the next three vehicle passes) + P(Only one of the next three passes)
We have calculated the P(Only one of the next three passes) in the above part of this question;
Hence, P(at most one of the next three vehicles inspected passes) =
(0.2 * 0.2 * 0.2) + (0.8 * 0.2 * 0.2) + (0.2 * 0.8 * 0.2) + (0.2 * 0.2 * 0.8)
= 0.008 + 0.096 = 0.104 .
(e) Probability that all three pass given that at least one of the next three vehicles passes inspection is given by;
P(All three passes/At least one of the next three vehicles passes inspection)
= P( All next three passes [tex]\bigcap[/tex] At least one of next three passes) /
P(At least one of next three passes)
= P( All next three passes ) / P(At least one of next three passes)
= P( All next three passes ) / 1 - P(none of the next three passes)
= [tex]\frac{0.8*0.8*0.8}{1-(0.2*0.2*0.2)}[/tex] = 0.516 .
Therefore, Probability that all three pass given that at least one of the next three vehicles passes inspection = 0.516 .
a) P(all of the next three vehicles pass) = 0.512
(b) P(at least one of the next three inspected fails) = 0.488
(c) P(exactly one of the next three inspected passes) = 0.096
(d) P(at most one of the next three vehicles inspected passes) = 0.104
(e) Given that at least one of the next three vehicles passes inspection, the probability that all three pass is approximately 1.049 (rounded to three decimal places).
Let's calculate the probabilities step by step:
(a) To find the probability that all of the next three vehicles pass, we can use the probability of a single vehicle passing (0.80) raised to the power of 3 (because the events are independent).
P(all pass) = (0.80)^3 = 0.512
(b) To find the probability that at least one of the next three vehicles fails, we can use the complement rule. It's easier to find the probability that none of them fail and then subtract that from 1.
P(at least one fails) = 1 - P(all pass) = 1 - 0.512 = 0.488
(c) To find the probability that exactly one of the next three vehicles passes, we need to consider three cases: Pass-Fail-Fail, Fail-Pass-Fail, and Fail-Fail-Pass. Each of these cases has a probability of (0.80) * (0.20) * (0.20).
P(exactly one passes) = 3 * (0.80) * (0.20) * (0.20) = 0.096
(d) To find the probability that at most one of the next three vehicles passes, we sum the probabilities of exactly one passing and none passing.
P(at most one passes) = P(none pass) + P(exactly one passes) = (0.20)^3 + 0.096 = 0.104
(e) To calculate the conditional probability that all three pass given that at least one passes, we use the formula for conditional probability:
P(all three pass | at least one passes) = P(all three pass and at least one passes) / P(at least one passes)
P(all three pass and at least one passes) = P(all pass) = 0.512 (from part a)
So, P(all three pass | at least one passes) = 0.512 / 0.488 ≈ 1.049 (rounded to three decimal places).
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According to the manufacturer of the candy Skittles, 25% of the candy produced are green. If we take a random sample of 10 bags of Skittles, what is the probability that the proportion in our sample of green candies will be more than 25%?
Answer:
49.32% probability that the proportion in our sample of green candies will be more than 25%.
Step-by-step explanation:
For each candy, there are only two possible outcomes. Either it is green, or it is not. The probabilities of each candy being green are independent from each other. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
According to the manufacturer of the candy Skittles, 25% of the candy produced are green. This means that [tex]p = 0.25[/tex]
If we take a random sample of 10 bags of Skittles, what is the probability that the proportion in our sample of green candies will be more than 25%?
10 bags, so [tex]n = 10[/tex]
This is [tex]P(X > 0.25*10) = P(X > 2.5)[/tex]
The number of bags is a discrete number, so [tex]P(X > 2.5) = P(X \geq 3)[/tex].
Either the number of green bags is lower than 3, or it is greater or equal than 3. The sum of these probabilities is decimal 1. Mathematically, this means that:
[tex]P(X \leq 2) + P(X \geq 3) = 1[/tex]
We want [tex]P(X \geq 3)[/tex]. So
[tex]P(X \geq 3) = 1 - P(X \leq 2)[/tex]
In which
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{10,0}.(0.25)^{0}.(0.75)^{10} = 0.0563[/tex]
[tex]P(X = 1) = C_{10,1}.(0.25)^{1}.(0.75)^{9} = 0.1689[/tex]
[tex]P(X = 2) = C_{10,2}.(0.25)^{2}.(0.75)^{8} = 0.2816[/tex]
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0563 + 0.1689 + 0.2816 = 0.5068[/tex]
Then
[tex]P(X \geq 3) = 1 - P(X \leq 2) = 1 - 0.5068 = 0.4932[/tex]
49.32% probability that the proportion in our sample of green candies will be more than 25%.
