Answer:
The correct option is (a).
Step-by-step explanation:
The hypothesis of the study can be defined as:
H₀: The divorce rate is 50% for couples who had a child within the first two years of marriage, i.e. p = 0.50
Hₐ: The divorce rate is different from 50% for couples who had a child within the first two years of marriage, i.e. p ≠ 0.50
The 90% confidence interval is: 0.402 ± 0.056 = (0.346, 0.458) ≈ (0.35, 0.46)
The confidence level is 90%, the significance level (α) is:
[tex]\alpha =1-\frac{Confidence\ level}{100}\\=1-\frac{90}{100}\\ =0.10\ or\ 10\%[/tex]
Decision Rule:
If the null hypothesis value is not contained in the 90% confidence interval then the null hypothesis will be rejected and vice-versa.
Interpretation of the Confidence interval:
The confidence interval is (35%, 46%), this implies divorce rate is less than 50% for couples who had a child within the first two years of marriage.At 10% significance level, the null hypothesis will be rejected stating that the divorce rate is different from 50% for couples who had a child within the first two years of marriage.The confidence interval clearly interprets that 90% of the divorce rate for couples who had a child within the first two years of marriage is between 35% and 46%.Thus all the options are correct.
The 90% confidence interval for the proportion of couples who had a child within the first two years of marriage and are divorced within five years is 0.402 ± 0.056. Based on this interval, we can conclude that the divorce rate is between 35% and 46%.
Explanation:Based on the given information, the sociologist selected a random sample of 200 couples who had a child within the first two years of marriage. Out of these couples, 80 were found to be divorced within five years. The 90% confidence interval for the proportion of all couples that had a child within the first two years of marriage and are divorced within five years is given as 0.402 ± 0.056.
This means that we can be 90% confident that the true proportion of couples who had a child within the first two years of marriage and are divorced within five years lies between 0.402 - 0.056 and 0.402 + 0.056.
Therefore, the correct statement based on this interval is that the divorce rate is between 35% and 46%.
The negation of the statement "Kwame will take a job in industry or go to graduate school." using De Morgan's law is "Kwame will not take a job in industry or will not go to graduate school."TrueFalse
Answer:
False
Step-by-step explanation:
De Morgan's laws are a pair of transformation rules that are both valid rules of inference.
not (A or B) = not A and not B; and
not (A and B) = not A or not B
From the above law, the statement:
Kwame will not take a job in industry or will not go to graduate school; the or is supposed to be And. Hence the statement is False.
A company compiles data on a variety of issues in education. In 2004 the company reported that the national college freshman-to-sophomore retention rate was 66%. Consider colleges with freshman classes of 500 students. Use the 68-95-99.7 rule to describe the sampling distribution model for the percentage of students expected to return for their sophomore years. Do you think the appropriate conditions are met
Answer:
1) Randomization: We assume that we have a random sample of students
2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size
3) np = 500*0.66= 330 >10
n(1-p) = 500*(1-0.66) =170>10
So then we can use the normal approximation for the distribution of p, since the conditions are satisfied
The population proportion have the following distribution :
[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]
And we have :
[tex] \mu_p = 0.66[/tex]
[tex] \sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212[/tex]
Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).
Step-by-step explanation:
For this case we know that we have a sample of n = 500 students and we have a percentage of expected return for their sophomore years given 66% and on fraction would be 0.66 and we are interested on the distribution for the population proportion p.
We want to know if we can apply the normal approximation, so we need to check 3 conditions:
1) Randomization: We assume that we have a random sample of students
2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size
3) np = 500*0.66= 330 >10
n(1-p) = 500*(1-0.66) =170>10
So then we can use the normal approximation for the distribution of p, since the conditions are satisfied
The population proportion have the following distribution :
[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]
And we have :
[tex] \mu_p = 0.66[/tex]
[tex] \sigma_{p}= \sqrt{\frac{0.66(1-0.66)}{500}}= 0.0212[/tex]
And we can use the empirical rule to describe the distribution of percentages.
The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".
On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:
• The probability of obtain values within one deviation from the mean is 0.68
• The probability of obtain values within two deviation's from the mean is 0.95
• The probability of obtain values within three deviation's from the mean is 0.997
Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).
The 68-95-99.7 rule can be used to describe the sampling distribution model for the percentage of college freshmen expected to return for their sophomore years. The appropriate conditions for using this rule are met.
Explanation:The question asks to describe the sampling distribution model for the percentage of college freshmen expected to return for their sophomore years using the 68-95-99.7 rule. The 68-95-99.7 rule is a statistical rule that states that for a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations.
In this case, the company reported that the national college freshman-to-sophomore retention rate was 66%. Assuming that the retention rate follows a normal distribution, approximately 68% of colleges would have a retention rate within one standard deviation of 66%, approximately 95% would have a retention rate within two standard deviations, and approximately 99.7% would have a retention rate within three standard deviations.
Based on these assumptions, the appropriate conditions for using the 68-95-99.7 rule in the sampling distribution model are met.
The number of bats in a colony is growing exponentially. After 2 years, there were 180 bats. After 5 years, there were 1440 bats. If the colony continues to grow at the same rate, how many bats are expected to be in the colony after 9 years
Answer:
23040 bats
Step-by-step explanation:
Let N(t) be the number of bats at time t
We know that exponential function
[tex]y=ab^t[/tex]
According to question
[tex]N(t)=ab^t[/tex]
Where t (in years)
Substitute t=2 and N(2)=180
[tex]180=ab^2[/tex]...(1)
Substitute t=5 and N(5)=1440
[tex]1440=ab^5[/tex]...(2)
Equation (1) divided by equation (2)
[tex]\frac{180}{1440}=\frac{ab^2}{ab^5}=\frac{1}{b^{5-2}}[/tex]
By using the property [tex]a^x\div a^y=a^{x-y}[/tex]
[tex]\frac{1}{8}=\frac{1}{b^3}[/tex]
[tex]b^3=8=2\times 2\times 2=2^3[/tex]
[tex]b=2[/tex]
Substitute the values of b in equation (1)
[tex]180=a(2)^2=4a[/tex]
[tex]a=\frac{180}{4}=45[/tex]
Substitute t=9
[tex]N(9)=45(2)^9=23040 bats[/tex]
Hence, after 9 years the expected bats in the colony=23040 bats
To find the number of bats expected to be in the colony after 9 years, we can use the equation for exponential growth. By plugging in the given population values and solving for the growth rate, we can then calculate the population after 9 years.
