# Help with Homework

Homework #4: Chapters 4.12-4.14, Due 10/3/22 by Midnight PT
Question 1:
Question 2:
In the library on a university campus, there is a sign in the elevator that indicates a weight limit of 2500
pounds. Assume the average weight of students, faculty and staff on campus is right-skewed, with a
mean of 150 pounds, and standard deviation 27 pounds. A random sample of 16 persons from the
campus is selected.
a. Describe the sampling distribution of the sample mean weight.
b. What is the probability that the average weight of the 16 people in the sample is less than 160
pounds?
c. Suppose the sample of 16 people is placed in the library elevator. What is the probability that the
total weight of the 16 persons on the elevator will exceed the weight limit of 2500 pounds?
Question 3:
Suppose individuals with a certain gene have a 0.70 probability of eventually contracting a certain
disease. Using normal approximation to the Binomial, answer the following questions:
a. If 100 individuals with the gene participate in a lifetime study, what is the distribution of the random
variable, X, describing the number of individuals who will contract the disease?
b. Suppose in the study from the problem above you found 78 of the individuals contracted the disease.
Does this seem too high? Justify your answer by finding the probability that at least 78 individuals
contract the disease.
Question 4:
Question 5: MINITAB PROBLEM
The IQ scores of a certain city follows a bell-shaped curve with a mean of 100 and variance of 225.
(a) Using Minitab, generate a random sample of size 100 from this population (do not copy and paste
the whole raw data to your submission). Draw a histogram of your sample values and calculate the
mean, standard deviation, and variance.
(c) Using the Empirical Rule, estimate the percentage of your sample values that fall within 1, 2, and 3
standard deviations of your sample mean.
(d) Draw a Normal Probability Plot of your sample values and determine whether your sample
distribution follows a normal distribution.
Stat 350A: Chapter 4.12
Part 1: Sampling Distributions
• Population: Entire collection of items or individuals you wish to study
• Sample: A subset of the population that has been selected to study or measure
• Statistics: Measurements made on sample data (Ȳ, s, π)
• Parameters: Measurements made on population data (μ, σ, π)
• A point estimate of a population parameter is a sample statistic that represents a feasible value
of the parameter of interest.
• An unbiased estimator is a sample statistic whose mean value is equal to the value of the
population parameter being estimated.
• The sample mean Ȳ is an unbiased estimator of the population mean μ, but Ȳ varies from sample
to sample (sampling variation).
EXAMPLE 1: Suppose we wish to estimate the mean height of Stat 350A students. The following
samples of size 2 were collected.
Random Sample
1
2
3
4
5
Height 1
74”
65”
66”
64”
63”
Height 2
76”
69”
68”
70”
73”
ȳ
EXAMPLE 2: Tossing a die
(A) Tossing a single die 10,000 times.
(B) Tossing a pair of dice 10,000 times and calculating the average of each pair.
(C) Tossing twenty dice 10,000 times and calculating the averages of each toss.
Part 2: The Central Limit Theorem and Sampling Distribution of the Sample Mean
• The Central Limit Theorem (CLT): When drawing a random sample of size n from any nonnormal population with a mean μ and the standard deviation σ is known, then the sample mean,
Ȳ, has a sampling distribution that is approximately normal as long as n is large enough (rule of
thumb: n > 30).
• Assumptions and Conditions:
◦1) The data values must be sampled randomly.
◦2) The sampled values must be independent of one another.
◦3) Sample size should be less than 10% of the population size.
• The Sampling Distribution Model for a Sample Mean Ȳ
◦The mean of the sample averages is μ = μ
Ȳ
◦The standard deviation of the sample averages is σ = σ/√n
Ȳ
◦If a population is normal, then the sampling distribution of the sample
2
mean is normal: Ȳ ~ N(μ, σ /n)
◦If the a population is non-normal, then the sampling distribution of the
sample mean is approximately normal according to the CLT as long as n
is large enough: Ȳ ~ AN(μ, σ2 /n)
◦The Z-score formula for the sample mean is de ned as
Z=
Ȳ-μ
σ/√n
EXAMPLE: Suppose the weights of men are normally distributed with a mean of 173 lbs. and
variance of 900.
