Final answer:
The probability of a student making more than 20 mistakes on the exam is 0.19, making 6 or more mistakes is 0.72, and making at most 20 mistakes is 0.81. Events making at most 20 mistakes and making more than 20 mistakes are complementary.
Explanation:
Given a driving exam consisting of 30 multiple-choice questions, we have the following probabilities:
The probability a student makes fewer than 6 mistakes: 0.28The probability a student makes from 6 to 20 mistakes (inclusive): 0.53Now, we can find the probabilities for each scenario using these probabilities:
A student makes more than 20 mistakes: We calculate this by subtracting the sum of the other two given probabilities from 1. The probability is 1 - (0.28 + 0.53) = 0.19.A student makes 6 or more mistakes: This is the sum of the probabilities of making from 6 to 20 mistakes and making more than 20 mistakes. The probability is 0.53 + 0.19 = 0.72.A student makes at most 20 mistakes: This is the complement of making more than 20 mistakes, so we already have this probability as the sum of making fewer than 6 mistakes and making from 6 to 20 mistakes. The probability is 0.28 + 0.53 = 0.81.The two complementary events are:
Event c: A student makes at most 20 mistakesEvent a: A student makes more than 20 mistakesThese events are complementary because their probabilities sum up to 1.
In a study of pain relievers, 50 people were given product A, and all but 11 experienced relief. In the same study, 100 people were given product B, and all but 14 experienced relief. Fill in the blanks of the statement below to make the statement the most reasonable possible. Produ. V ? performed worse in the study because % failed to get relief with this product, whereas only 6 failed to get relief with Product ?
Product A performed worse in the study because 22% failed to get relief with it, whereas only 14% failed to get relief with Product B.
In the given study of pain relievers, we need to determine which product performed worse based on the percentage of people who did not experience relief. For Product A, 50 people were given the product and all but 11 experienced relief. This means that 11 out of 50 people did not experience relief, so we calculate the failure rate as follows: (11/50) * 100 = 22%. For Product B, 100 people were given the product and all but 14 experienced relief, therefore the failure rate is: (14/100) * 100 = 14%.
With these failure rates, we can now fill in the blanks of the statement:
Product A performed worse in the study because 22% failed to get relief with this product, whereas only 14% failed to get relief with Product B.
The data below are the number of absences and the final grades of 9 randomly selected students from a literature class. Find the equation of the regression line for the given data.What would be the predicted final grade if a student was absent 14 times? Round the regression line values to the nearest hundredth. Round the predicted grade to the nearest whole number Number of absences X 0,3,6, 4,9,2, 15,8,5 Final grade Y 98,86, 80,82, 71,92, 55,76,82
Answer:
The regression equation is:
Final Grade = 96.14 - 2.76 Number of absence
A student who was absent for 14 days received a final grade of 58.
Step-by-step explanation:
The general form a regression equation is:
[tex]y=\alpha +\beta x[/tex]
Here,
y = dependent variable = Final grade
x = independent variable = Number of absence
α = intercept
β = slope
The formula to compute the intercept and slope are:
[tex]\alpha =\frac{\sum Y. \sum X^{2}-\sum X.\sum XY}{n.\sum X^{2}-(\sum X)^{2}}[/tex]
[tex]\beta =\frac{n.\sum XY-\sum X.\sum Y}{n.\sum X^{2}-(\sum X)^{2}}[/tex]
The value of α and β are computed as follows:
[tex]\alpha =\frac{\sum Y. \sum X^{2}-\sum X.\sum XY}{n.\sum X^{2}-(\sum X)^{2}}=\frac{(722\times460-(52\times3732)}{(9\times460)-(52)^{2}} =96.139\approx96.14[/tex]
[tex]\beta =\frac{n.\sum XY-\sum X.\sum Y}{n.\sum X^{2}-(\sum X)^{2}}=\frac{(9\times3732-(52\times722)}{(9\times460)-(52)^{2}} =-2.755\approx-2.76[/tex]
The regression equation is:
Final Grade = 96.14 - 2.76 Number of absence
For the value of Number of absence = 14 compute the value of Final grade as follows:
[tex]Final\ Grade = 96.14 - 2.76\ Number\ of\ absence\\=96.14-(2.76\times14)\\=57.5\\\approx58[/tex]
Thus, a student who was absent for 14 days received a final grade of 58.
To find the predicted final grade for 14 absences, calculate the slope and y-intercept of the regression line for the given data set to form the equation y=mx+b. With x as 14, solve the equation.
Explanation:To answer this question, we first need to find the equation of the regression line using the given number of absences (x) and final grades (y). This is achieved by calculating the slope and y-intercept of the best fit line for the data set. The formula for the slope (m) is given by the expression [N(Σxy) - (Σx)(Σy)] / [N(Σx^2) - (Σx)^2] and the y-intercept (b) by (Σy - m(Σx)) / N. After calculating these values, you can form the equation y = mx + b. Using the equation, input the absent times (14) into the x-variable to predict the final grade.
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Discrete or Continuous? Identify the random variables in Exercises 2, 3, 4, 5, 6, 7, 8, 9, 10, and 11 as either discrete or continuous. Total number of points scored in a football game
Answer:
Discrete variable
Step-by-step explanation:
We are given the following in the question:
Variable:
Total number of points scored in a football game
Discrete and continuous data:
Discrete data is the data that can be expressed in whole numbers. They cannot take all the values within an interval.Discrete variables are usually counted and not measured.Continuous variable can be expressed in fractions and can take any value within an interval.Continuous variable are usually measured and not counted.Since, total number of points score in a foot game are expressed in whole numbers and cannot be expressed in decimals, they are discrete variable. They cannot take all the values within an interval and they are usually counted.
Thus,
Total number of points scored in a football game is a discrete variable.
A major department store chain is interested in estimating the mean amount its credit card customers spent on their first visit to the chain's new store in the mall. Fifteen credit card accounts were randomly sampled and analyzed with the following results: X = $50.50 and S = 20.
Construct a 95% confidence interval for the mean amount its credit card customers spent on their first visit to the chain's new store in the mall assuming that the amount spent follows a normal distribution.