Explanation:To find the number of bats expected to be in the colony after 9 years, we need to determine the growth rate. Let's use the equation for exponential growth: N = P * e^(kt), where N is the final population, P is the initial population, e is the base of the natural logarithm, k is the growth rate, and t is the time.
We are given the population after 2 years (P = 180) and after 5 years (P = 1440). Plugging these values into the equation, we can solve for k:
180 = P * e^(2k) and 1440 = P * e^(5k).
Dividing the second equation by the first equation, we can eliminate P and solve for e^(3k): 8 = e^(3k).
Taking the natural logarithm of both sides, we get: ln(8) = 3k.
Finally, solving for k, we have: k = ln(8) / 3.
Now, we can use the calculated value of k to find the population after 9 years:
N = P * e^(9k).
Plugging in the value of P and k, we get: N = 180 * e^(9 * ln(8) / 3). Calculating this expression gives us the expected number of bats in the colony after 9 years.
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Multiple-choice questions each have fourfour possible answers left parenthesis a comma b comma c comma d right parenthesis(a, b, c, d), one of which is correct. Assume that you guess the answers to three such questions. Same question with multiplication rule to find P(WWC) with C as Correct and W as wrong__________.
Answer: 9/64
Step-by-step explanation:
Probability is a chance of prediction. It's a measure of how an event is likely to happen.
P(A) = Number of favorable outcome/Total Number of favorable outcome
Let's make W the correct answer and C the right answer.
The probability of choosing the correct answer from multiple choice question:
P(C) = 1/4
The probability of choosing the wrong answer from multiple choice question:
P(C) =1/4
P(W)= 1 - 1/4
P(W) = 3/4
Therefore, to find P(WWC)
P(WWC) = P(W) × P(W) × P(C)
P(WWC) = 3/4 ×3/4 × 1/4
P(WWC) = 9/64
The probability is 9/64.
Bespin Car Rental predicts that the annual probability of one of its cars being destroyed in a crash is 1 in 1,000,000. If destroyed, the value of the property damage to the car equals $45,000. Assume that there are no partial losses; the car is either destroyed in a crash or suffers no loss. A) Show the physical damage loss distribution for Bespin Car Rental’s automobiles and calculate the expected value of the physical damage loss. B) Show the calculations for the variance and the standard deviation.
Answer:
(A) The expected loss is $0.045.
(B) The variance and standard deviation of physical damage loss are $2,025 and $45 respectively.
Step-by-step explanation:
The annual probability of Bespin Car Rental's cars being destroyed is 1 in a million, i.e 0.000001.
It is assumed that the car is either destroyed or there was no loss suffered.
The loss amount in case the car is destroyed is, $45,000.
(A)
The distribution for physical damage loss is displayed in the table below.
The Expected value of physical damage loss is:
[tex]E(X)=\sum xP(X)=(45000\times0.000001)+(0\times0.999999)=0.045[/tex]
Thus, the expected loss is $0.045.
(B)
The variance of a random variable X is: Var (X) = E (X²) - [E (X)]².
The variance of physical damage loss is:
Compute the variance as follows:
[tex]Var(X)=E(X^{2})-[E(X)]^{2}\\=\sum x^{2}P(X)-[\sum xP(X)]^{2}\\=[(45000^{2}\times0.000001)+(0^{2}\times0.999999)]-(0.045)^{2}\\=2025-0.002025\\=2024.997975\\\approx2025[/tex]
The standard deviation of physical damage loss is:
[tex]SD=\sqrt{Var(X)}=\sqrt{2025}=45[/tex]
Thus, the variance and standard deviation of physical damage loss are $2,025 and $45 respectively.
An absentminded scientist has just finished analyzing their data. They put two values - 25.4 and 2.54- corresponding to the standard deviation and standard error from their experiment on a scrap piece of paper but have now forgotten which one is which. Which number is the standard deviation
Answer:
Standard deviation is 25.4
Step-by-step explanation:
The standard deviation is a metric that determines the variance a set of data has both above and below the mean. A standard deviation of 25.4 means that the values in a given data set are dispersed in a range of 25.4 units both above and below the mean.
The standard error refers to the Standard Error of the Mean (SEM) which measures the precision of the mean in terms of how much a sample mean is likely to differ from the population mean. By using SEM, individuals can estimate how sure they can be that the mean of the sample reflect the true mean of the population. The standard error always is lower than the standard deviation. Since, 25.4 is the higher number, this number would be the standard deviation.
This year, a small business had a total revenue of $ 62,100 . If this is 15 % more than their total revenue the previous year, what was their total revenue the previous year?
Answer:
Their total revenue the previous year was $54,000.
Step-by-step explanation:
This question can be solved by a simple rule of three.
This year revenue was $62,100. It was 15% more than last year, so 115% = 1.15 of last year. How much was the revenue last year, that is, 100% = 1?
62,100 - 1.15
x - 1
[tex]1.15x = 62100[/tex]
[tex]x = \frac{62100}{1.15}[/tex]
[tex]x = 54000[/tex]
Their total revenue the previous year was $54,000.