Final answer:
The subject matter revolves around the concept of probability in mathematics, particularly involving calculations using the binomial and Poisson distributions, and statistical methods for analyzing sample data.
Explanation:
The question posed relates to the concept of probability in mathematics, specifically the calculation of the probability of obtaining a certain number of items of a particular color in a random sample from a given population. This field of study typically involves understanding and applying theories of probability distributions, such as the binomial distribution and the Poisson distribution, as well as using statistical methods to estimate population parameters and test hypotheses.
For example, to find the probability of drawing exactly 3 green marbles from a bag containing 50 marbles of which 15 are green, one would need to use the binomial distribution formula P(X = k) = C(n, k) p^k (1-p)^(n-k), where 'C(n, k)' is the combination of 'n' items taken 'k' at a time, 'p' is the probability of success on a single trial, and 'X' is the random variable representing the number of successes in 'n' trials. Similar concepts are applicable to the drawing of candies from a bag and calculating the proportion of colors in a sample to match or compare with theoretical distributions.
Exercises that involve calculating mean and standard deviations for repeated sampling, such as in the exercise with M&M's, are instances of applying statistical methods to analyze data. This process of analysis provides a deeper understanding of the underlying distributions and the behavior of random variables.
List more than two events (i.e., categorical events) that might describe the outcome of each situation given below (a) A student applies for admission to Oxnard University. (You may select more than one answer. Click the box with a check mark for the correct answer and double click to empty the box for the wrong answer.) a) Admitted unconditionally b) Awarded a degree c) Not admitted d) Granted a visa e) Admitted conditionally
Answer:
Step-by-step explanation: From the above question, the possible outcome are:
a) Admitted unconditionally √
b) Awarded a degree √
c) Not admitted √
d) Granted a visa ×
e) Admitted conditionally √
A typical person has an average heart rate of 70.0 70.0 beats/min. Calculate the given questions. How many beats does she have in 6.0 6.0 years? How many beats in 6.00 6.00 years? And finally, how many beats in 6.000 6.000 years? Pay close attention to significant figures in this question.
Answer:
a) 2.2 × 10⁸ beats
b) 2.20 × 10⁸ beats
c) 2.207 × 10⁸ beats
Step-by-step explanation:
Data provided in the question:
Average heart rate of a typical person = 70.0 beats/min
Now,
In the given cases, the significance is on the significant figures after the decimal
Therefore,
the answer is will be provided accordingly
Now,
a) Time = 6.0 years
[since 1 significant figure after decimal. answer will be give in 1 significant figure after decimal ]
time in minutes = 6.0 × 365 × 24 × 60
= 3.1 × 10⁶ minutes
Total beats = Average heart rate × Time
= 70 × 3.1 × 10⁶
= 2.2 × 10⁸ beats
b) Time = 6.00 years
[since 2 significant figure after decimal. answer will be give in 2 significant figure after decimal ]
time in minutes = 6.00 × 365 × 24 × 60
= 3.15 × 10⁶ minutes
Total beats = Average heart rate × Time
= 70 × 3.15 × 10⁶
= 2.20 × 10⁸ beats
c) Time = 6.000 years
[since 3 significant figure after decimal. answer will be give in 3 significant figure after decimal ]
time in minutes = 6.000 × 365 × 24 × 60
= 3.154 × 10⁶ minutes
Total beats = Average heart rate × Time
= 70 × 3.154 × 10⁶
= 2.207 × 10⁸ beats
G and H are mutually exclusive events.
• P(G) = 0.5
• P(H) = 0.3
1.) Explain why the following statement MUST be false: P(H | G) = 0.4. (Select a letter for your answer)
(A.)The events are mutually exclusive, which makes P(H AND G) = 0; therefore, P(H | G) = 0.
(B.)The statement is false because P(H | G) = P(H)/P(G)= 0.6
(C.) The events are mutually exclusive, which means they can be added together, and the sum is not 0.4.
(D.)To find conditional probability, divide P(G AND H) by P(H), which gives 0.5.
2.) Find P(H OR G).
3.) Are G and H independent or dependent events? Explain. (Select a letter for your answer)
(A.) G and H are dependent events because P(G OR H) ≠ 1.
