Question

Among American women aged 20 to 29 years, mean height = 64.2 inches, with a standard deviation of 2.7 inches.

A)Use Chebyshev’s Inequality to construct an interval guaranteed to contain at least 75% of all such women`s heights. Include appropriate units.

B)Suppose that a random sample size of n = 50 is selected from this population of women. What is the probability that the sample mean height will be less than 63.2 inches?

Answer 1

Given that, mean = 64.2 inches and

standard deviation = 2.7 incheses

A) According to Chebyshev's inequality

at least [ 1 - (1/k²) * 100% ] of the data fall within k standard deviations of the mean.

For k = 2

1 - (1/2²) * 100% = 0.75 * 100% = 75%

Hence, according to Chebyshev's inequality at least 75% of all women's heights is between 58.8 inches to 69.6 inches

B) sample size ( n ) = 50

we want to find,

Therefore, the probability that the sample mean height will be less than 63.2 inches is 0.0044

Note: using standard normal z-table we get,

P(Z < -2.62) = 1 - P(Z < 2.62) = 1 - 0.9956 = 0.0044