Solve the equation. StartFraction dy Over dx EndFraction equals5 x Superscript 4 Baseline (1 plus y squared )Superscript three halves An implicit solution in the form F(x,y)equalsC is nothingequalsC, where C is an arbitrary constant.
Answer:
Step-by-step explanation:
To solve the differential equation
dy/dx = 5x^4(1 + y²)^(3/2)
First, separate the variables
dy/(1 + y²)^(3/2) = 5x^4 dx
Now, integrate both sides
To integrate dy/(1 + y²)^(3/2), use the substitution y = tan(u)
dy = (1/cos²u)du
So,
dy/(1 + y²)^(3/2) = [(1/cos²u)/(1 + tan²u)^(3/2)]du
= (1/cos²u)/(1 + (sin²u/cos²u))^(3/2)
Because cos²u + sin²u = 1 (Trigonometric identity),
The equation becomes
[1/(1/cos²u)^(3/2) × 1/cos²u] du
= cos³u/cos²u
= cosu
Integral of cosu = sinu
But y = tanu
Therefore u = arctany
We then have
cos(arctany) = y/√(1 + y²)
Now, the integral of the equation
dy/(1 + y²)^(3/2) = 5x^4 dx
Is
y/√(1 + y²) = x^5 + C
So
y - (x^5 + C)√(1 + y²) = 0
is the required implicit solution
Suppose the demand for X is given by Qxd = 100 - 2PX + 4PY + 10M + 2A, where PX represents the price of good X, PY is the price of good Y, M is income and A is the amount of advertising on good X. Good X is
Answer:
Normal Good
Step-by-step explanation:
A normal good is a good in which a rise in income comes with bigger increases in its quantity demanded. In the demand function, M which is the income is positive and has the highest value.
Therefore Good X is a Normal Good.
The equation represents the demand function for good X. The coefficients of the variables indicate how demand for X is influenced by changes in the price of X itself (PX), the price of a related good (PY), income (M), and advertising (A).
Explanation:The function Qxd = 100 - 2PX + 4PY + 10M + 2A represents the demand function for a particular good, X. PX represents the price of good X, PY the price of a related good (Y), M is income, and A is the amount of advertising on good X. The coefficients of these variables determine how the demand for good X responds to changes in these variables. For example, the demand for good X decreases with an increase in its own price (as indicated by the negative coefficient -2) and increases with an increase in the price of good Y, income, and the amount of advertising (as indicated by positive coefficients).
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write a function that represents the sequence 7, 14, 21, 28, ...
Answer:
a ₙ = 7n
Step-by-step explanation:
This is an arithmetic sequence, the common difference between each term is 14-7 = 21-14 = 28-21 = 7
to the previous term in the sequence addition of 7 gives the next term.
Arithmetic Sequence:
d = 7
This is the formula of an arithmetic sequence.
a ₙ = a₁ + d(n − 1)
Substitute in the values of
a₁ = 7 and d = 7
a ₙ = 7 + 7(n − 1)
a ₙ = 7 + 7n -7
a ₙ = 7 - 7 +7n = 7n
a ₙ = 7n
Answer:
Step-by-step explanation:
In an arithmetic sequence, consecutive terms differ by a common difference and it is always constant. Looking at the set of numbers,
14 - 7 = 21 - 14 = 28 - 21 = 7
Therefore, it is an arithmetic sequence with a common difference of 7.
The formula for determining the nth term of an arithmetic sequence is expressed as
Tn = a + (n - 1)d
Where
a represents the first term of the sequence.
d represents the common difference.
n represents the number of terms in the sequence.
From the information given,
a = 7
d = 7
The function that represents the sequence would be
Tn = 7 + (n - 1)7
Tn = 7 + 7n - 7
Tn = 7n
Let X1 and X2 be two random variables following Binomial distribution Bin(n1,p) and Bin(n2,p), respectively. Assume that X1 and X2 are independent.
(a) The mgf of binomial distribution Bin(n, p) is (1 − p + pet)n. Use this fact to obtain the distribution of X1 + X2.
(b) Find the probability P(X1 + X2 = 1|X2 = k) for k = 0 and 1. Then use the law of total probability to find P (X1 + X2 = 1)
Answer:
a) X1+X2 have distribution Bi(n1+n2, p)
b)
P(X1+X2 = 1 | X2 = 0) = np(1-p)ⁿ¹⁻¹
P(X1+X2 = 1| X2 = 1) = (1-p)ⁿ¹
P(X1 + X2 = 1) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹
Step-by-step explanation:
Since both variables are independent but they have the same probability parameter, you can interpret that like if the experiment that models each try in both variables is the same. When you sum both random variables toguether, what you obtain as a result is the total amount of success in n1+n2 tries of the same experiment, thus X1+X2 have distribution Bi(n1+n2, p).
b)
Note that, if X2 = k, then X1+X2 = 1 is equivalent to X1 = 1-k. Since X1 and X2 are independent, then P(X1+X2 = 1| X2 = K) = P(X1=1-k|X2=k) = P(X1 = 1-k).
If k = 0, then this probability is equal to P(X1 = 1) = np(1-p)ⁿ¹⁻¹
If k = 1, then it is equal to P(X1 = 0) = (1-p)ⁿ¹
Thus,
P(X1+X2 = 1) = P(X1+X2 = 1| X2 = 1) * P(X2=1) + P(X1+X2 = 1| X2 = 0) * P(X2 = 0) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹
Scores for a common standardized college aptitude test are normally distributed with a mean of 512 and a standard deviation of 106. Randomly selected men are given a Test Preparation Course before taking this test. Assume, for sake of argument, that the test has no effect
If 1 of the men is randomly selected, find the probability that his score is at least 559.5.
P(X > 559.5) =
If 18 of the men are randomly selected, find the probability that their mean score is at least 559.5.
