Question

A random sample of 30 receipts for individuals shopping at the Community Drug Store showed the sample mean to be`x = $28. 19 with sample standard deviation s = $4.06 a. Use Chevyshev’s theorem to find the smallest interval centered on the mean in which we can expect at least 75% of the data to fall.

Answer 1

Chebyshev's inequality states that


`\mathbb P(|X-\mu|\ge k\sigma)\le\frac1{k^2}\implies\mathbb P(|X-\mu|\le k\sigma)\ge1-\frac1{k^2}`

We assume
`\mu=\bar x=28.19` and
`\sigma=s=4.06`, so that the inequality is


`\mathbb P(|X-28.19|\le4.06k)\ge1-\frac1{k^2}`

and since we want the probability to be at least 0.75, we have


`1-\frac1{k^2}=\frac34\implies k=2`

which means we get from Chebyshev's inequality that


`\mathbb P(|X-28.19|\le8.12)=\mathbb P(20.07\le X\le36.31)\ge0.75`