Answer:
95% Confidence interval: (39.43, 61.58)
Step-by-step explanation:
We are given the following in the question:
Sample mean, [tex]\bar{x}[/tex] = $50.50
Sample size, n = 15
Alpha, α = 0.05
Sample standard deviation = 20
95% Confidence interval:
[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]t_{critical}\text{ at degree of freedom 14 and}~\alpha_{0.05} = \pm 2.1447[/tex]
[tex]=50.50 \pm 2.1447(\dfrac{20}{\sqrt{15}} ) \\\\= 50.50 \pm 11.0751 \\= (39.4249,61.5751)\\\approx (39.43, 61.58)[/tex]
95% Confidence interval: (39.43, 61.58)
The 95% confidence interval for the mean amount spent on their first visit to the chain's new store in the mall by credit card customers is approximately $41.99 to $59.01.
Explanation:We can construct the 95% confidence interval using the sample mean ( X = $50.50) and the standard deviation ( S = $20). Since we know that the distribution is normal, we can use the z-score for a 95% confidence level, which is approximately 1.96.
The formula for a confidence interval is given by: X ± Z*(S/√n). By substituting the given values into the formula, we get: 50.5 ± 1.96*(20/√15).
After calculating, we find that the 95% confidence interval is approximately $41.99 - $59.01. Thus, we are 95% confident that the true mean amount spent by the department store's credit card customers on their first visit is between $41.99 and $59.01.
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To better understand the financial burden students are faced with each term, the statistics department would like to know how much their ST201 students are spending on school materials on average. Let’s use our class data to calculate a 95% confidence interval to estimate the average amount ST201 students spend on materials each term. The average from our student survey is $248 and the number of students sampled is 90. Assume . State the question of interest. On average, how much do ST201 students spend on school materials each term? a. (1 point) Identify the parameter. b. Check the conditions. a. (2 points) Does the data come from a random sample? What are some potential biases about the way the data was collected? (1 point) Is the sample size large enough for distribution of the sample mean to be normal according to the rules for Central Limit Theorem?
Answer:
Answer:
a).
The amount spent on school materials for each term of all ST201students
b).
a).
It is not a random sample. This looks like a convenience sampling and there is sampling bias. This sample is not representative of the entire population. Since it is not a random sample it is not appropriate to generalize the results to all students.
b).
The sample size is 80 which is greater than 30. It is large enough to assume normal distribution according to central limit theorem.
c).
mean: $617
z critical value at 95%: 1.96
standard error = σ/sqrt(n) =500/sqrt(80) = 55.9017
lower limit= mean-1.96*se = 617-1.96*55.9017=507.43
upper limit= mean+1.96*se = 617+1.96*55.9017=726.57
d).
The amount spent on school materials for each term for the 80 ST201students is $617. We are 95% confident that amount spent on school materials for each term of all ST201students falls in the interval ($507.43, $726.57).
Step-by-step explanation:
A cupcake stand has 40 chocolate, 30 coconut and 20 banana cupcakes. Alice chooses 20 cupcakes at random to create a box as a present for her friend.
What is the probability that she chose:
(a) Eight banana and 6 coconut cupcakes?
(b) At least 2 chocolate cupcakes?
(c) All cupcakes of the same kind?
Answer:
a) 0.00563
b) 1
c) 0
Step-by-step explanation:
Total = 40+30+20 =90
a) (20C8×30C6×40C6)/90C20
= 0.00563
b) 1 - (no chocolate + 1 chocolate)
1 - [(50C20) + (40C1×50C19)]/90C20
1 - 0.00002478
= 0.9999752187
c) [40C20+20C20+30C20]/90C20
= 0.0000000027045
This is about permutations and combinations.
a) Probability = 0.00563
b) Probability = 0.99997522
c) Probability = 0.0000000027045
We are told cupcakes at the stand are;Chocolate = 40
Coconut = 30
Banana = 20
Total number of chocolates = 40 + 30 + 20
Total number of chocolates = 90
a) Probability that she will choose 8 banana and 6 chocolate cakes if she chooses 20 cupcakes at random will be;
(20C₈ × 30C₆ × 40C₆)/90C₂₀
(125970 × 593775 × 3838380)/50980740277700939310
This gives us 0.00563
b) Probability of at least 2 chocolate cupcakes is;
1 - [P(no chocolate) + P(1 chocolate)]
P(no chocolate) = (50C₂₀)/90C₂₀
P(1 chocolate) = (40C₁ × 50C₁₉)/90C₂₀
Thus;
1 - [P(no chocolate) + P(1 chocolate)] = 1 - [(40C₁ × 50C₁₉) + 50C₂₀]/90C₂₀
This gives us; 0.99997522
c) Probability of getting all cupcakes of same kind is;
(40C₂₀ + 20C₂₀ + 30C₂₀)/90C₂₀
⇒ 0.0000000027045
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onsider a random number generator designed for equally likely outcomes. If numbers between 0 and 99 are chosen, determine which of the following is not correct. a. If 100 numbers are generated comma each integer between 0 and 99 must occur exactly once. b. For each random number generated comma each integer between 0 and 99 has probability 0.01 of being selected. c. If a very large number of random numbers are generated comma then each integer between 0 and 99 would occur close to 1 % of the time. d. The cumulative proportion of times that a 0 is generated tends to get closer to 0.01 as the number of random numbers generated gets larger and larger.
Answer:
The following option is not correct:
(a) If 100 numbers are generated comma each integer between 0 and 99 must occur exactly once.
Step-by-step explanation:
This is not correct, as there is a possibility of an integer being generated twice when 100 numbers are generated.
This can be explained with an example such as stated below:
3 numbers are to be generated.
The number generated can either be 1, 2, or 3.
The probability for all three numbers to be generated once when the generator is run 3 times is:
(1/3)*(1/3)*(1/3) * Number of ways to arrange the three numbers
Thus this probability will be:
Probability = (1/3)^3 * 3!
Probability = 0.222
Since the probability here is not equal to 1, the probability for the same thing happening at a larger scale will also not be 1.
Final answer:
In a random number generator for numbers between 0 and 99, each integer should occur exactly once when 100 numbers are generated.
