Question

Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 238 feet and a standard deviation of 40 feet. We randomly sample 49 fly balls.

Part (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. X ~ , . ( , )

Part (b) What is the probability that the 49 balls traveled an average of less than 226 feet? (Round your answer to four decimal places.)

Sketch the graph. Scale the horizontal axis for X. Shade the region corresponding to the probability.

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

Answer 1

Solution :

(a)

= / n = 40 / 49 = 5.71

N (238 , 5.71)

(b)

P( < 226) = P(( - ) / < (226 - 238) / 5.71)

= P(z < -2.10)

= 0.0179

Probability = 0.0179

(c)

Using standard normal table,  

P(Z < z) = 60%

P(Z < 0.25) = 0.6

z = 0.25

Using z-score formula,

= z * + = 0.25 * 5.71 + 238 = 239.43

239.43 ft