5. Assume that the mean height of American women is 63 inches with a standard de

Question

5. Assume that the mean height of American women is 63 inches with a standard deviation of 2.5 inches.

If you chose a woman at random, what is the probability she is taller than 65 inches?

What is the probability that she is between 64 and 66 inches tall? What is the 90th percentile for women’s heights?
What is the interquartile range for women’s heights?

6. (continued from problem #5) Now assume you take a random sample of 4 women and calculate a sample mean height.

What is the probability that the mean height of the women is greater than 65 inches? What is the probability that the mean height is between 64 and 66 inches tall?

6. (continued from problem #5) Now assume you take random samples of 16 women and calculate a sample mean height.

What is the mean of the sampling distribution of sample means in this case?

What is the standard deviation of the sampling distribution of sample means in this case?

Into what interval will such a sample mean fall 95% of the time?

Find the probability that your sample mean is greater than 66 inches.

Explanation / Answer

5. Mean = 63 inches

Standard deviation = 2.5 inches

P(X < A) = P(Z < (A - mean)/standard deviation)

P(she is taller than 65 inches) = P(X > 65)

= 1 - P(X < 65)

= 1 - P(Z < (65 - 63)/2.5)

= 1 - P(Z < 0.8)

= 1 - 0.7881

= 0.2119

Probability that she is between 64 and 66 inches tall = P(X < 66) - P(X < 64)

= P(Z < (66 - 63)/2.5) - P(Z < (64-63)/2.5)

= P(Z < 1.2) - P(Z < 0.4)

= 0.8849 - 0.6554

= 0.2295

Let A denote the 90th percentile

P(X < A) = 0.9

P(Z < (A - 63)/2.5) = 0.9

(A - 63)/2.5 = 1.28

A = 66.2 inches

Let Q1 denote first quartile

P(X < Q1) = 0.25

P(Z < (Q1 - 63)/2.5) = 0.25

(Q1 - 63)/2.5 = -0.67

Q1 = 62.325

Interquartile range IQR= (63-62.325)x2 = 1.35 inches

P.S: Please post different questions separately

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.