A large university offers STEM (science, technology, engineering. And mathematics) internships to women in STEM majors at the university. A woman must be 20 years or older to meet the age requirement for the internships. 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 .005 .107 .111 .252 .213 .249 0.063 (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 internship.
The correct answer and explanation is:
To determine the probability that at least 30% of the women in the sample will not meet the age requirement for the internship, we need to calculate the following:
- Identify the probability that a woman does not meet the age requirement: Women who do not meet the age requirement are those aged 17, 18, or 19. Summing their probabilities: P(not eligible)=P(17)+P(18)+P(19)=0.005+0.107+0.111=0.223P(\text{not eligible}) = P(17) + P(18) + P(19) = 0.005 + 0.107 + 0.111 = 0.223
- Define the problem in terms of sample proportions: Let XX be the number of women in the sample who do not meet the age requirement. XX follows a binomial distribution: X∼Binomial(n=100,p=0.223)X \sim \text{Binomial}(n=100, p=0.223) The sample proportion of women not meeting the age requirement is p^=X/100\hat{p} = X / 100. We are tasked with finding: P(p^≥0.30)=P(X≥30)P(\hat{p} \geq 0.30) = P(X \geq 30)
- Use the normal approximation to the binomial distribution: For large sample sizes (n>30n > 30), the binomial distribution can be approximated by a normal distribution: X∼N(μ,σ2)whereμ=npandσ=np(1−p)X \sim N(\mu, \sigma^2) \quad \text{where} \quad \mu = np \quad \text{and} \quad \sigma = \sqrt{np(1-p)} Compute the mean and standard deviation: μ=100⋅0.223=22.3,σ=100⋅0.223⋅(1−0.223)≈4.182\mu = 100 \cdot 0.223 = 22.3, \quad \sigma = \sqrt{100 \cdot 0.223 \cdot (1 – 0.223)} \approx 4.182
- Standardize the variable: To find P(X≥30)P(X \geq 30), use the z-score formula: Z=X−μσZ = \frac{X – \mu}{\sigma} For X=30X = 30: Z=30−22.34.182≈1.84Z = \frac{30 – 22.3}{4.182} \approx 1.84
- Find the probability: Using standard normal tables or a calculator: P(Z≥1.84)≈1−0.9671=0.0329P(Z \geq 1.84) \approx 1 – 0.9671 = 0.0329
Final Answer:
The probability that at least 30% of the women in the sample will not meet the age requirement is approximately 0.033 (3.3%).
Explanation:
This solution applies statistical concepts to determine the likelihood of a proportion in a sample exceeding a threshold. The problem is modeled as a binomial distribution since each woman either meets or does not meet the requirement. With a sufficiently large sample size (n=100n=100), the distribution is approximated using the normal distribution for simplicity. Calculations involve computing the mean and standard deviation of the sample count, standardizing the variable, and finding the corresponding tail probability. This approach is both efficient and accurate for problems involving large sample sizes and proportions.