Explanation:
The correct statement among the options is:
a. If 100 numbers are generated, each integer between 0 and 99 must occur exactly once. This statement holds true in a random number generator designed for equally likely outcomes.Let's break it down:
Option a: To ensure each number between 0 and 99 appears once in 100 numbers, the generator should evenly distribute the outcomes.Option b: The probability of selecting each integer should indeed be 0.01 in an equally likely random number generator.Option c: With a large number of random numbers, each integer between 0 and 99 would indeed occur close to 1% of the time due to the even distribution.Option d: The cumulative proportion of selecting 0 getting closer to 0.01 is a characteristic of equally likely outcomes over a large number of trials.A recent highway safety study found that in 65% of all accidents a driver was wearing a seatbelt. Accident reports indicated that 83% of those drivers escaped serious injury (defined as hospitalization or death), but only 49% of the non-belted drivers were so fortunate. Find the probability that a randomly selected driver was wearing a seatbelt, if this driver was not seriously injured. Show your work (if using notations, make sure to identify them). (Round your answer to 2 places after the decimal point).
Answer:
The probability that a randomly selected driver was wearing a seatbelt, if this driver was not seriously injured, that is, P(B|E) = 0.76
Step-by-step explanation:
Probability of wearing a seatbelt in an accident = P(B) = 65% = 0.65
Probability of not wearing a seatbelt in an accident = P(B') = 1 - 0.65 = 0.35
Probability of escaping hospitalization and/or death given that one is wearing a seatbelt = P(E|B) = 83% = 0.83
Probability of escaping hospitalization and/or death given that one isn't wearing a seatbelt = P(E|B') = 0.49
Find the probability that a randomly selected driver was wearing a seatbelt, if this driver was not seriously injured, that is, P(B|E)
The probability of P(X|Y) is given mathematically as P(X n Y)/P(Y)
P(B|E) = P(B n E)/P(E)
But P(E) is unknown at the moment.
But P(E) = P(B n E) + P(B' n E) mathematically,.
P(B n E) can be obtained using P(E|B) and P(B)
P(E|B) = P(B n E)/P(B)
P(B n E) = P(E|B) × P(B) = 0.83 × 0.65 = 0.5395
And
P(B' n E) can be obtained using P(E|B') and P(B')
P(E|B') = P(B' n E)/P(B')
P(B' n E) = P(E|B') × P(B') = 0.49 × 0.35 = 0.1715
P(E) = P(B n E) + P(B' n E) = 0.5395 + 0.1715 = 0.711
The probability that a randomly selected driver was wearing a seatbelt, if this driver was not seriously injured, that is, P(B|E)
P(B|E) = P(B n E)/P(E) = 0.5395/0.711 = 0.76
A light bulb has a lifetime that is exponential with a mean of 200 days. When it burns out a janitor replaces it immediately. In addition there is a handyman who comes at times of a Poisson process at rate .01 and replaces the bulb as "preventive maintenance." (a) How often is the bulb replaced? (b) In the long run what fraction of the replacements are due to failure?
Answer:
(a) The number of bulbs often replaces is 66.67.
(b) The fraction of the replacements that are due to failure, in the long run, is [tex]\frac{1}{3}[/tex].
Step-by-step explanation:
Let X = lifetime of a bulb and Y = time after which the bulb is replaced.
It is provided that X follows Exponential distribution with mean lifetime of a bulb is, 200 days.
And the rate at which the bulb is replaced is, 0.01 also following an Exponential distribution.
(a)
A bulb is replaced only after it burns out or a handyman comes at times of a Poisson process and replaces it.
Then min (X, Y) follows an Exponential distribution with parameter [tex](\frac{1}{200}+0.01)[/tex].
The mean of an Exponential distribution with parameter θ is:
[tex]Mean=\frac{1}{\theta}[/tex]
Compute the mean of min (X, Y) as follows:
[tex]Mean =\frac{1}{(\frac{1}{200}+0.01)} =\frac{1}{0.015}= 66.67[/tex]
Thus, the number of bulbs often replaces is 66.67.
(b)
Compute the probability of the event (X < Y) as follows:
[tex]P(X<Y)=\frac{0.005}{0.015} =\frac{1}{3}[/tex]
Thus, the fraction of the replacements that are due to failure, in the long run, is [tex]\frac{1}{3}[/tex].
Answer:
(a) The number of bulbs often replaces is 66.67.
(b) The fraction of the replacements that are due to failure, in the long run, is .
Step-by-step explanation:
Most analysts focus on the cost of tuition as the way to measure the cost of a college education. But incidentals, such as textbook costs, are rarely considered. A researcher at Drummand University wishes to estimate the textbook costs of first-year students at Drummand. To do so, she monitored the textbook cost of 250 first-year students and found that their average textbook cost was $300 per semester. Identify the population of interest to the researcher.
Answer:
The population of interest to the researcher were the 250 first-year students that were monitored.
Step-by-step explanation:
In descriptive statistics, the portion of the cost of college education to be determined and has been selected for analysis is calle d "sample", the sample the researcher is interested in, considers the textbooks cost of first-year students, therefore the 250 first-year students is the researcher´s population of interest. This method involved the collection, presentation, and characterization.
Linear differential equations sometimes occur in which one or both of the functions p(t) and g(t) for y′+p(t)y=g(t) have jump discontinuities. If t0 is such a point of discontinuity, then it is necessary to solve the equation separately for t < t0 and t > t0. Afterward, the two solutions are matched so that y is continuous at t0; this is accomplished by a proper choice of the arbitrary constants. The following problem illustrates this situation. Note that it is impossible also to make y′ continuous at t0.
Solve the initial value problem.
y' + 6y = g(t), y(0) = 0
where
g(t) = 1, 0 ≤ t ≤ 1,
= 0, t > 0.