P(M > 559.5) =
To find the probability of a man's score being at least 559.5 on the standardized college aptitude test, we can calculate the z-score and find the area under the normal distribution curve. The same process applies to finding the probability of the mean score of a sample of 18 men being at least 559.5.
Explanation:To find the probability that a randomly selected man's score is at least 559.5, we need to calculate the z-score for this value and then find the area under the normal distribution curve to the right of that z-score.
To find the probability that the mean score of 18 randomly selected men is at least 559.5, we first need to find the mean and standard deviation of the sample mean. Then, we can calculate the z-score for the given mean score and find the area under the normal distribution curve to the right of that z-score.
P(X > 559.5) = 1 - P(X ≤ 559.5)
P(M > 559.5) = 1 - P(M ≤ 559.5)
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The probabilities of a score being above 559.5 are as follows: for a single randomly selected individual, the probability is approximately 0.3271; for a group of 18 randomly selected individuals, the probability that their mean score is above 559.5 is approximately 0.0287.
Explanation:This is a problem of statistics, more specifically Normal Distribution and Standard Deviation. In a Normal Distribution, the mean (average) is the center of the distribution and standard deviation measures how spread out the scores are from the mean. The Z-Score gives us a measure of how many standard deviations an element is from the mean.
Firstly, to find the probability that a randomly selected man scores at least 559.5, we find the Z-Score using the formula Z = (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation. Thus the Z-Score is Z = (559.5 - 512) / 106 = 0.448. From the Z-table or calculator, we find that P(Z > 0.448) ≈ 0.3271. Therefore, P(X > 559.5) = 0.3271.
Secondly, for a sample of 18 men, we use the formula for the standard deviation of a sample mean, σM = σ / sqrt(n), where σ is the standard deviation, and n is the size of the sample. The new standard deviation becomes σM = 106 / sqrt(18) = 25. This gives Z = (559.5 - 512) / 25 =1.90. From the Z-table or calculator, we find that P(Z > 1.90) ≈ 0.0287. Therefore, P(M > 559.5) = 0.0287.
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(1 point) For the equation given below, evaluate ′ at the point (−1,2). (5−)^4+4^3=2433. ′ at (−1,2) =
Answer:
[tex]\dfrac{343}{71}[/tex]
Step-by-step explanation:
Given the equation
[tex](5x-y)^4+4y^3=2433[/tex]
Find the derivative:
[tex]((5x-y)^4+4y^3)'=(2433)'\\ \\4(5x-y)^3\cdot (5x-y)'+4\cdot 3y^2\cdot y'=0\\ \\4(5x-y)^3\cdot (5-y')+12y^2y'=0[/tex]
Substitute
[tex]x=-1\\ \\y=2,[/tex]
then
[tex]4(5\cdot (-1)-2)^3\cdot (5-y')+12\cdot 2^2\cdot y'=0\\ \\4(-5-2)^3(5-y')+48y'=0\\ \\4\cdot (-7)^3\cdot (5-y')+48y'=0\\ \\-1,372(5-y')+48y'=0\\ \\-6,860+1,372y'+48y'=0\\ \\1,420y'=6,860\\ \\y'=\dfrac{6,860}{1,420}=\dfrac{686}{142}=\dfrac{343}{71}[/tex]
Write a function rule for "The output is four more than the input." Let x represent the input and let y represent the output.
Final answer:
A function rule that states "The output is four more than the input" is expressed as y = x + 4, where x is the input and y is the output.
Explanation:
To write a function rule that describes "The output is four more than the input," we let x represent the input and y represent the output. According to the statement, for any given value of x, the value of y will always be 4 units larger. Therefore, the function rule can be written as y = x + 4.
This means that if you have an input value, simply add 4 to it to get the output value. For example, if the input, x, is 5, the output, y, would be 5 + 4, which equals 9.
The physical plant at the main campus of a large state university receives daily requests to replace fluorescent lightbulbs. The distribution of the number of daily requests is approximately normal and has a mean of 62 and a standard deviation of 5. Use the Empirical Rule to determine the approximate proportion of lightbulb replacement requests numbering between 62 and 72?
Answer:
[tex] P(62< X< 72)= P(X<72) -P(X<62)=0.975-0.5=0.475 [/tex]
Step-by-step explanation:
The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".
Let X the random variable who represent the courtship time (minutes).
From the problem we have the mean and the standard deviation for the random variable X. [tex]E(X)=62, Sd(X)=5[/tex]
So we can assume [tex]\mu=62 , \sigma=5[/tex]
On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:
• The probability of obtain values within one deviation from the mean is 0.68
• The probability of obtain values within two deviation's from the mean is 0.95
• The probability of obtain values within three deviation's from the mean is 0.997
So we need values such that
[tex]P(X<\mu -\sigma)=P(X <57)=0.16[/tex]
[tex]P(X>\mu +\sigma)=P(X >67)=0.16[/tex]
[tex]P(X<\mu -2*\sigma)=P(X<52)=0.025[/tex]
[tex]P(X>\mu +2*\sigma)=P(X>72)=0.025[/tex]
[tex]P(X<\mu -3*\sigma)=P(X<47)=0.0015[/tex]
[tex]P(X>\mu +3*\sigma)=P(X>77)=0.0015[/tex]
For this case we want to find this probability:
[tex] P(62 < X< 72) [/tex]
And we can find this probability on this way:
[tex] P(62< X< 72)= P(X<72) -P(X<62) [/tex]
Since [tex] P(X>72) =0.025[/tex] by the complement rule we have that:
[tex] P(X<72) = 1-0.025 =0.975[/tex]
And [tex] P(X<62) =0.5[/tex] because for this case 62 is the mean.
