Solved Examples on Sampling Distributions

Assume all the requirements are met.
Solve all questions.
Show all work.

Unless otherwise specified, round: all probability values to four decimal places as needed; and all other values to three decimal places as needed.

The names of the towns (in italics) are written to make you smile.
Yes, those towns exist. 😊
I want to make this topic fun in any way I can.

(1.) Several students enrolled in the Introductory Statistics class in the town of Zigzag, Oregon took a Statistics exam.
The average score on that exam is normally distributed with a mean of 70% and a standard deviation of 8%

(a.) One student who took the exam is randomly selected.
Determine the probability that the student's score is at least 74%

(b.) A sample of 36 students who took the exam are randomly selected.
Determine the probability that the average score of that sample is at least 74%

At least 74% means: Greater than OR Equal to 74%

$ \mu = 70 \\[3ex] \sigma = 8 \\[3ex] (a.) \\[3ex] n = 1 \\[3ex] x = 74 \\[3ex] z = \dfrac{x - \mu}{\dfrac{\sigma}{\sqrt{n}}} \\[7ex] x - \mu = 74 - 70 = 4 \\[3ex] \sqrt{n} = \sqrt{1} = 1 \\[3ex] \dfrac{\sigma}{\sqrt{n}} = \dfrac{8}{1} = 8 \\[5ex] \rightarrow z = \dfrac{4}{8} = 0.5 \\[5ex] P(x \lt 74) = P(z \lt 0.50) = 0.691462461274013 \\[3ex] P(x \ge 74) = P(z \ge 0.50) = 1 - 0.691462461274013 \\[3ex] \therefore P(x \ge 74) = 0.308537538725987 \\[5ex] (b.) \\[3ex] n = 36 \\[3ex] x = 74 \\[3ex] z = \dfrac{\bar{x} - \mu}{\dfrac{\sigma}{\sqrt{n}}} \\[7ex] \bar{x} - \mu = 74 - 70 = 4 \\[3ex] \sqrt{n} = \sqrt{36} = 6 \\[3ex] \dfrac{\sigma}{\sqrt{n}} = \dfrac{8}{6} = \dfrac{4}{3} \\[5ex] \rightarrow z = 4 \div \dfrac{4}{3} = 4 * \dfrac{3}{4} = 3 \\[5ex] P(\bar{x} \lt 74) = P(z \lt 3.00) = 0.9986501019683698 \\[3ex] P(\bar{x} \ge 74) = P(z \ge 3.00) = 1 - 0.9986501019683698 \\[3ex] \therefore P(\bar{x} \ge 74) = 0.0013498980316302144 $

(2.) The letter, C is thought to make up an estimated 12% of Igbo language.
A random sample of 700 letters is taken from a randomly selected Igbo literature book and the C’s are counted.
Calculate the probability that the random sample of 700 letters will contain:
(a.) At most 10% of C’s

(b.) At least 10% of C’s

At most 10% means: Less than OR Equal to 10%
At least 10% means: Greater than OR Equal to 10%

$ p = 12\% = \dfrac{12}{100} = 0.12 \\[5ex] q = 1 - p = 1 - 0.12 = 0.88 \\[3ex] n = 700 \\[3ex] \mu_{\hat{p}} = p = 0.12 \\[3ex] \mu_{\hat{p}} = p = 0.12 \\[3ex] \sigma_{\hat{p}} = \sqrt{\dfrac{p * q}{n}} \\[5ex] = \sqrt{\dfrac{0.12 * 0.88}{700}} \\[5ex] = 0.0122823916 \\[3ex] \hat{p} = 10\% = 0.1 \\[3ex] z = \dfrac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} \\[5ex] = \dfrac{0.1 - 0.12}{0.0122823916} \\[5ex] = -1.628347365 \\[3ex] z \approx -1.63 \\[3ex] (a.) \\[3ex] P(z \le -1.63) = 0.0516 \\[3ex] (b.) \\[3ex] P(z \ge -1.63) = 1 - 0.051725625 = 0.9484 $

(3.) Some statisticians in the town of Surprise, Arizona claimed that the mean number of chickens consumed by the population in a month is 200 chickens.
A sample of 144 residents is taken from the population, and the mean number of chickens consumed by that sample in a month is calculated to be 205 chickens.
Assume that somehow, it is known that the standard deviation of the population of interest is 15 chickens

(a.) Determine the probability of taking a sample of n = 144 and getting a sample mean of 205 or more if the actual population mean is 200 and the actual population standard deviation is 15.