(B.) G and H are independent events because they are mutually exclusive.
(C.) G and H are dependent events because they are mutually exclusive.T
(D.) There is not enough information to determine if G and H are independent or dependent events
Answer:
(1) Correct option is (A)
(2) P (H or G) = 0.80
(3) Correct option is (C)
Step-by-step explanation:
Mutually exclusive events are those events that cannot occur at same time.
If events A and B are mutually exclusive then, P (A and B) = 0.
Given: P (G) = 0.50 and P (H) = 0.30
(1)
The statement is: P (H|G) = 0.40.
The conditional probability of event B given event A is:
[tex]P(B|A)=\frac{P(A\cap B)}{P(A)}[/tex]
The probability statement P (H|G) can be written as:
[tex]P(H|G)=\frac{P(H\cap G)}{P(G)}[/tex]
But as H and G are mutually exclusive, P (H ∩ G) = 0.
Hence, P (H|G) = 0.
Thus, the provided statement is false because events H and G are mutually exclusive, which makes P(H ∩ G) = 0.
Option (A) is correct.
(2)
The addition rule of probability states that:
[tex]P(A\cup B)=P(A)+P(B)-P(A\cap B)[/tex]
Compute the value of P (H or G) as follows:
[tex]P(H\cup G)=P(H)+P(G)-P(H\cap G\\=0.30+0.50-0\\=0.80[/tex]
Thus, the value of P (H or G) is 0.80.
(3)
Independent events are those events that are not affected by the occurrence of other events.
Events H and G are mutually exclusive events, i.e. occurrence of one affects the occurrence of other.
Thus, events H and G are dependent events.
Option (C) is correct.
Mutually exclusive events have a probability of 0 when they occur together. To find the probability of G OR H, use P(H) + P(G) - P(H AND G). G and H are independent events because they are mutually exclusive.
Explanation:G and H are mutually exclusive events, which means they cannot occur at the same time. The probability of two mutually exclusive events happening together, P(H AND G), is always 0.
Therefore, the statement P(H | G) = 0.4 must be false. The correct answer is (A.) The events are mutually exclusive, which makes P(H AND G) = 0; therefore, P(H | G) = 0.
To find P(H OR G), you can use the formula P(H OR G) = P(H) + P(G) - P(H AND G). Plugging in the given values, P(H OR G) = 0.3 + 0.5 - 0 = 0.8.
G and H are mutually exclusive events, which means that the occurrence of one event does not affect the probability of the other event happening.
Therefore, G and H are independent events.
The correct answer is (B.) G and H are independent events because they are mutually exclusive.
Solve 5x − 6y = −38
3x + 4y = 0
(4, 3)
(−4, 3)
(4, −3)
(−4, −3)
Answer:
The answer is :- ( -4 , 3 )
List and explain the elements that make up the structure of a fugue, then list and explain at least four techniques Bach utilized when writing fugues to create musical development.
Final answer:
The structure of a fugue includes elements such as the subject, counter-subject, expositions, and episodes. Bach used techniques like inversion, retrograde, augmentation, and diminution to develop his fugues.
Explanation:
The structure of a fugue traditionally includes a subject, which is the main theme introduced at the beginning, a counter-subject, which complements the subject, expositions where the subject is introduced in different voices, and episodes, which are free sections providing contrast. Johann Sebastian Bach used various techniques to create musical development within his fugues, some of which include:
Inversion, where the subject is mirrored vertically.
Retrograde, where the subject is played backwards.
Augmentation, where the subject is presented with longer note values, thus slower.
Diminution, where the subject is presented with shorter note values, making it faster.
These techniques helped to create complexity and interest, ensuring that the fugue evolves musically throughout the piece.