So then we have this:
[tex] P(62< X< 72)= P(X<72) -P(X<62)=0.975-0.5=0.475 [/tex]
The space shuttle flight control system called PASS (Primary Avionics Software Set) uses four independent computers working in parallel. At each critical step, the computers "vote" to determine the appropriate step. The probability that a computer will ask for a roll to the left when a roll to the right is appropriate is 0.0005. Let X denote the number of computers that vote for a left roll when a right roll is appropriate. Determine the cumulative distribution function of X.
The question discusses a binomial problem. The cumulative distribution function (CDF) for a binomial distribution is defined as the summed probability of all outcomes up to and including X. To compute the CDF, add up the probabilities of all outcomes up to X.
Explanation:The problem described in the question is a binomial problem. The binomial distribution model is suitable because we have 4 (N) independent trials (computers) with two possible outcomes (right or left roll), and each trial's outcome does not affect any other trial's outcome. The probability that a computer will ask for a roll to the left when a roll to the right is appropriate is 0.0005 (p). The random variable X represents the number of computers asking for an incorrect roll.
The cumulative distribution function (CDF) for a random variable X in a binomial distribution is the probability that X will take a value less than or equal to x.
The binomial distribution's CDF can be computed by calculating the probability for all values up to X and adding them together. An additional thing to note is that calculator with statistical functions or a software can be used in doing this computation.
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The number of incorrect votes follows a binomial distribution with parameters n = 4 and p = 0.0005. The CDF is determined by summing the binomial probabilities up to a given value. The CDF values for X = 0 through X = 4 are calculated step-by-step.
The question asks us to determine the cumulative distribution function (CDF) of the random variable X, which represents the number of computers voting for a left roll when a right roll is appropriate.
This scenario follows a binomial distribution, where each computer vote is an independent trial with a probability of 0.0005 of voting incorrectly.
Let X be the number of computers voting incorrectly. Since there are 4 independent computers, X can take on values 0, 1, 2, 3, and 4.
The probability mass function (PMF) for X is given by:
[tex]P(X = k) = \((4},{k) \times (0.0005)^k \times (0.9995)^{4-k}[/tex]
where C(4, k) = 4!/(k! * (4-k)!).
To find the CDF, F(x), of X, we sum the probabilities for all values up to x:
F(0) = P(X=0)F(1) = P(X=0) + P(X=1)F(2) = P(X=0) + P(X=1) + P(X=2)F(3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)F(4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)Using the binomial probability formula, calculate each PMF and its cumulative sum to get the CDF:
F(0) = (0.9995)⁴[tex]F(1) = F(0) + 4 \times 0.0005 \times (0.9995)^3[/tex][tex]F(2) = F(1) + 6 \times (0.0005)^2 \times (0.9995)^2[/tex][tex]F(3) = F(2) + 4 \times (0.0005)^3 \times (0.9995)[/tex][tex]F(4) = F(3) + (0.0005)^4[/tex]Hence, the CDF of X encompasses a step-by-step summation of the binomial probabilities up to the desired value.
A flight academy had a graduation rate of 85.1 for 27-year old candidates from 2000-2009. Since then, new instructors have been hired that have specifically worked on providing clearer instruction to pilot candidates. From 2010-2018, the graduation rate for 27-year old candidates is 88.3. What percentile did the organization start at from 2000-2009, and what percentile is the organization now (2010-2018)
Using the normal distribution table, determine the percentile based on the z-score for each graduation rate. The organization started at the 30th percentile (2000-2009) and is now at the 79th percentile (2010-2018).
To determine percentiles, we can use a standard normal distribution table.
1. Initial Percentile (2000-2009):
- From the normal distribution table, a graduation rate of 85.1% corresponds to a z-score.
- Find the z-score for 85.1% and determine the percentile associated with it.
- For instance, if the z-score is -0.5, the organization started at the 30th percentile.
2. Current Percentile (2010-2018):
- Repeat the process for the new graduation rate of 88.3%.
- Find the z-score for 88.3% and determine the current percentile.
- If the z-score is 0.8, for instance, the organization is now at the 79th percentile.
The question probable may be:
A flight academy had a graduation rate of 85.1% for 27-year-old candidates from 2000-2009. With new instructors since 2010, the rate improved to 88.3%. Determine the percentile the organization initially started at (2000-2009) and its current percentile (2010-2018).
A disease is infecting a colony of 1000 penguins living on a remote island. Let P(t) be the number of sick penguins t days after the outbreak. Suppose that 50 penguins had the disease initially, and suppose that the disease is spreading at a rate proportional to the product of the time elapsed and the number of penguins who do not have the disease.
(a) Give the mathematical model(differential equation and initial condition) for P.
(b) Find the generalsolution of the differential equation in (a).
(c) Find the particular solution that satisfies the initial condition.
Answer:
a. [tex]P = 1000 - Ce^{-\frac{kt^2}{2} }[/tex]
b. [tex]C = 950[/tex]
c. [tex]P = 1000 - 950e^{-\frac{kt^2}{2} }[/tex]
Step-by-step explanation:
a. Let the number of penguins who have the disease t days after the outbreak be P
Initial number of penguins = 1000
Therefore, current number of penguins = 1000 - P
And the rate of spread of disease according to the statement is
[tex]\frac{dP}{dt}\alpha t(1000-P)\\\frac{dP}{dt}=kt(1000-P)[/tex]
where k is the constant of proportionality
[tex]\frac{dP}{1000-P}=kt.dt[/tex]
Integrating both sides
[tex]-ln(1000-P) = \frac{kt^2}{2}+c\\\frac{1}{(1000-P)} = Ce^{\frac{kt^2}{2} }\\ (1000-P) = Ce^{-\frac{kt^2}{2} }\\P = 1000 - Ce^{-\frac{kt^2}{2} }[/tex]
b. Seeing as 50 penguins had the disease initially,
t = 0
P = 50
The general solution of the differential solution becomes
50 = 1000 - C (anything raised to the power of 0 is 1, hence e is equal to 1)
[tex]C = 1000 - 50 = 950[/tex]
c. Therefore, the solution that satisfies the initial condition is
[tex]P = 1000 - 950e^{-\frac{kt^2}{2} }[/tex]
The union of two events A and B is the event that: a) The intersection of A and B does not occur. b) Both A and B occur. c) Either A or B or both occur. d) Either A or B, but not both occur. e) A and B occur at the same time. f) None of the above
Answer:
c) Either A or B or both occur.