(b.) Does this sample outcome support the claim that the mean of the population of interest is 200?

205 or more means Greater than OR Equal to $205$

$ \mu = 200 \\[3ex] \sigma = 15 \\[3ex] (a.) \\[3ex] n = 144 \\[3ex] \bar{x} = 205 \\[3ex] z = \dfrac{\bar{x} - \mu}{\dfrac{\sigma}{\sqrt{n}}} \\[7ex] \bar{x} - \mu = 205 - 200 = 5 \\[3ex] \sqrt{n} = \sqrt{144} = 12 \\[3ex] \dfrac{\sigma}{\sqrt{n}} = \dfrac{15}{12} = \dfrac{5}{4} \\[5ex] \implies \\[3ex] z = 5 \div \dfrac{5}{4} = 5 * \dfrac{4}{5} = 4 \\[5ex] P(\bar{x} \lt 205) = P(z \lt 4.00) = 0.9999683287581669 \approx 1 \\[3ex] P(\bar{x} \ge 205) = P(z \ge 4.00) = 1 - 0.9999683287581669 = 0.00003167124183311998 \approx 0 \\[3ex] \therefore P(\bar{x} \ge 205) = 0 \\[3ex] $ This is an impossible event

Because this is an impossible event, the sample outcome of $144$ does not support the claim that the mean of the population of interest is $200$

(4.) Do you think juries should have the same racial distribution proportion as their demographics?
About 21% of residents in the town of Rough and Ready, California identify as biracial or multiracial.
A local court randomly selects 150 residents to participate in the jury pool.
Use the Central Limit Theorem and the Empirical Rule to find the probability of selecting a proportion of residents that are:
(a.) Less than three standard errors below 21%
(b.) More than three standard errors below 21%
(c.) Less than three standard errors above 21%
(d.) More than three standard errors above 21%
(e.) Within three standard errors of 21%

Remember the definition of the z score.
The z score tells us the number of standard deviations (or in this case, standard errors) from the mean.
From the Central Limit Theorem, $\mu_{\hat{p}} = p = 21\% = 0.21$
There are at least two approaches to solving this question.
Use any approach you prefer.

1st Approach: (Less Work)
(a.) Less than 3 standard errors below the mean = $P(z \le -3) = 0.0013$

(b.) More than 3 standard errors below the mean = $P(z \ge -3) = 1 - 0.0013 = 0.9987$

(c.) Less than 3 standard errors above the mean = $P(z \le 3) = 0.9987$

(d.) More than 3 standard errors above the mean = $P(z \le 3) = 1 - 0.9987 = 0.0013$

(e.) Within 3 standard errors from the mean =
$P(-3 \le z \le 3)$
= $P(z \le 3) - P(z \le -3)$
= 0.9986501 − 0.0013499
= 0.9973

2nd Approach: (More Work)

$ p = 21\% = 0.21 \\[3ex] q = 1 - 0.21 = 0.79 \\[3ex] n = 150 \\[3ex] \sigma_{\hat{p}} = \sqrt{\dfrac{p * q}{n}} \\[5ex] $ 3 standard errors = 3(0.033256578) = 0.09976973489

3 standard errors below 0.21 = 0.21 − 0.09976973489 = 0.1102302651

$ \hat{p} = 0.1102302651 \\[3ex] z = \dfrac{\hat{p} - p}{\sigma_{\hat{p}}} \\[5ex] z = \dfrac{0.1102302651 - 0.21}{0.0332565783} \\[5ex] z = -\dfrac{0.09976973489}{0.0332565783} \\[5ex] z = -3 \\[3ex] (a.) \\[3ex] P(z \le -3) = 0.0013 \\[3ex] (b.) \\[3ex] P(z \ge -3) = 1 - 0.0013 = 0.9987 \\[3ex] $ 3 standard errors above 0.21 = 0.21 + 0.09976973489 = 0.3097697349

