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

Answer:

$CI=189.5,194.5$

Step-by-step 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

 $CI =\=x \pm z_(a/2) \frac{\sigma}{\sqrt{n}}$

 $CI =192 \pm 1.96 \frac{8}{\sqrt{40}}$

 $CI=192 \pm 2.479$

 $CI=189.5,194.5$