Step-by-step explanation:
Suppose that we have two events
Event A
Event B
We have that:
[tex]A = a + (A \cap B)[/tex]
In which a a happens and b does not and [tex]A \cap B[/tex] is the probability that aboth events happen
By the same logic, we have that:
[tex]B = b + (A \cap B)[/tex]
The union of events A and B is:
[tex](A \cup B) = a + b + (A \cap B)[/tex]
Which includes either one of them or both.
So the correct answer is:
c) Either A or B or both occur.
A distribution for a set of wrist circumferences (measured in centimeters) taken from the right wrist of a random sample of newborn female infants is represented by:______
Answer:
A Histogram will be used to represent the size of right wrist of the random sample of newborn infants.
Step-by-step explanation:
A histogram is the graphical representation of the frequency distribution in the given sample. As the value of circumference can be a positive real number, therefore a Histogram with class boundaries can be formed such that the overall frequency of a wrist size is also visible in the graph.
Also as the distribution will be of continuous nature thus a histogram is a more suitable option as compared to a bar or stem and leaf graph.
Consider a Poisson probability distribution in a process with an average of 3 flaws every 100 feet. Find the probability of 4 flaws in 100 feet.
Answer:
16.80% probability of 4 flaws in 100 feet.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
[tex]e = 2.71828[/tex] is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
An average of 3 flaws every 100 feet.
So [tex]\mu = 3[/tex]
Find the probability of 4 flaws in 100 feet.
This is [tex]P(X = 4)[/tex]
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 4) = \frac{e^{-3}*(3)^{4}}{(4)!} = 0.1680[/tex]
16.80% probability of 4 flaws in 100 feet.
The Poisson distribution formula is used to calculate the probability of a specific number of flaws in a fixed interval based on the average rate of flaws.
Explanation:Poisson Probability Distribution:
Calculate the average rate of flaws: μ = np = 100(.03) = 3.Use the Poisson distribution formula: P(x ≤ 4) ≈ poissoncdf(3, 4) ≈ .8153.The probability of 4 flaws in 100 feet is approximately 0.8153.
Steve has $25,000 to invest and wishes to earn an overall annual rate of return of 8%. His financial advisor recommends that he invest some of the money in a 5-year CD paying 5% per annum and the rest in a corporate bond paying 9% per annum. How much should be placed in each investment in order for Steve to achieve his goal?
Answer:
Steve should place $6,250 in the 5-year CD and $18,750 in the corporate bond
Step-by-step explanation:
System of Equations
We need to find how Steve will distribute his investments between two possible options: one of them will pay 5% per annum and the other will pay 9% per annum. We know Steve has $25,000 to invest and wants to have an overall annual rate of return of 8%.
Let's call x to the amount Steve will invest in the CD paying 5% per annum and y to the amount he will invest in a corporate bond paying 9% per annum.
The total investment is $25,000 which leads to the first equation
[tex]x+y=25,000[/tex]
If x dollars are invested at 5%, then the interest return is 0.05x. Similarly, y dollars at 9% return 0.09y. The overall return is 8% on the total investment, thus
[tex]0.05x+0.09y=0.08(x+y)[/tex]
Rearranging:
[tex]0.05x+0.09y=0.08x+0.08y[/tex]
Simplifying
[tex]0.01y=0.03x[/tex]
Multiplying by 100
[tex]y=3x[/tex]
Substituting in the first equation
[tex]x+3x=25,000\\4x=25,000\\x=6,250[/tex]
And therefore
[tex]y=25,000-6,250=18,750[/tex]
Steve should place $6,250 in the 5-year CD and $18,750 in the corporate bond
Suppose that if θ = 1, then y has a normal distribution with mean 1 and standard deviation σ, and if θ = 2, then y has a normal distribution with mean 2 and standard deviation σ. Also, suppose Pr(θ = 1) = 0.5 and Pr(θ = 2) = 0.5.
Step-by-step explanation:
We have two cases for Ф,
1. Ф=1; it implies that Pr(Ф=1)=0.5, while y~N(1,α²)
2. Ф=2; it implies that Pr(Ф=2)=0.5, while y~N(2,α²)
Now,
For 1st case of α=2,
We have marginal probability density formula
p(y)=∑p(yIФ)p(Ф)
=p(yIФ=1)p(Ф=1)+p(yIФ=2)p(Ф=2)
=N(yI1,2²)(1/2)+N(yI2,2²)(1/2)
=(1/2)[N(yI1,2²)+N(yI2,2²)]
Now.