For [tex]0\le t\le1[/tex], we have
[tex]y'+6y=1\implies e^{6t}y'+6e^{6t}=(e^{6t}y)'=e^{6t}\implies y=\dfrac16+Ce^{-6t}[/tex]
Given that [tex]y(0)=0[/tex], we have
[tex]0=\dfrac16+C\implies C=-\dfrac16[/tex]
so that
[tex]y=\dfrac16(1-e^{-6t})[/tex]
For [tex]t>1[/tex] (I think you mistakenly wrote [tex]t>0[/tex], which overlaps with the first definition of [tex]g(t)[/tex]), we have
[tex]y'+6y=0\implies e^{6t}y'+6e^{6t}y=(e^{6t}y)'=0\implies y=Ke^{-6t}[/tex]
We want this to be a continuation of the previously found solution [tex]y[/tex] at [tex]t=1[/tex], which means we need to pick [tex]K[/tex] such that
[tex]\dfrac16(1-e^{-6})=Ke^{-6}\implies K=\dfrac16(e^6-1)[/tex]
Then the solution to the IVP is
[tex]y(t)=\begin{cases}\frac16(1-e^{-6t})&\text{for }0\le t\le1\\\frac{e^6-1}6e^{-6t}&\text{for }t>1\end{cases}[/tex]
Alternatively, we can get around treating [tex]g(t)[/tex] piecemeal and resorting to the Laplace transform. Write
[tex]g(t)=\begin{cases}1&\text{for }0\le t\le1\\0&\text{for }t>1\end{cases}=u(t)-u(t-1)[/tex]
where
[tex]u(t-c)=\begin{cases}0&\text{for }t<c\\1&\text{for }t\ge c\end{cases}[/tex]
is the unit step function.
Take the Laplace transform of both sides of the ODE:
[tex]y'+6y=g(t)\overset{\text{L.T.}}{\implies}(sY-y(0))+6Y=\mathcal L_s\{g(t)\}[/tex]
where [tex]Y=Y(s)[/tex] is the Laplace transform of [tex]y(t)[/tex].
We have
[tex]\mathcal L_s\{g(t)\}=\displaystyle\int_0^\infty g(t)e^{-st}\,\mathrm dt=\int_0^1e^{-st}\,\mathrm dt=\dfrac{1-e^{-s}}s[/tex]
so that
[tex](s+6)Y=\frac{1-e^{-s}}s\implies Y=\dfrac{1-e^{-s}}{s(s+6)}=\dfrac{1-e^{-s}}6\left(\dfrac1s-\dfrac1{s+6}\right)[/tex]
Taking the inverse transform yields
[tex]y=\dfrac{1-u(t-1)}6-\dfrac{e^{-6t}}6(e^tu(t-1)-1)[/tex]
[tex]y=\dfrac{1-e^{-6t}}6+\dfrac{e^{6-6t}-1}6u(t-1)[/tex]
which is equivalent to the same solution found earlier; for [tex]0\le t\le1[/tex], [tex]u(t-1)=0[/tex], so that [tex]y=\frac{1-e^{-6t}}6[/tex]; for [tex]t>1[/tex], [tex]u(t-1)=1[/tex], and [tex]y=\frac{1-e^{-6t}}6+\frac{e^{6-6t}-1}6=\frac{(e^6-1)e^{-6t}}6[/tex].
The given differential equation needs to be solved separately for two time ranges because of the piecewise-defined function g(t). Solution for the corresponding equations are founded using the techniques of homogeneous equation solutions and the integrating factor method. These solutions are then matched at the point of continuity.
Explanation:The given differential equation is a first-order linear differential equation of the form y′+p(t)y=g(t). We need to solve this equation considering two cases due to the piecewise definition of g(t).
Case 1: For 0 ≤ t ≤ 1, g(t) = 1. The corresponding homogeneous equation is y' + 6y = 0, with the solution being y(t) = Ce-6t. We find the particular solution using the integrating factor method, yielding y(t) = t/6 - 1/36 + Ce-6t. Substituting the initial condition y(0) = 0 helps us solve for C, giving the final solution for this range as y(t) = t/6 - 1/36.
Case 2: For t > 1, g(t) = 0. The homogeneous solution is the same as in Case 1, but in this case, no particular solution needs to be added, so the solution is y(t) = Ce-6t. The constant is determined by making the function continuous at t=1. We ultimately get y(t) = (1-e-6(t-1))/36.
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A student wanted to construct a 95% confidence interval for the mean age of students in her statistics class. She randomly selected nine students. Their average age was 19.1 years with a sample standard deviation of 1.5 years. What is the best point estimate for the population mean? A. 1.5 years B. 19.1 years C. 9 years D. 2.1 years
Answer:
Option B) 19.1 years
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 9
Sample mean, [tex]\bar{x}[/tex] = 19.1 years
Alpha, α = 0.05
Population standard deviation, σ = 1.5 years
We have to approximate best point estimate for population mean.
The best point estimate for population mean is the sample mean.
Thus, we can write
[tex]\mu = \bar{x} = 19.1[/tex]
Thus, the correct answer is
Option B) 19.1 years
Suppose the probability that a company will be awarded a certain contract is .25, the probability that it will be awarded a second contract is .21 and the probability that it will get both contracts is .13. What is the probability that the company will win at least one of the two contracts?
Answer:
33% probability that the company will win at least one of the two contracts
Step-by-step explanation:
We solve this problem building the Venn's diagram of these probabilities.
I am going to say that:
A is the probability that a company is awarded the first contract.
B is the probability that a company is awarded the second contract.
We have that:
[tex]A = a + (A \cap B)[/tex]
In which a is the probability that a company is awarded the first contract but not the second and [tex]A \cap B[/tex] is the probability that a company is awarded both contract.
By the same logic, we have that:
[tex]B = b + (A \cap B)[/tex]
The probability that it will get both contracts is .13.
This means that [tex]A \cap B = 0.13[/tex]
The probability that it will be awarded a second contract is .21
This means that [tex]B = 0.21[/tex]
[tex]B = b + (A \cap B)[/tex]
[tex]0.21 = b + 0.13[/tex]
[tex]b = 0.08[/tex]
The probability that a company will be awarded a certain contract is .25
This means that [tex]A = 0.25[/tex]
[tex]A = a + (A \cap B)[/tex]
[tex]0.25 = a + 0.13[/tex]
[tex]a = 0.12[/tex]
What is the probability that the company will win at least one of the two contracts?
[tex]A \cup B = a + b + A \cap B = 0.12 + 0.08 + 0.13 = 0.33[/tex]
33% probability that the company will win at least one of the two contracts
The probability that the company will win at least one of the two contracts is [tex]\(\frac{17}{50}\) or 0.34.[/tex]
To find the probability that the company will win at least one of the two contracts, we can use the principle of inclusion-exclusion. The principle states that the probability of the union of two events (in this case, winning either contract) is the sum of the probabilities of each event occurring individually, minus the probability of both events occurring together.