$ \hat{p} = 0.3097697349 \\[3ex] z = \dfrac{\hat{p} - p}{\sigma_{\hat{p}}} \\[5ex] z = \dfrac{0.3097697349 - 0.21}{0.0332565783} \\[5ex] z = \dfrac{0.09976973489}{0.0332565783} \\[5ex] z = 3 \\[3ex] (c.) \\[3ex] P(z \le 3) = 0.9987 \\[3ex] (d.) \\[3ex] P(z \ge 3) = 1 - 0.9987 = 0.0013 \\[3ex] $ Within 3 standard errors from the mean = $P(-3 \le z \le 3)$

$ (e.) \\[3ex] P(-3 \le z \le 3) \\[3ex] = P(z \le 3) - P(z \le -3) \\[3ex] = 0.9986501 - 0.0013499 \\[3ex] = 0.9973 \\[3ex] $ NOTE: These answers are comparable to the Empirical Rule (68 – 95 – 99.7% Rule).
Approximately 99.7% of the data lie within 3 standard deviations from the mean.

(5.) A survey of eating habits showed that approximately 5% of people in the City of Weed, California are vegans.
Vegans do not eat meat, poultry, fish, seafood eggs, or milk.
A restaurant in the city expects 350 people on opening night, and the chef is deciding on the menu.
Assume the patrons to be a simple random sample from the city and the surrounding area, which has a population of about 600,000.
If 18 vegan meals are available, what is the approximate probability that there will not be enough vegan meals: that is, the probability that 19 or more vegans will come to the restaurant?
Assume the vegans are independent and there are no families of vegans.

$ p = 5\% = 0.05 \\[3ex] q = 1 - p = 1 - 0.05 = 0.95 \\[3ex] \mu_{\hat{p}} = p = 0.05 \\[3ex] x = 19 \\[3ex] n = 350 \\[3ex] \hat{p} = \dfrac{x}{n} \\[5ex] \hat{p} = \dfrac{19}{350} = 0.0542857143 \\[5ex] \sigma_{\hat{p}} = \sqrt{\dfrac{p * q}{n}} \\[5ex] \sigma_{\hat{p}} = \sqrt{\dfrac{0.05 * 0.95}{350}} \\[5ex] \sigma_{\hat{p}} = 0.0116496475 \\[3ex] z = \dfrac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} \\[5ex] z = \dfrac{0.0542857143 - 0.05}{0.0116496475} \\[5ex] z = 0.3678836033 \\[3ex] z \approx 0.37 \\[3ex] P(z \le 0.37) = 0.6443087 \\[3ex] P(\hat{p} \gt 0.0542857143) = P(z \gt 0.37) = 1 - 0.6443087 = 0.3556913 \\[3ex] P(\hat{p} \gt 0.0542857143) \approx 0.3557 $

(6.) A True/False test has 40 questions. Assume the passing grade to be 58% or more correct answers.
(a.) Determine the probability that a student will guess correctly on one question.
(b.) Determine the probability that a student will guess incorrectly on one question.
(c.) Determine the probability that a person who guesses will pass the test.
(d.) If a similar test were given with multiple-choice questions with four choices for each question, would the approximate probability of passing the test by guessing be higher or lower than the approximate probability of passing the true/false test? Why?

For each question on the True/False test;
n(S) = n(options) = 2
n(correct answer) = 1
n(incorrect answer) = 1

$ (a.) \\[3ex] P(correct\;\;answer) = \dfrac{1}{2} = 0.5 \\[5ex] (b.) \\[3ex] P(incorrect\;\;answer) = \dfrac{1}{2} = 0.5 \\[5ex] p = P(correct\;\;answer) = 0.5 \\[3ex] \mu_{\hat{p}} = p = 0.5 \\[3ex] q = P(incorrect\;\;answer) = 0.5 \\[3ex] \hat{p} = 58\% = 0.58 \\[3ex] n = 40 \\[3ex] \sigma_{\hat{p}} = \sqrt{\dfrac{p * q}{n}} \\[5ex] \sigma_{\hat{p}} = \sqrt{\dfrac{0.5 * 0.5}{40}} \\[5ex] \sigma_{\hat{p}} = 0.0790569415 \\[3ex] z = \dfrac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} \\[5ex] z = \dfrac{0.58 - 0.5}{0.0790569415} \\[5ex] z = 1.011928851 \approx 1.01 \\[3ex] P(z \le 1.01) = 0.84375235 \\[3ex] (c.) \\[3ex] P(\hat{p} \gt 0.58) \\[3ex] = P(z \gt 1.01) \\[3ex] = 1 - 0.84375235 \\[3ex] = 0.15624765 \\[3ex] $ (d.)
For each question on the multiple-choice test;
n(S) = n(options) = 4
n(correct answer) = 1
n(incorrect answers) = 3