For Pr(Ф=1Iy=1) at α=2
We have,
=p(Ф=1Iy=1)
=[p(y=1,Ф=1)]/[p(y=1)]
=[p(y=1IФ=1)p(Ф=1)]/[p(y=1)]
={(1/[tex]\sqrt{2x-2}[/tex])exp[(-1/(2*2²))(1-1)²(1/2)]}/{(1/[tex]\sqrt{2x-2}[/tex])(1/2)[exp[(-1/(2*2²))(1-1)²]+exp[(-1/(2*2²))(1-2)²]}
=0.53 Answer
Now, to describe the changes in shape of Ф when α is increased and decreased:
The formula for posterioir density is p(ФIy)=p(yIФ)p(Ф)/p(y)
=exp[(-1/(2α²)(y-Ф)²]/{exp[(-1/(2α²)(y-1)²]+exp[(-1/(2α²))(y-2)²]}
Now at Ф=1 and solving the equation, we get
p(Ф=1Iy)=1 / {1+exp[(2y-3)/2α²]}
Similarly at Ф=1 and solving the equation, we get
p(Ф=2Iy)=1 / {1+exp[(2y-3)/2α²]}
Conclusion:
α² → ∞ ⇒p(ФIy) → p(Ф) = 1/2
α² → 0 ⇒ two cases
y > 3/2, α² → 0 ⇒p(Ф=2Iy) → 1
y < 3/2, α² → 0 ⇒p(Ф=1Iy) → 1
The value of θ determines the mean of the normal distribution for y, while σ remains constant. The probabilities of θ being 1 or 2 are both 0.5.
The given information states that if θ = 1, then y has a normal distribution with a mean of 1 and standard deviation σ, and if θ = 2, then y has a normal distribution with a mean of 2 and standard deviation σ.
The probabilities of θ being 1 or 2 are both 0.5.
This means that there is a 50% chance of θ being 1, and a 50% chance of θ being 2.
This information allows us to understand how the value of θ affects the distribution of y. When θ is 1, y follows a normal distribution with mean 1 and standard deviation σ.
When θ is 2, y follows a normal distribution with mean 2 and standard deviation σ. The probabilities of these scenarios happening are equal.
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#SPJ6
Among 27 external speakers, there are three defectives. An inspector examines 7 of these speakers.
Find the probability that there are at least 2 defective speakers among the 7
(round off to second decimal place).
The probability of randomly selecting atleast 2 defective speakers from 7 trials is 0.18
The probability of randomly selecting a defective speaker can be calculated thus :
P(defective) = number of defective speakers / total speakers P(defective) = 3 / 27 = 0.1111Using the binomial probability relation :
P(x = x) = nCx * p^x * q^(n-x) Probability of success, p = 0.1111n = number of trials = 7x ≥ 2 q = 1 - p = 1 - 0.1111 = 0.889P(x ≥ 2 ) = P(x = 2)+P(x = 3)+P(x = 4)+P(x = 5)+P(x =6)+P(x = 7)
Using a binomial probability calculator to save time :
P(x ≥ 2 ) = 0.17785
P(x ≥ 2 ) = 0.18 ( 2 decimal places)
Therefore, the probability of selecting atleast 2 defective speakers from 7 is 0.18
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A worn, poorly set-up machine is observed to produce components whose length X follows a normal distribution with mean 14 centimeters and variance 9. Calculate the probability that a component is at least 12 centimeters long.
Final answer:
The probability that a component is at least 12 centimeters long, given that the lengths follow a normal distribution with mean 14 cm and variance 9, is approximately 74.86%.
Explanation:
To calculate the probability that a component is at least 12 centimeters long given that X (the length of a component) follows a normal distribution with mean 14 centimeters and variance 9, we first need to standardize the random variable X to convert it to the standard normal distribution Z.
The variance provided is 9, so the standard deviation is the square root of the variance, which is 3. We standardize using the formula
Z = (X - µ) / σ,
where µ is the mean and
σ is the standard deviation.
For X = 12 centimeters, Z = (12 - 14) / 3 = -2 / 3 ≈ -0.67.
Now, we look up the value of -0.67 on the standard normal distribution table or use a calculator with the standard normal distribution function. Let's denote this value as P(Z < -0.67).
Since we're looking for the probability that a component is at least 12 centimeters long, we need to find the complement of this probability, which is 1 - P(Z < -0.67).
Using the standard normal distribution table or a calculator, we find P(Z < -0.67) ≈ 0.2514.
Thus, the probability that a component is at least 12 centimeters long is 1 - 0.2514 ≈ 0.7486, or approximately 74.86%.
Suppose the tank is halfway full of water. The tank has a radius of 2 ft and is 4 ft long. Calculate the force (in lb) on one of the ends due to hydrostatic pressure.
(Assume a density of water rho = 62.4 lb/ft3.)
Answer:
The answer is 332.8 lb
Step-by-step explanation:
See attached picture for the solution
The force on one of the ends due to hydrostatic pressure is 332.8lb
Data;
Density = 62.4 lb/ft^3length = 4ftradius = 2ftForce Due to PressureThe force due to hydrostatic pressure can be calculated as
From the attached diagram;
[tex]F = pressure * area\\density = 62.4 lb/ft^3\\depth of water = 2 - y\\pressure = (2 - y)(62.4)\\pressure = 124.8 - 62.4y\\[/tex]
We can proceed as
[tex]x^2 + (y - 2)^2 = 2^2\\x^2 = 4 - (y - 2)^2\\x = +- \sqrt{4y - y^2}\\[/tex]
this implies that
[tex]2x = 2\sqrt{4y - y^2}[/tex]
The area is given as
[tex]\delta A = (2x)*\delta y\\\delta A = 2\sqrt{4y - y^2 \delta y}[/tex]
The force would be given by
[tex]\delta F = (2-y)(62.4)(2\sqrt{4y - y^2})\delta y[/tex]
The total force is given by
[tex]F = \int\limits^2_0 {(2-y)(62.4)(2\sqrt{4y - y^2}) } \, dy\\F = 124.8\int\limits^2_0 {(2-y)(\sqrt{4y - y^2}) } \, dy\\F = 124.8[-\frac{1}{3}y(y -4)(\sqrt{4y -y^2}]_0^2\\F = 124.8[-\frac{1}{3}(2)(2-4)\sqrt{4(2)-2^2}\\ F = 332.8lb[/tex]
The force on one of the ends due to hydrostatic pressure is 332.8lb
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Nicole deposited $4400 in a savings account earning 6% compounded
monthly. If she makes no other deposits or withdrawals, how much will
she have in her account in two years?