Let [tex]\(P(A)\)[/tex] be the probability that the company will win the first contract, [tex]\(P(B)\)[/tex] be the probability that the company will win the second contract, and [tex]\(P(A \cap B)\)[/tex] be the probability that the company will win both contracts. We are given:
[tex]\(P(A) = 0.25\), \(P(B) = 0.21\), \(P(A \cap B) = 0.13\).[/tex]
The probability that the company will win at least one contract, [tex]\(P(A \cup B)\)[/tex], is given by:
[tex]\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\][/tex]
Substituting the given probabilities:
[tex]\[P(A \cup B) = 0.25 + 0.21 - 0.13\] \[P(A \cup B) = 0.46 - 0.13\] \[P(A \cup B) = 0.33\][/tex]
To express this probability as a fraction, we can write 0.33 as [tex]\(\frac{33}{100}\)[/tex], which simplifies to[tex]\(\frac{17}{50}\)[/tex] when reduced to its simplest form.
Therefore, the probability that the company will win at least one of the two contracts is [tex]\(\frac{17}{50}\)[/tex] or 0.34
If the volume of the square pyramid is 40 cubic millimeters, and the length of s is 2 millimeters, what is the length of the altitude, h?
Answer: the length of the altitude is 10 mm
Step-by-step explanation:
The formula for determining the volume of a square base pyramid is expressed as
Volume = Ah
Where
A represents the area of the square base.
h represents the height or altitude of the pyramid.
From the information given,
Length of each side of the square base = 2 millimeters
Volume of the square pyramid is 40 cubic.
Area of square base = 2² = 4 mm²
Therefore,
40 = 4h
h = 40/4
h = 10 mm
Suppose you want to have $700,000 for retirement in 35 years. Your account earns 9% interest. How much would you need to deposit in the account each month?
Answer: you should deposit $236.2 each month.
Step-by-step explanation:
We would apply the formula for determining future value involving deposits at constant intervals. It is expressed as
S = R[{(1 + r)^n - 1)}/r][1 + r]
Where
S represents the future value of the investment.
R represents the regular payments made(could be weekly, monthly)
r = represents interest rate/number of payment intervals.
n represents the total number of payments made.
From the information given,
there are 12months in a year, therefore
r = 0.09/12 = 0.0075
n = 12 × 35 = 420
S = $700000
Therefore,
700000 = R[{(1 + 0.0075)^420 - 1)}/0.0075][1 + 0.0075]
700000 = R[{(1.0075)^420 - 1)}/0.0075][1.0075]
700000 = R[{(23.06 - 1)}/0.0075][1.0075]
700000 = R[{22.06}/0.0075][1.0075]
700000 = R[2941.3][1.0075]
700000 = 2963.36R
R = 700000/2963.36
R = 236.2
Three students work independently on a homework problem. The probability that the first student solves the problem is 0.95. The probability that the second student solves the problem is 0.85. The probability that the third student solves the problem is 0.80. What is the probability that all are able to solve the problem
The probability that all three students solve the problem is calculated by multiplying their individual success probabilities together. The total probability in this case is 64.6%.
Explanation:Your question pertains to probability, a topic in Mathematics. When three students independently attempt to solve a problem, and you have the probabilities of their success, the probability that all three will successfully solve the problem is determined by the product of their respective probabilities.
Therefore, the probability that all three students - the first with a probability of 0.95, the second with 0.85, and the third with 0.80 - will successfully solve the problem is calculated as follows:
0.95 * 0.85 * 0.80 = 0.646
Hence, there is a 64.6% probability that all three students will successfully solve the problem.
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Hamid has selected one middle manager from each department that will be affected by the updated system and one lower-level manager from each department, along with a few senior staff as well as the project sponsor for a JAD session. He is trying to_____________.
Answer:
Have a broad mix of organizational levels in the JAD session
Explanation:
It is not possible for Hamid to include every employee in the JAD session, what Hamid needed to do is to select participants from the different departments and other key important people to ensure every one is well represented at the JAD session. Selecting lower-level and mid-level managers from the affected departments as well as the some senior staff and the project sponsor for the JAD session will ensure everyone's interest is well represented at the session.
A study was made of seat belt use among children who were involved in car crashes that caused them to be hospitalized. It was found that children not wearing any restraints had hospital stays with a mean of 7.37 days and a standard deviation of 1.25 days with an approximately normal distribution. (a) Find the probability that their hospital stay is from 5 to 6 days, rounded to five decimal places. .10756 (b) Find the probability that their hospital stay is greater than 6 days, rounded to five decimal places.
Answer:
a) [tex]P(5<X<6)=P(\frac{5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{6-\mu}{\sigma})=P(\frac{5-7.37}{1.25}<Z<\frac{6-7.37}{1.26})=P(-1.90<z<-1.10)[/tex]
And we can find this probability with this difference:
[tex]P(-1.90<z<-1.10)=P(z<-1.10)-P(z<-1.90)[/tex]
And using the norma standard distribution or excel we got:
[tex]P(-1.90<z<-1.10)=P(z<-1.10)-P(z<-1.90)=0.136-0.029=0.107[/tex]
b) [tex]P(X>6) =P(Z> \frac{6-7.37}{1.25}) = P(Z>-1.096)[/tex]
And using the complement rule we got:
[tex]P(Z>-1.096) =1-P(Z<-1.096) = 1-0.137= 0.863[/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".
Part a
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(7.37,1.25)[/tex]
Where [tex]\mu=7.37[/tex] and [tex]\sigma=1.25[/tex]
We are interested on this probability
[tex]P(5<X<6)[/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(5<X<6)=P(\frac{5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{6-\mu}{\sigma})=P(\frac{5-7.37}{1.25}<Z<\frac{6-7.37}{1.26})=P(-1.90<z<-1.10)[/tex]
And we can find this probability with this difference:
[tex]P(-1.90<z<-1.10)=P(z<-1.10)-P(z<-1.90)[/tex]
And using the norma standard distribution or excel we got:
[tex]P(-1.90<z<-1.10)=P(z<-1.10)-P(z<-1.90)=0.136-0.029=0.107[/tex]
Part b
For this case we want this probability:
[tex] P(X>6)[/tex]
And we can use the z score and we got:
[tex]P(X>6) =P(Z> \frac{6-7.37}{1.25}) = P(Z>-1.096)[/tex]
And using the complement rule we got:
[tex]P(Z>-1.096) =1-P(Z<-1.096) = 1-0.137= 0.863[/tex]
To find the probability that the hospital stay is from 5 to 6 days, we need to standardize the values using the z-score formula. The probability that their hospital stay is from 5 to 6 days is approximately 0.10756. The probability that their hospital stay is greater than 6 days is approximately 0.86301.