$ P(correct\;\;answer) = \dfrac{1}{4} = 0.25 \\[5ex] P(incorrect\;\;answer) = \dfrac{3}{4} = 0.75 \\[5ex] $ The probability of passing a multiple-choice test would be lower than passing a true/false test because the probability of guessing correctly on a multiple-choice test, 0.25 is lower than guessing correctly on a true/false question test, 0.5

(13.) The school district in the town of Sandwich, Massachusetts has 1000 teachers.
50% of those teachers are men, and 50% are women.
In this district, administrators are promoted from among the teachers.
Presently, there are 50 administrators.
60% of the administrators are men.

(a.) If administrators are selected randomly from the faculty, determine the probability that the percentage of male administrators will be 60% or more.

(b.) If administrators are selected randomly from the faculty, determine the probability that the percentage of female administrators will be 40% or less.

$ (a.) \\[3ex] \underline{Male\;\;Teachers} \\[3ex] p = 50\% = 0.5 \\[3ex] q = 1 - p = 1 - 0.5 = 0.5 \\[3ex] n = 50 \\[3ex] \underline{Male\;\;Administrators} \\[3ex] \hat{p} = 60\% = 0.6 \\[3ex] \mu_{\hat{p}} = p = 0.6 \\[3ex] \sigma_{\hat{p}} = \sqrt{\dfrac{p * q}{n}} \\[5ex] \sigma_{\hat{p}} = \sqrt{\dfrac{0.5 * 0.5}{50}} \\[5ex] \sigma_{\hat{p}} = 0.0707106781 \\[3ex] z = \dfrac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} \\[5ex] z = \dfrac{0.6 - 0.5}{0.0707106781} \\[5ex] z = 1.414213562 \approx 1.41 \\[3ex] P(z \le 1.41) = 0.92073016 \\[3ex] P(\hat{p} \gt 0.6) = P(z \gt 1.41) = 1 - 0.92073016 = 0.07926984 \\[3ex] P(\hat{p} \gt 0.6) \approx 0.0793 \\[3ex] $ Selecting 60% or more male administrators is the same as selecting 40% or less female administrators.
Our answers for both questions should be the same. However, let us solve it.

$ (b.) \\[3ex] \underline{Female\;\;Teachers} \\[3ex] p = 50\% = 0.5 \\[3ex] q = 1 - p = 1 - 0.5 = 0.5 \\[3ex] n = 50 \\[3ex] \underline{Female\;\;Administrators} \\[3ex] \hat{p} = 100\% - 60\% = 40\% = 0.4 \\[3ex] \mu_{\hat{p}} = p = 0.4 \\[3ex] \sigma_{\hat{p}} = \sqrt{\dfrac{p * q}{n}} \\[5ex] \sigma_{\hat{p}} = \sqrt{\dfrac{0.5 * 0.5}{50}} \\[5ex] \sigma_{\hat{p}} = 0.0707106781 \\[3ex] z = \dfrac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} \\[5ex] z = \dfrac{0.4 - 0.5}{0.0707106781} \\[5ex] z = -1.414213562 \approx -1.41 \\[3ex] P(z \le -1.41) = 0.07926984 \\[3ex] P(\hat{p} \le 0.4) = P(z \le -1.41) \approx 0.0793 $

(19.) The weights of full-term newborn babies in the town of No Name, Colorado are normally distributed with a mean of 8 pounds and a standard deviation of 0.6 pounds.

(a.) Determine the probability that the average of four randomly selected babies will be less than 0.6 pounds less than the mean.

(b.) Determine the probability that the average of four randomly selected babies will be more than 0.6 pounds less than the mean.

(c.) Determine the probability that the average of four randomly selected babies will be less than 0.6 pounds more than the mean.

(d.) Determine the probability that the average of four randomly selected babies will be more than 0.6 pounds more than the mean.

(e.) Determine the probability that the average of four randomly selected babies will be within 0.6 pounds of the mean.

(f.) Determine the probability that the average of a randomly selected baby will be within 0.6 pounds of the mean.