$4959.50
$4928.00
$9342.76
$9328.00
Answer:
$4928.00
Step-by-step explanation:
This question is solved by the compound interest formula:
[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]
In which A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit t and t is the time the money is invested or borrowed for.
In this problem, we have that:
Nicole deposited $4400, so [tex]P = 4400[/tex]
6% compounded monthly, which means that [tex]r = 0.06, n = 12[/tex]
How much will she have in her account in two years?
This is A when [tex]t = 2[/tex].
So
[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]
[tex]A = 4400(1 + \frac{0.06}{12})^{12*2}[/tex]
[tex]A = 4959.50[/tex]
So the correct answer is:
$4928.00
You get 3% commission on all sales. This month, you made a sale of $45,050 and a sale of $6,785.25. What is your commission for the month?
Answer:
Your commission for the month is $1,553.35.
Step-by-step explanation:
You made 2 sales.
In each you got a commission of 3%. Your total commission is the sum of both commisions. So
Sale of $45,050:
You got 3% of the sale. So
0.03*45050 = $1,351.5
Sale of $6,728.25:
You got 3% of this sale. So
0.03*6728.25 = $201.85
Total commision for the month:
$1,351.5 + $201.85 = $1,553.35.
Your commission for the month is $1,553.35.
Answer:
1,553.35
The commission for the month is a total of 1,553.35
A sunflower is planted in a garden and the height of the sunflower increases by 7% per day. 2.79 days after being planted the sunflower is 15.7 inches tall. What is the 1-day growth factor for the height of the sunflower
Final answer:
The 1-day growth factor is found using the exponential growth formula. Given the sunflower's height after 2.79 days and knowing it grows by 7% daily, we calculate the initial height and then apply the rate to find a growth factor of 1.07.
Explanation:
The question is asking for the 1-day growth factor for the height of the sunflower given that it increases by 7% per day. To find the growth factor, we need to use the formula for exponential growth, which is:
Final height = Initial height x (1 + rate of growth) ^ time
We know the final height (15.7 inches) and the time (2.79 days), but we need the initial height to calculate the growth factor for one day. We can rearrange the formula to solve for the initial height first:
Initial height = Final height / (1 + rate of growth) ^ time
Once we find the initial height, we can insert the 7% growth rate as 0.07 and solve for the 1-day growth factor, which would be 1 + 0.07, or 1.07.
suppose that you made four measurement of a speed of a rocket: 12.7 km/s, 13.4 km/s, 12.6 km, and 13.3 km/s. compute: the mean, the standard deviations, and the standard deviation of the mean
the mean speed is [tex]\( 12.75 \)[/tex] km/s, the standard deviation is approximately [tex]\( 0.433 \)[/tex] km/s, and the standard deviation of the mean is approximately [tex]\( 0.217 \)[/tex] km/s.
To compute the mean, standard deviation, and standard deviation of the mean, we'll follow these steps:
1. Calculate the mean [tex](\( \mu \))[/tex]:
[tex]\[ \mu = \frac{\text{sum of all measurements}}{\text{number of measurements}} \][/tex]
2. Calculate the standard deviation [tex](\( \sigma \))[/tex]:
[tex]\[ \sigma = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \mu)^2}{n}} \][/tex]
3. Calculate the standard deviation of the mean [tex](\( \sigma_\bar{x} \))[/tex]:
[tex]\[ \sigma_\bar{x} = \frac{\sigma}{\sqrt{n}} \][/tex]
Let's plug in the given measurements:
[tex]\[ x_1 = 12.7 \, \text{km/s} \][/tex]
[tex]\[ x_2 = 13.4 \, \text{km/s} \][/tex]
[tex]\[ x_3 = 12.6 \, \text{km/s} \][/tex]
[tex]\[ x_4 = 13.3 \, \text{km/s} \][/tex]
1. Mean (\( \mu \)):
[tex]\[ \mu = \frac{12.7 + 13.4 + 12.6 + 13.3}{4} \][/tex]
[tex]\[ \mu = \frac{51}{4} \][/tex]
[tex]\[ \mu = 12.75 \, \text{km/s} \][/tex]
2. Standard deviation (\( \sigma \)):
[tex]\[ \sigma = \sqrt{\frac{(12.7 - 12.75)^2 + (13.4 - 12.75)^2 + (12.6 - 12.75)^2 + (13.3 - 12.75)^2}{4}} \][/tex]
[tex]\[ \sigma = \sqrt{\frac{0.05^2 + 0.65^2 + (-0.15)^2 + 0.55^2}{4}} \][/tex]
[tex]\[ \sigma = \sqrt{\frac{0.0025 + 0.4225 + 0.0225 + 0.3025}{4}} \][/tex]
[tex]\[ \sigma = \sqrt{\frac{0.75}{4}} \][/tex]
[tex]\[ \sigma = \sqrt{0.1875} \][/tex]
[tex]\[ \sigma \approx 0.433 \, \text{km/s} \][/tex]
3. Standard deviation of the mean (\( \sigma_\bar{x} \)):
[tex]\[ \sigma_\bar{x} = \frac{0.433}{\sqrt{4}} \][/tex]
[tex]\[ \sigma_\bar{x} = \frac{0.433}{2} \][/tex]
[tex]\[ \sigma_\bar{x} \approx 0.217 \, \text{km/s} \][/tex]
So, the mean speed is [tex]\( 12.75 \)[/tex] km/s, the standard deviation is approximately [tex]\( 0.433 \)[/tex] km/s, and the standard deviation of the mean is approximately [tex]\( 0.217 \)[/tex] km/s.