Explanation:To find the probability that the hospital stay is from 5 to 6 days, we first need to standardize the values using the z-score formula.
z = (x - µ) / σ
Let's calculate the z-scores for x = 5 and x = 6.
For x = 5:
z = (5 - 7.37) / 1.25 = -1.896
For x = 6:
z = (6 - 7.37) / 1.25 = -1.096
Next, we can use the standard normal distribution table or a calculator to find the probabilities associated with these z-scores:
P(z < -1.896) = 0.02999
P(z < -1.096) = 0.13699
To find the probability that the hospital stay is from 5 to 6 days, we subtract P(z < -1.096) from P(z < -1.896):
P(5 < x < 6) = P(z < -1.896) - P(z < -1.096) = 0.02999 - 0.13699 = 0.10756
Therefore, the probability that their hospital stay is from 5 to 6 days is approximately 0.10756, rounded to five decimal places.
To find the probability that their hospital stay is greater than 6 days, we can use the standard normal distribution table or a calculator to find the probability associated with the z-score for x = 6:
P(z > -1.096) = 1 - P(z < -1.096) = 1 - 0.13699 = 0.86301
Therefore, the probability that their hospital stay is greater than 6 days is approximately 0.86301, rounded to five decimal places.
A 30% solution of fertilizer is to be mixed with a 70% solution of fertilizer in order to get 40 gallons of a 60% solution. How many gallons of the 30% solution and 70%
solution should be mixed?
lion of the 30% solution should be mixed?
Answer:10 gallons of 30% solution and 30 gallons of 70% solution should be mixed.
Step-by-step explanation:
Let x represent the number of gallons of the 30% solution that should be mixed.
Let y represent the number of gallons of the 70% solution that should be mixed.
The total number of gallons of the mixture to be made is 40. This means that
x + y = 40
The 30% solution of fertilizer is to be mixed with a 70% solution of fertilizer in order to get 40 gallons of a 60% solution. This means that
0.3x + 0.7y = 0.6 × 40
0.3x + 0.7y = 24- - - - - - - - - - - - - -1
Substituting x = 40 - y into equation 1, it becomes
0.3(40 - y) + 0.7y = 24
12 - 0.3y + 0.7y = 24
- 0.3y + 0.7y = 24 - 12
0.4y = 12
y = 12/0.4
y = 30
x = 40 - y = 40 - 30
x = 10
A fair coin is continually flipped until heads appears for the 10th time. Let X denote the number of tails that occur. Compute the probability mass function of X.
Answer:
The probability mass function is expressed as:
P(x) = [(x+r-1)C(r-1)]*[p^r]*[(1-p)^x]
Step-by-step explanation:
This is not a binomial distribution. It is actually a negative binomial distribution. The probability mass function is expressed below:
P(x) = [(x+r-1)C(r-1)]*[p^r]*[(1-p)^x]
where:
x = number of failures
r-1 = number of successes (10 in this scenario)
p = probability of a success
nCr = n!/[r!(n-r)!]
The main formula difference in the positive binomial versus negative binomial is this: With respect to the negative binomial, it is obviously known that the last event will be: when we reach our 10th "head", we stop .
Thus, the last flip will ALWAYS be a "head".
Your broker recommends that you purchase XYZ Inc. at $60. The stock pays a $2.40 dividend which (like its per share earnings) is expected to grow annually at 6.5 percent. If you want to earn 11.5 percent on your funds, is this a good buy
Answer:
XYZ is NOT a good buy.
Step-by-step explanation:
Calculate the market price of stock:
[tex]\frac{Next year's Dividend}{Reqd.return - Growth rate}[/tex]
[tex]= \frac{(2.4)(1.065)}{0.115-0.065}[/tex]
[tex]= 51.12[/tex]
The Market price of the stock is $51. Therefore, buying the stock at $60 is overpriced and is NOT a good buy.
Terri and Donna both sell crafts at two different craft shows each weekend. Terri is charged a 5% commission on the amount of money she earns and pays $35 for her booth. Donna is charged a 3% commission on the amount of money she earns and pays $55 for her booth. On the last weekend in November, Terri and Donna both earned the same amount of money at their craft shows. They both paid their respective craft shows the same total amount of money for their booths and commission.
Set up a system of equations to model the amount of money Terri and Donna pay each weekend at the craft shows. Let x represent the money earned from sales, let T represent the total amount Terri pays in one weekend, and let D represent the total amount Donna pays in one weekend.
What is the solution to the system of equations found in Part A? Give your answer as an ordered pair.
What does the solution of the system of equations found in Part B represent in the context of this situation? Be sure to explain the meaning of the values in the solution.
Answer:
(x, T) = (x, D) = (1000, 85)each booth pays $85 in fees on rental and sales of $1000Step-by-step explanation:
A. Given
T = 0.05x +35 . . . . Terri's cost of operating a craft booth
D = 0.03x +55 . . . . Donna's cost of operating a craft boot
T = D
where x is the dollar amount of sales.
__
B. Solution
Subtracting the equation for D from that of T, we get ...
T - D = 0
(0.05x +35) -(0.03x +55) = 0 = 0.02x -20
0 = x -1000 . . . . . divide by 0.02
x = 1000
T = D = 0.05(1000) +35 = 85
(x, T) = (x, D) = (1000, 85)
__
C. Meaning
According to the given definitions of the variables, each booth pays a total of $85 in fees for sales of $1000.
The system of equations is T = 0.05x + 35 and D = 0.03x + 55. The solution is (1000, 85), meaning Terri and Donna each earned $1000 from sales and paid $85 total in booth and commission fees.
Explanation:The system of equations to model the amount of money Terri and Donna pay each weekend at the craft shows can be written as follows: T = 0.05x + 35 and D = 0.03x + 55.
