Question

g From a distribution with mean 38 and variance 52, a sample of size 16 is taken. Let X be the mean of the sample. Show that the probability is at least 0.87 that X is in (33, 43)

Answer 1

`P=8.869`

Explanation:

From the question we are told that:

Mean
`\=x =38`

Variance
`\sigma=52`

Sample size
`n=16`

`X=(33, 43)`

Generally the equation for Chebyshev's Rule is mathematically given by

`A=(1-(1)/(k^2))*100\%`

Where

`k=(\=x-\mu)/((\sigma)/(\sqrt n))}}`

`k=(43-38)/((52)/(\sqrt 16))}}`

`k=2.77`

Therefore

Probability

`P=(1-(1)/(2.77^2))`

`P=8.869`