(3 pts) Given sample statistics for two data sets: _ Set A: x 23 Med 22 S 1.2 _ Set B: x 24 Med 29 S 3.1 a) Calculate the Pearson’s Index of Skewness for both sets. 3 b) Based on your findings, what type of distribution each set has (circle correct answer): Set A: symmetric skewed left skewed right uniform Set B: symmetric skewed left skewed right uniform c) Which set, A or B, can be considered and analyzed as symmetric _______
Answer:
Step-by-step explanation:
Given Data:
Set A : x = 23, Med = 22, S = 1.2
Set B : x = 24, Med = 29, S = 3.1
The Formula for Pearson's Index of Skewness (for Median in given data) is:
[tex]Sk_{2} = 3(\frac{x - Med}{S} )[/tex]
where,
[tex]Sk_{2} =[/tex] Pearson's Coefficient of Skewness
[tex]Med =[/tex] Median of Distribution
[tex]x=[/tex] Mean of Distribution
[tex]S=[/tex] Standard Deviation of Distribution
a) Finding Skewness:
For Set A:
[tex]Sk_{2_{A}} = 3(\frac{23 - 22}{1.2} )\\\\Sk_{2_{A}} = (\frac{3}{1.2} )\\\\Sk_{2_{A}} = 2.5[/tex]
For Set B:
[tex]Sk_{2_{B}} = 3(\frac{24 - 29}{3.1} )\\\\Sk_{2_{B}} = 3(\frac{-5}{3.1} )\\\\Sk_{2_{B}} = -4.84[/tex]
b) Type of Distribution:
For Set A:
As the value of skewness is a positive value (i.e. 2.5). Hence, Set A is right (positively) skewed.
For Set B:
As the value of skewness is a negative value (i.e. -4.84). Hence, Set B is skewed left (or negatively skewed).
c) Which Set can be considered as symmetric?
As the Pearson's Coefficient of skewness for Set A (2.5) is closer to 0 as compared to that of Set B (-4.84). Set A is more closer to that of a symmetric distribution and therefore can be considered as one.
Answer:
Is there a picture?
Step-by-step explanation:
It would be better if i could see a image!
The diameter of a brand of ping-pong balls is approximately normally distributed, with a mean of 1.32 inches and a standard deviation of 0.08 inch. A random sample of 4 ping-pong balls is selected. Complete parts (a) through (d). a. What is the sampling distribution of the mean? A. Because the population diameter of Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 will also be approximately normal. B. Because the population diameter of Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 can not be found. C. Because the population diameter of Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 will not be approximately normal. D. Because the population diameter of Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 will be the uniform distribution.
Answer:
a) A. Because the population diameter of Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 will also be approximately normal.
b) [tex]P(\bar X <1.28)=P(Z<\frac{1.28-1.32}{\frac{0.08}{\sqrt{4}}}=-1)[/tex]
And using a calculator, excel or the normal standard table we have that:
[tex]P(Z<-1)=0.159[/tex]
c) [tex]P(1.28< \bar X <1.34)=P(\frac{1.28-1.32}{\frac{0.08}{\sqrt{4}}}<Z<\frac{1.34-1.32}{\frac{0.08}{\sqrt{4}}})= P(-1< Z<0.5)[/tex]
And using a calculator, excel or the normal standard table we have that:
[tex]P(-1<Z<0.5)=P(Z<0.5)-P(Z<-1) = 0.691-0.159=0.532 [/tex]
Step-by-step explanation:
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".
Solution to the problem
Part a
Let X the random variable that represent the diameter of a brand of ping pong of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(1.32,0.08)[/tex]
Where [tex]\mu=1.32[/tex] and [tex]\sigma=0.08[/tex]
And we select a sample of size n =4 and we want to find the distribution for [tex]\bar X[/tex]
Since the distribution for X is normal then the distribution for the sample mean [tex]\bar X[/tex] is normal and given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
So the best option for this case would be:
A. Because the population diameter of Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 will also be approximately normal.
Part b
[tex] P(\bar X<1.28)[/tex]
We can use the z score given by:
[tex]z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]P(\bar X <1.28)=P(Z<\frac{1.28-1.32}{\frac{0.08}{\sqrt{4}}}=-1)[/tex]
And using a calculator, excel or the normal standard table we have that:
[tex]P(Z<-1)=0.159[/tex]
Part c
[tex] P(1.28< \bar X<1.34)[/tex]
[tex]P(1.28< \bar X <1.34)=P(\frac{1.28-1.32}{\frac{0.08}{\sqrt{4}}}<Z<\frac{1.34-1.32}{\frac{0.08}{\sqrt{4}}})= P(-1< Z<0.5)[/tex]
And using a calculator, excel or the normal standard table we have that:
[tex]P(-1<Z<0.5)=P(Z<0.5)-P(Z<-1) = 0.691-0.159=0.532 [/tex]
The sampling distribution of the mean diameter of the ping-pong balls, given that the population diameter is normally distributed, will also be normally distributed. The expected mean is 1.32 inches and the standard deviation will be 0.04 inches.
Explanation:The correct answer to this question should be A. Because the population diameter of Ping-Pong balls is approximately normally distributed, the sampling distribution of samples of 4 will also be approximately normal. This is grounded on the Central Limit Theorem which states that the distribution of sample means tends to be normal regardless of the shape of the population distribution, especially when the sample size is large.
