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?

P(X≥112) = 0.9495

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

Step-by-step explanation:

Given mean of the Population = 130

Given standard deviation of the Population = 11

Let X= 112

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%

step-by-step explanation:

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step-by-step explanation:

what grade level is this? need to know so i can look in my old textbooks!