Since we know that Terri and Donna both paid the same total amount of money for their booths and commission, it means T = D. Or, we can equate the two equations: 0.05x + 35 = 0.03x + 55. Solving for x gives us x = 1000. Substituting x = 1000 into T = 0.05x + 35 equation, we get T (and D) = 85. So, the solution to the system of equations is (1000, 85).
In the context of this situation, the solution means that Terri and Donna both earned $1000 from sales, and each paid $85 total for their booth and commission fees.
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What point is between 4,16 and 16,16
The point between (4,16) and (16,16) is (10,16) as calculated using the midpoint formula. This point is exactly halfway between the given points.
To determine a point between the two points (4,16) and (16,16), we need to calculate the midpoint.
The formula for finding the midpoint M between two points (x1, y1) and (x2, y2) is:[tex]M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)[/tex]
Putting the coordinates (4,16) and (16,16) into the midpoint formula:[tex]M = \left(\frac{4 + 16}{2}, \frac{16 + 16}{2}\right) = (10, 16)[/tex]
Therefore, the point that lies between (4,16) and (16,16) is (10,16).
The probability density function of the weight of packages delivered by a post office is f(x) = 70/69x^2 for 1 < x < 70 pounds.
a) Determine the mean and variance of weight. Round your answers to two decimal places (e.g. 98.76).
Mean = pounds
Variance = pounds2
b) If the shipping cost is $2.50 per pound, what is the average shipping cost of a package? Round your answer to two decimal places (e.g. 98.76).
pounds
c) Determine the probability that the weight of a package exceeds 59 pounds. Round your answer to four decimal places (e.g. 98.7654).
Answer:
(a) The mean is 4.31 pounds. The variance is 51.42 pounds.
(b) The average shipping cost of a package is $10.78.
(c) The probability that the weight of a package exceeds 59 pounds is 0.0027.
Step-by-step explanation:
The probability density function of the weight of packages is:
[tex]f(x) = \frac{70}{69x^{2}};\ 1 < x < 70[/tex]
(a)
The formula for expected value (or mean) of X is:
[tex]E(X)=\int\limits^a_b {x\times f(x)} \, dx[/tex]
Compute the expected value of X as follows:
[tex]E(X)=\int\limits^{70}_{1} {x\times \frac{70}{69x^{2}}} \, dx=\frac{70}{69} \int\limits^{70}_{1} {x \times x^{-2}} \, dx\\=\frac{70}{69} \int\limits^{70}_{1} {x^{-1}} \, dx=\frac{70}{69} |\ln x|^{70}_{1}\\=\frac{70}{69}\times\ln 70\\=4.31[/tex]
Thus, the mean is 4.31 pounds.
The formula to compute the variance is:
[tex]V(X)=E(X^{2})-[E(X)]^{2}[/tex]
Compute the E (X²) as follows:
[tex]E(X^{2})=\int\limits^{70}_{1} {x^{2}\times \frac{70}{69x^{2}}} \, dx=\frac{70}{69} \int\limits^{70}_{1} {x^{2} \times x^{-2}} \, dx\\=\frac{70}{69} \int\limits^{70}_{1} {1} \, dx=\frac{70}{69} | x|^{70}_{1}\\=\frac{70}{69}\times69\\=70[/tex]
The variance is:
[tex]V(X)=E(X^{2})-[E(X)]^{2}\\=70-(4.31)^{2}\\=51.4239\\\approx51.42[/tex]
Thus, the variance is 51.42 pounds.
(b)
It is provided that the shipping cost for per pound is, C = $2.50.
Compute the average shipping cost of a package as follows:
[tex]Average\ cost=Cost\ per\ pound\times E(X)\\=2.50\times4.31\\=10.775\\\approx10.78[/tex]
Thus, the average shipping cost of a package is $10.78.
(c)
Compute the probability that the weight of a package exceeds 59 pounds as follows:
[tex]P(59<X<70)=\int\limits^{70}_{59} {\frac{70}{69x^{2}}} \, dx=\frac{70}{69} \int\limits^{70}_{59} {x^{-2}} \, dx\\=\frac{70}{69} |-\frac{1}{x}|^{70}_{59}=\frac{70}{69} [-\frac{1}{70}+\frac{1}{59}]\\=\frac{70}{69}\times0.0027\\=0.0027[/tex]
Thus, the probability that the weight of a package exceeds 59 pounds is 0.0027.
The weights of the package follows a probability density function
The mean is 4.31 and the variance is 51.42, respectively.The average cost of shipping a package is $10.78The probability a package weighs over 59 pounds is 0.0027The probability density function is given as:
[tex]\mathbf{f(x) = \frac{70}{69x^2},\ 1 < x < 70}[/tex]
(a) The mean and the variance
The mean is calculated as:
[tex]\mathbf{E(x) = \int\limits^a_b {x \cdot f(x)} \, dx }[/tex]
So, we have:
[tex]\mathbf{E(x) = \int\limits^{70}_1 {x \cdot \frac{70}{69x^2} } \, dx }[/tex]
[tex]\mathbf{E(x) = \int\limits^{70}_1 {\frac{70}{69x} } \, dx }[/tex]
Rewrite as:
[tex]\mathbf{E(x) = \frac{70}{69}\int\limits^{70}_1 {x^{-1} } \, dx }[/tex]
Integrate
[tex]\mathbf{E(x) = \frac{70}{69} {ln(x)}|\limits^{70}_1 } }[/tex]
Expand
[tex]\mathbf{E(x) = \frac{70}{69} \cdot {(ln(70) - ln(1)) }}[/tex]
[tex]\mathbf{E(x) = 4.31 }}[/tex]
The variance is calculated as:
[tex]\mathbf{Var(x) = E(x^2) - (E(x))^2}[/tex]
Where:
[tex]\mathbf{E(x^2) = \int\limits^a_b {x^2 \cdot f(x)} \, dx }[/tex]
So, we have:
[tex]\mathbf{E(x^2) = \int\limits^{70}_1 {x^2 \cdot \frac{70}{69x^2} } \, dx }[/tex]
[tex]\mathbf{E(x^2) = \int\limits^{70}_1 {\frac{70}{69} } \, dx }[/tex]
Rewrite as:
[tex]\mathbf{E(x^2) = \frac{70}{69}\int\limits^{70}_1 {1 } \, dx }[/tex]
Integrate
[tex]\mathbf{E(x^2) = \frac{70}{69} x|\limits^{70}_1 } }[/tex]
Expand
[tex]\mathbf{E(x^2) = \frac{70}{69} \cdot {(70 - 1) }}[/tex]
[tex]\mathbf{E(x^2) = 70 }}[/tex]
So, we have:
[tex]\mathbf{Var(x) = E(x^2) - (E(x))^2}[/tex]
[tex]\mathbf{Var(x) = 70 - 4.31^2}[/tex]
[tex]\mathbf{Var(x) = 51.42}[/tex]
Hence, the mean is 4.31 and the variance is 51.42, respectively.