$ n = 4 \\[3ex] \mu = 8 \\[3ex] \sigma = 0.6 \\[3ex] $ 0.6 pounds less than the mean ⇒ x̄ = 8 − 0.6 = 7.4

$ z = \dfrac{\bar{x} - \mu_{\bar{x}}}{\dfrac{\sigma}{\sqrt{n}}} \\[7ex] z = \dfrac{7.4 - 8}{\dfrac{0.6}{\sqrt{4}}} \\[7ex] z = \dfrac{7.4 - 8}{\dfrac{0.6}{2}} \\[7ex] z = -\dfrac{0.6}{0.3} \\[5ex] z = -2.00 \\[3ex] (a.) \\[3ex] P(\bar{x} \lt -0.6) = P(z \lt -2.00) = 0.02275013 \approx 0.0228 \\[3ex] (b.) \\[3ex] P(\bar{x} \gt -0.6) = P(z \gt -2.00) = 1 - 0.02275013 = 0.97724987 \approx 0.9772 \\[3ex] $ 0.6 pounds less than the mean ⇒ x̄ = 8 + 0.6 = 8.6

$ z = \dfrac{\bar{x} - \mu_{\bar{x}}}{\dfrac{\sigma}{\sqrt{n}}} \\[7ex] z = \dfrac{8.6 - 8}{\dfrac{0.6}{\sqrt{4}}} \\[7ex] z = \dfrac{8.6 - 8}{\dfrac{0.6}{2}} \\[7ex] z = \dfrac{0.6}{0.3} \\[5ex] z = 2.00 \\[3ex] (c.) \\[3ex] P(\bar{x} \lt 0.6) = P(z \lt 2.00) = 0.97724987 \approx 0.9772 \\[3ex] (d.) \\[3ex] P(\bar{x} \gt 0.6) = P(z \gt 2.00) = 1 - 0.97724987 = 0.02275013 \approx 0.0228 \\[3ex] $ 0.6 pounds within the mean ⇒ that x̄ is between 7.4 and 8.6

$ (e.) \\[3ex] P(7.4 \le x \le 8.6) \\[3ex] = P(-2 \le z \le 2) \\[3ex] = P(z \le 2.00) - P(z \le -2.00) \\[3ex] = 0.97724987 - 0.02275013 \\[3ex] = 0.95449974 \\[3ex] \approx 0.9544 \\[3ex] $ Compare with the Empirical Rule (68 – 95 – 99.7% Rule).
About 95% lie within 2 standard deviations from the mean.

$ n = 1 \\[3ex] \bar{x} = 7.4 \\[3ex] \mu_{\bar{x}} = 8 \\[3ex] \sigma = 0.6 \\[3ex] z = \dfrac{\bar{x} - \mu_{\bar{x}}}{\dfrac{\sigma}{\sqrt{n}}} \\[7ex] z = \dfrac{7.4 - 8}{\dfrac{0.6}{\sqrt{1}}} \\[7ex] z = \dfrac{7.4 - 8}{\dfrac{0.6}{1}} \\[7ex] z = -\dfrac{0.6}{0.6} \\[5ex] z = -1.00 \\[5ex] n = 1 \\[3ex] \bar{x} = 8.6 \\[3ex] \mu_{\bar{x}} = 8 \\[3ex] \sigma = 0.6 \\[3ex] z = \dfrac{\bar{x} - \mu_{\bar{x}}}{\dfrac{\sigma}{\sqrt{n}}} \\[7ex] z = \dfrac{8.6 - 8}{\dfrac{0.6}{\sqrt{1}}} \\[7ex] z = \dfrac{8.6 - 8}{\dfrac{0.6}{1}} \\[7ex] z = \dfrac{0.6}{0.6} \\[5ex] z = 1.00 \\[5ex] P(7.4 \le x \le 8.6) \\[3ex] = P(-1 \le z \le 1) \\[3ex] = P(z \le 1.00) - P(z \le -1.00) \\[3ex] = 0.8413447 - 0.1586553 \\[3ex] = 0.6826894 \\[3ex] \approx 0.6827 \\[3ex] $ Compare with the Empirical Rule (68 – 95 – 99.7% Rule).
About 68% lie within 1 standard deviation from the mean.

The difference between a sample size of 4 and a sample size of for the same population mean, population standard deviation, and the value of the variable is that the distribution of means is taller and narrower for the sample size of 4 than the sample size of 1.
This implies that the distribution of means will have more observations located closer to the center of the distribution.

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