In this case, though our sample size (4) is fairly small due, because the distribution of the population is specified as normal, the sampling distribution will remain normal. The important numbers to know would be the expected value (mean which is the same as population mean i.e., 1.32 inches) and the standard deviation (which is the population standard deviation divided by square root of sample size i.e., 0.08/sqrt(4)).
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It is proposed to build a raft of pine logs for carrying a cargo on a river. The cargo will weigh 500 kg, and it must be kept entirely above the water level. How many kilograms of pine logs must we use to make the raft, if the logs may be entirely submerged, and they have SG
Answer:
They will need 200kg of pine logs.
Gauge pressure in the tank = 24045.29 Pa
Absolute pressure= 125370.29Pa
Step-by-step explanation:
Full question:sp.gr=0.8.The fluid in the nanometer is ethyl iodide with sp.gr =1.93. The manometric fluid height is 50 inches. What is the gauge pressure and absolute pressure in the tank?
Converting 50inches to metres. Multiply by 0.0254
50×0.0254=1.27m
Gauge pressure in tank is given by:
P = pgh
Where p= density,g= acceleration due to gravity, h= height
P= 1.93× 1000× 9.81× 1.27= 24045.29Pa
Absolute pressure=Ptotal= Pliquid + Patm
Absolute pressure =24045.29 + 101325=125370.29pa
Normal Distribution Problem. Red Blood Cell Counts are expressed millions per cubic millimeter of whole blood. For healthy females, x has a approximately normal distribution with mu = 4.8 and sigma =.3. Note: my probabilities are exact probabilities. Solutions using the standard normal table will be close.What is the probability that a healthy female has a red blood count between 3.9 and 5.0?a..4787b..2475c..7248d. .7462
Answer:
[tex]P(3.9<X<5)=P(\frac{3.9-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{5-\mu}{\sigma})=P(\frac{3.9-4.8}{0.3}<Z<\frac{5-4.8}{0.3})=P(-3<z<0.67)[/tex]
[tex]P(-3<z<0.67)=P(z<0.67)-P(z<-3)[/tex]
[tex]P(-3<z<0.67)=P(z<0.67)-P(z<-3)=0.748571-0.00135=0.74722[/tex]
d. .7462
Step-by-step explanation:
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".
Solution to the problem
Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(4.8,0.3)[/tex]
Where [tex]\mu=4.8´[/tex] and [tex]\sigma=0.3[/tex]
We are interested on this probability
[tex]P(3.9<X<5.0)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(3.9<X<5)=P(\frac{3.9-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{5-\mu}{\sigma})=P(\frac{3.9-4.8}{0.3}<Z<\frac{5-4.8}{0.3})=P(-3<z<0.67)[/tex]
And we can find this probability with this difference:
[tex]P(-3<z<0.67)=P(z<0.67)-P(z<-3)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-3<z<0.67)=P(z<0.67)-P(z<-3)=0.748571-0.00135=0.74722[/tex]
And on this case the most accurate answer would be:
d. .7462
Answer: option d is the correct answer.
Step-by-step explanation:
For healthy females, x has a approximately normal distribution. We would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = red blood counts.
µ = mean count
σ = standard deviation
From the information given,
µ = 4.8
σ = 0.3
We want to find the that a healthy female has a red blood count between 3.9 and 5.0. It is expressed as
P(3.9 ≤ x ≤ 5) = P(x ≤ 5) - P(x ≤ 3.9)
For x = 3.9,
z = (3.9 - 4.8)/0.3 = - 3
Looking at the normal distribution table, the probability corresponding to the z score is 0.00135
For x = 5,
z = (5 - 4.8)/0.3 = 0.67
Looking at the normal distribution table, the probability corresponding to the z score is 0.7486
P(3.9 ≤ x ≤ 5) = = 0.7486 - 0.00135
= 0.747
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 L of a dye solution with a concentration of 1 g/L. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 L/min, the well-stirred solution flowing out at the same rate.
a. Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.
Answer: t= 460.52 minutes
Step-by-step explanation:Q'=Q/100
Q'= rate in and out of water
Finding the differential equation
Let Q'(t)= The quantity of dye in the tank for t time
But rate in=0 Q/200 ×2=Q'
Q'/Q=-1/100
Dividing by Q gives
Ln/Q/ + c = -1/100 + c1
Integrating both sides gives
Ln/Q/ = -(1/100)t + c2
But c+c1=C2= A constant
Q=C2e(-t/100)
200e-(t/100)
t= ln200
t=460.52minutes
Final answer:
To find the time it takes for the dye concentration in the tank to reach 1% of its original value, we can use the concept of dilution. By applying the dilution formula, we can calculate that it would take approximately 166.67 hours for the concentration to reduce to 1% of its original value.
