Question

When Easton goes bowling, his scores are normally distributed with a mean of 130

and a standard deviation of 11. What percentage of the games that Easton bowls does
he score higher than 112, to the nearest tenth?

Answer 1

P(X≥112) = 0.9495

The percentage of the games that Easton bowls does he score higher than 112 = 94.95%

Explanation:

Given mean of the Population = 130

Given standard deviation of the Population = 11

Let X= 112

`Z =(x-mean)/(S.D) =(112-130)/(11)=-1.636`

P(X≥112) = P(Z≥ z)

= 1-P(Z≤z)

= 1- ( P(Z≤-1.636)

= 1- (0.5-A(-1.636)

= 0.5 +A(1.636)

= 0.5 + 0.4495

P(X≥112) = 0.9495

The percentage of the games that Easton bowls does

he score higher than 112 = 94.95%