Question

A group of 40 bowlers showed that their average score was 192. Assume the population standard deviation is 8. Find the 95% confidence interval of the mean score of all bowlers.

Answer 1

`CI=189.5,194.5`

Explanation:

From the question we are told that:

Sample size
`n=40`

Mean
`\=x =192`

Standard deviation
`\sigma=8`

Significance Level
`\alpha=0.05`

From table

Critical Value of
`Z=1.96`

Generally the equation for momentum is mathematically given by

`CI =\=x \pm z_(a/2) (\sigma)/(√(n))`

`CI =192 \pm 1.96 (8)/(√(40))`

`CI=192 \pm 2.479`

`CI=189.5,194.5`