(A) What is the probability a randomly selected men weighs more than 200 lbs.?
(B) What is the probability that the mean weight of 9 randomly selected men is more than 200 lbs.?
Part 3: More Examples
EXAMPLE 1: The times of the nishers in a 10km run are normally distributed with a mean of 61
minutes and a variance of 81. A random sample of 30 runners is selected.
(A) Describe the sampling distribution of the average 10km nishing times for this sample.
(B) Find the probability that the average time of the above sample will be more than 65 minutes?
EXAMPLE 2: A rental car company has noticed that the distribution of the number of miles
customers put on rental cars per day is right-skewed. The distribution has a mean of 60 miles and a
standard deviation of 25 miles. A random sample of 120 rental cars is selected.
(A) Describe the sampling distribution of the average number of miles per day for this sample.
(B) What is the probability that the mean number of miles driven per day for this sample is less than
54?
(C) What is the probability that the total number of miles driven per day for this sample exceeds
7400?
Stat 350A: Chapter 4.13
Part 1: Normal Approximation to Binomial
• Let Y ~ Bin(n, π). If n has a very large sample size, calculations (by hand) of the Binomial
distribution can be strenuous. For these large number of trials for Binomial experiments, what we
can do instead is use the Normal distribution to approximate these Binomial probabilities.
• The Normal approximation to the Binomial distribution is an application of the Central Limit
Theorem. Why?
‣ a) Let X ~ Bin(1, π) = Bernoulli(π), then μ = π, σ = √π(1-π). And suppose we run n
Bernoulli trials.
‣ b) X̄ = (X1 + X2 + … + Xn)/n ~ AN(π, π(1-π)/n)
if n is large enough
‣ c) Let Y = X1 + X2 + … + Xn = nX̄ , then Y ~ AN(nπ, nπ(1-π))
• Assumptions:
◦1) Random sample.
◦2) Trials are independent.
◦3) If sampling without replacement, n < 10% of N ◦4) Rule of Thumb for su ciently large sample size: nπ > 5 and n(1-π) > 5
• Approximating Probabilities for Binomial Random Variables
◦To estimate probabilities for a binomial random variable, Z-scores can be used.
◦P(Y < y) ≈ P(Z < z) y - nπ Z= nπ(1-π) Part 2: Examples EXAMPLE 1: Eighty percent of all patrons at a local restaurant request water with their meal. (a) Suppose we sample 7 customers. What is the probability that at least 5 of the 7 customers selected will request water with their meal? (b) Suppose we now take a larger random sample of 119 customers. What is the probability that at most 100 of this sample will request water with their meal? EXAMPLE 2: Many toothpaste commercials that 3 out of 4 dentists recommend their brand of toothpaste. A random survey of 400 dentists is taken. Assuming the commercials are correct, what is the probability that at least 320 dentists from this sample will recommend Brand X toothpaste? Part 3: Continuity Correction EXAMPLE 1: Suppose 30% of the population have 20/20 vision. What is the probability of having 5 people with 20/20 vision in a sample of 20? EXAMPLE 2: Can we use the normal approximation for the previous problem? EXAMPLE 3: Referring to the above two examples. What is the probability that at least 8 of 20 have 20/20? Use both Binomial and a Normal approximation. EXAMPLE 4: Referring to the above examples. What is the probability that 5 to 7 people out of 20 have 20/20 vision? Use both Binomial and a Normal approximation. • Summary: ◦1) If nπ < 5 or n(1 - π) < 5, Normal approximation cannot be used. ◦2) If nπ > 20 and n(1 – π) > 20, Normal approximation with no continuity correction can be
used.

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