Question

Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 240 feet and a standard deviation of 58 feet. We randomly sample 49 fly balls.

A) If X= average distance in feet for 49 fly balls, then give the distribution of X. Round your standard deviation to two decimal places.

B) What is the probability that the 49 balls traveled an average of less than 232 feet? (Round your answer to four decimal places.)

C) Find the 60th percentile of the distribution of the average of 49 fly balls. (Round your answer to two decimal places.)

Answer #1

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Its a normal distribution with

Mean = 240 ft

Stdev = 58 ft

n = 49

a. Stdev = population dev/sqrt(n) = 58/sqrt(49) = 58/7 = 8.2857 or 8.29

**Answer: 8.29**

b. P(Xbar < 232) = ?

Standardizing using formula:

Z = (X-Mu)/Stdev

P(Z< (232-240)/8.29)

= P(Z<-0.9650)

= 0.1673

**Answer: 0.1673**

c. 60th percentile is a Z score of 0.2533 ( used Z tables for this)

So, 60th percentile score is :

Mean + Z*stdev

= 240 + .2533*8.29 = 242.10

**Answer: 242.10**

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