Question

A large university offers STEM (science, technology,engineenng. and mathematics) intemshups to women in STEM majors at the university. A woman must be 20 years or older to meet the age requirement for the intemships. The table shows the probability distribution of the ages of the women in STEM majors at the university.
Age (years) 17 18 19 20 21 22 23 or older
Probability 0.005 0.107 0.111 0.252 0.249 0.213 0.063

a. Suppose one woman is selected at random from the women in STEM majors at the university. What is the probahility that the woman selected will not meet the age requirement for the internships?

The university will select a sample of 100 women in STEM majors to participate in a focus group about the intemships.

b. Suppose a simple random sampling process is used to select the sample of 100 women. What is the probability that at least 30 percent of the women in the sample will not meet the age requirement for the intersdhips?
c. Suppose a stratified random sampling design is used to select a sample of 30 women who do not meet the age requirement and a sample of 70 women who do meet the age requirement. Based on the probability distribution, is a woman who does not meet the age requirement more likely, less likely, or equally likely to be selected with a stratified random sample than with a simple random sample?

Answer 1

Answer:

a) Probability of a randomly sampled women not being qualified for the internship = 0.223

b) Probability that at least 30 percent of the women in the sample will not meet the age requirement for the internships = 0.03216

c) A woman who does not meet the age requirement is more likely to be selected with a stratified random sample than with a simple random sample.

Step-by-step explanation:

Age | Probability

17 | 0.005

18 | 0.107

19 | 0.111

20 | 0.252

21 | 0.249

22 | 0.213

23 or older | 0.063

a) Only 20+ year olds are qualified for the internship

So, probability of being qualified for the internship = P(x ≥ 20)

Probability of not being qualified for the internship = P(x < 20) = P(x=17) + P(x=18) + P(x=19) = 0.005 + 0.107 + 0.111 = 0.223

b) According to the Central limit theorem, a sampling distribution of sample size as large as 100 selected from this population distribution will approximate a normal distribution. It also has that

Mean proportion of sampling distribution of women who do not meet the internship requirements = Population proportion of women who do not meet the internship requirements = p = 0.223

The standard deviation of the is given by

σₓ = √[p(1-p)/n]

n = sample size = 100

σₓ = √[(0.223×0.777)/100] = 0.041625833 = 0.04163

So, to obtain the probability that at least 30 percent of the women in the sample will not meet the age requirement for the internships

P(x ≥ 0.30)

We first standardize 0.30

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (0.30 - 0.223)/0.04163 = 1.85

The required probability

P(x ≥ 0.30) = P(z ≥ 1.85)

We'll use data from the normal probability table for these probabilities

P(x ≥ 0.30) = P(z ≥ 1.85) = 1 - P(z < 1.85)

= 1 - 0.96784 = 0.03216

c) Probability of women not meeting the internship requirements = 0.223

Probability of women meeting the internship requirements = 1 - 0.223 = 0.777

Or

Probability of women meeting the internship requirements = P(x ≥ 20)

= P(x=20) + P(x=21) + P(x=21) + P(x ≥ 23) = 0.777

But as the stratified sample only contains women who do not meet the internship requirements, it is more likely that A woman who does not meet the age requirement is selected with a stratified random sample than with a simple random sample.

Hope this Helps!!!