(b) The average cost of shipping a package
In (a), we have:
[tex]\mathbf{E(x) = 4.31 }}[/tex] ---- Mean
So, the average cost of shipping a package is:
[tex]\mathbf{Average =4.31 \times 2.50}[/tex]
[tex]\mathbf{Average =10.78}[/tex]
Hence, the average cost of shipping a package is $10.78
(c) The probability a package weighs over 59 pounds
This is represented as: P(x > 59)
So, we have:
[tex]\mathbf{P(x > 59) = P(59 < x < 70)}[/tex]
So, we have:
[tex]\mathbf{P(x > 59) = \int\limits^{70}_{59} { \frac{70}{69x^2}} \, dx }[/tex]
Rewrite as:
[tex]\mathbf{P(x > 59) = \frac{70}{69}\int\limits^{70}_{59} { x^{-2}} \, dx }[/tex]
Integrate
[tex]\mathbf{P(x > 59) = \frac{70}{69} \cdot { -\frac 1x}|\limits^{70}_{59}}[/tex]
Expand
[tex]\mathbf{P(x > 59) = \frac{70}{69} \cdot (-\frac{1}{70} + \frac{1}{59})}[/tex]
[tex]\mathbf{P(x > 59) = 0.0027}[/tex]
Hence, the probability a package weighs over 59 pounds is 0.0027
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Lili has 20 friends. Among them are Kevin and Gerry, whoare husband and wive. Lili wants to invite 6 of her friends to her birthdayparty. If neither Kevin nor Gerry will go to a party without the other, howmany choices does Lili have?
Answer:
18
Step-by-step explanation:
if neither wants to go, from 20 it will be 18
A Randstad/Harris interactive survey reported that 25% of employees said their company is loyal to them. Suppose 9 employees are selected randomly and will be interviewed about company loyalty.
A. What is the probability that none of the 9 employees will say their company is loyal to them?
c. What is the probability that 4 of the 9 employees will say their company is loyal to them?
Answer:
(A) 0.999996
(B) 0.11680
Step-by-step explanation:
We are given that a Randstad Harris interactive survey reported that 25% of employees said their company is loyal to them.
And 9 employees are selected randomly and interviewed about company loyalty.
The Binomial probability distribution is given by;
[tex]P(X=r)= \binom{n}{r}p^{r}(1-p)^{n-r} for x = 0,1,2,3,....[/tex]
where, n = number of trials (samples) taken
r = number of success
p = probability of success
In our question; n = 9 , p = 0.25 (as employees saying their company is loyal to them is success to us)
(A) Probability that none of the 9 employees will say their company is loyal to them = 1 - Probability that all 9 employees will say their company is loyal to them
= 1 - P(X = 9) { As here number of success is 9 }
= 1 - [tex]\binom{9}{9}0.25^{9}(1-0.25)^{9-9}[/tex] = 1 - [tex]0.25^{9}[/tex] = 0.999996
(B) Probability that 4 of the 9 employees will say their company is loyal to them = P(X = 4)
P(X = 4) = [tex]\binom{9}{4}0.25^{4}(1-0.25)^{9-4}[/tex]
= [tex]126*0.25^{4}*0.75^{5}[/tex] = 0.11680
Ted is making trail mix for a party. He mixes 1 1/2 cups of nuts, 1/4 cup of raisins, and 1/4 cup of pretzels. How many cups of pretzels does Ted need to make 15 cups of trail mix?
Find the total of everything he is using:
1 1/2 + 1/4 + 1/4 = 2 total cups.
For every two cups of trail mix he uses 1/4 cup of pretzels.
15 cups/ 2 cups = 7.5
7.5 x 1/4 cup of pretzels = 1 7/8 cups of pretzels.
Answer: Ted needs 1.875 cups of pretzels to make 15 cups of trail mix
Step-by-step explanation:
He mixes 1 1/2 cups of nuts, 1/4 cup of raisins, and 1/4 cup of pretzels. This means that the ratio of ratio of the number of cups of nuts used to the number of cups of raisins used to the number of cups of pretzels used is
1.5 : 0.25 : 0.25
The total ratio is
1.5 + 0.25 + 0.25 = 2
Ted need to make 15 cups of trail mix. Therefore, the number of cups of pretzels that he needs to use is
0.25/2 × 15 = 1.875 cups of pretzels
What is the missing number in the table? 5 6 16 50
Answer:
60
Step-by-step explanation:
, $&7%"""%- &xgsfx,, 77$""$66"'++
A surfboard shaper has to limit the cost of development and production to $288 per surfboard. He has already spent $61,466.00 on equipment for the boards. The development and production costs are $142 per board. The cost per board is 142x /x+ 61,466 /x dollars. Determine the number of boards that must be sold to limit the final cost per board to $ 288.
How many boards must be sold to limit the cost per board to$288?
Answer:
At least 421 units of boards need to be sold to limit the cost per board to $288
Step-by-step explanation:
Let the number of surfboards made or sold be x
Total cost = fixed cost + variable cost
Fixed Cost = $61466
Variable Cost = 142 × x = $142x
Total cost = 61466 + 142x
Revenue = unit price × quantity = 288×x = 288x
The number of boards that needs to be sold to limit the cost off a board to $288 is the number of units at the point where the total cost matches the revenue.
61466 + 142x = 288x
288x - 142x = 61466
146x = 61466
x = 421 units.
Let Y be a normal random variable with mean μ and variance σ 2 . Assume that μ is known but σ 2 is unknown. Show that (( Y - μ )/ σ ) 2 is a pivotal quantity. Use this pivotal quantity to derive a 1- α confidence interval for σ 2 . (The answer should be left in terms of critical values for the appropriate distribution.)
Answer:
Step-by-step explanation:
answer is attached below