Question

The height of a person is a random variable with variance < 5 inches. According to Mr. Chebyshev, how many people do we need to sample to ensure that the sample mean is at most 1 inch away from the distribution mean with probability 2 95%?

Answer 1

P(|X - mean| < k ) > 1 - $\sigma^2$/k^2

here replace X with Xbar

P(|Xbar - mean| < k)>= 1 - $\sigma^2$/(n*k^2 )

put k = 1

$\sigma^2$= 5

1- $\sigma^2$/(n k^2) = 0.95

1-5/n = 0.95

5/n = 0.05

n = 100