Explanation:
To find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value, we can use the concept of dilution. The tank is being rinsed with fresh water flowing in at a rate of 2 L/min, and the well-stirred solution is flowing out at the same rate.
The concentration of dye in the tank after a certain amount of time can be calculated using the formula:
C1V1 = C2V2
Where:
C1 = initial concentration of dye (1 g/L)V1 = volume of dye solution (200 L)C2 = final concentration of dye (1% of the original value, which is 0.01 g/L)V2 = volume of water rinsed through the tank (unknown)Rearranging the formula to solve for V2:
V2 = (C1V1) / C2 = (1 g/L × 200 L) / (0.01 g/L) = 20000 L
So, it would take 20000 L / 2 L/min = 10000 min = 166.67 hours for the concentration of dye in the tank to reach 1% of its original value.
Are G and F mutually exclusive events? Explain. G and F are mutually exclusive because if you are not a finalist, then you cannot receive a Green Card. G and F are not mutually exclusive because you could be a finalist and also win a Green Card. G and F are mutually exclusive because becoming a finalist will not allow a person to receive a Green Card. G and F are mutually exclusive because becoming a finalist and receiving a Green Card cannot occur together.
Answer: G and F are mutually exclusive because they cannot occur together
Step-by-step explanation:
According to the definition of mutually exclusive events,
The events which can not occur together and probability of them occurring together is 0 are known as mutually exclusive events.
The first statement gives an implication that if one happens then other happens meaning they could both still happen so it is not true.
The second statement contradict the question about being mutually exclusive events.
The third statement also is a implication that if one event occurs then other does or does not occur.
The last statement is correct one that conforms with the question and obeys the definition of mutually exclusive events.
Suppose the reaction temperature X (in °C) in a certain chemical process has a uniform distribution with A = −8 and B = 8. (a) Compute P(X < 0). (b) Compute P(−4 < X < 4). (c) Compute P(−5 ≤ X ≤ 7).(d) For k satisfying−8 < k < k + 4 < 8,
Answer:
a) 1/2 (50%)
b) 1/2 (50%)
c) 3/4 (75%)
d) 3/4 (75%)
Step-by-step explanation:
for a uniform distribution
P(X=x)= x-A/(B-A) , for A≤x≤B
then
P(X=x)= (x+8)/16 , for -8≤x≤8
a) P(X<0)= interval chosen / total interval = (0-(-8))/(8-(-8)) = 1/2 (50%)
b) P(-4<X<4)= interval chosen / total interval =(4-(-4))/(8-(-8)) = 1/2 (50%)
c) P(-5<X<7)= interval chosen / total interval =(7-(-5))/(8-(-8)) = 3/4 (75%)
d) for −8 < k and k + 4 < 8 → k < 4 , then -8<k<4 , thus k
P(X=k)= P(-8<X<4) = interval chosen / total interval =(4-(-8))/(8-(-8)) = 3/4 (75%)
_____________ refers to the scenario in which the relationship between the dependent variable and one independent variable is different at different values of a second independent variable.
Answer: Interaction
Step-by-step explanation: Interaction in statistics occurs mostly during factorial experiments and analyzes of regression. When two variables interact, it means the
relationship between them and some other variable (dependent), depends on the value of the other interacting variable.Simply put, the relationship between the dependent variable and one independent variable is different at different values of a second independent variable.
Finding interacting variables involve having a factorial design. Here, independent variables are "crossed" with one another so that there are observations at every levels of combination of the independent variables.
Interaction effect refers to the scenario in which the relationship between the dependent variable and one independent variable is different at different values of a second independent variable.
Explanation:Interaction effect refers to the scenario in which the relationship between the dependent variable and one independent variable is different at different values of a second independent variable.
For example, let's say we are studying the effects of studying time and caffeine consumption on test scores. We find that the interaction effect between studying time and caffeine consumption is significant, indicating that the relationship between studying time and test scores depends on the level of caffeine consumption.
This concept is commonly explored in statistics and regression analysis.
hence, the final answer is Interaction effect .
Learn more about Interaction effect here:https://brainly.com/question/35699506
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