Question

Use the given values of n and p to find the minimum usual value μ−2σ and the maximum usual value μ+2σ. Round to the nearest hundredth unless otherwise noted. n=1139​; p=0.96

Answer 1

μ−2σ = 1,089.26

μ+2σ = 1,097.62

Explanation:

The standard deviation of a sample of size 'n' and proportion 'p' is:

`\sigma=\sqrt{(p*(1-p))/(n) }`

If n=1139 and p =0.96, the standard deviation is:

`\sigma=\sqrt{(p*(1-p))/(n)}\\\sigma = 0.001836`

The minimum and maximum usual values are:

`\mu-2\sigma = (p-2\sigma)*n\\\mu+2\sigma = (p+2\sigma)*n`

`\mu-2\sigma = (0.96-2*0.001836)*1139\\\mu-2\sigma = 1,089.26\\\mu+2\sigma = (0.96+2*0.001836)*1139\\\mu+2\sigma = 1,097.62`