Question

3. Assume that the body weights of American men are normally distributed and have a mean of 182 pounds and a standard deviation of 20 pounds.

a) If you select just one man at random, what is the probability that he weighs between 151 and 171 pounds? (Use Table 8.1)

b) According to the Rule of Sample Means, if multiple random samples of 25 men are collected and the sample mean weight is calculated for each sample, what should the standard deviation of the resulting distribution of sample means be?

c) Find the interval in which the sampling distribution of sample mean weights for samples of 25 men should fall 95% of the time using the Empirical Rule.

d) Using Table 8.1, find the probability that the mean weight of a single random sample of 25 men is greater than 184 pounds.

e) Using Table 8.1, find the probability that the mean weight of 25 randomly selected men is between 180 and 183 pounds.

Answer 1

This is a normal distribution question with

a) P(151.0 < x < 171.0)=?

This implies that

P(151.0 < x < 171.0) = P(-1.55 < z < -0.55) = P(Z < -0.55) - P(Z < -1.55)

P(151.0 < x < 171.0) = 0.29115968678834636 - 0.06057075800205901

b)

Since we know that

c)

Confidence interval(in %) = 95

Since we know that

Required confidence interval

Required confidence interval = (182.0-8.256, 182.0+8.256)

Required confidence interval = (173.744, 190.256)

d) P(x > 184.0)=?

The z-score at x = 184.0 is,

z = 0.5

This implies that

P(x > 184.0) = P(z > 0.5) = 1 - 0.6914624612740131

e) P(180.0 < x < 183.0)=?

This implies that

P(180.0 < x < 183.0) = P(-0.5 < z < 0.25) = P(Z < 0.25) - P(Z < -0.5)

P(180.0 < x < 183.0) = 0.5987063256829237 - 0.3085375387259869

PS: you have to refer z score table to find the final probabilities.

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