Question

Suppose that the distance of fly balls threw (hit) to the outfield (in baseball) is normally distributed with a mean of 246 feet and a standard deviation of 39 feet. Let X be the distance in feet for a fly ball. Round all answers to two decimal places.

A. X ~ N( ____ , _____ )
B. Find the probability that a randomly hit fly ball travels less than 200 feet.
C.Find the 70th percentile for the distribution of fly balls.

Answer 1

a)
`X\sim(246,1521)`

b) 0.1191

c)
`P_(70)=266.44`

Explanation:

We are given the following information in the question:

Mean, μ = 246 feet

Standard Deviation, σ = 39 feet

We are given that the distribution of distance of fly balls is a bell shaped distribution that is a normal distribution.

a) Distribution of X

Let X be the distance in feet for a fly ball. Then,

`X\sim (\mu, \sigma^2)\\X\sim(246,39^2)\\X\sim(246,1521)`

b) Probability that a randomly hit fly ball travels less than 200 feet.

`P( x < 200) = P( z < \displaystyle(200 - 246)/(39)) = P(z < -1.1794)`

Calculation the value from standard normal z table, we have,

`P(x < 200) = 0.1191`

c) 70th percentile for the distribution of fly balls.

We have to find the value of x such that the probability is 0.7

`P( X < x) = P( z < \displaystyle(x - 246)/(39))=0.7`

Calculation the value from standard normal z table, we have,

`\displaystyle(x - 246)/(39) = 0.524\\\\x = 266.436\approx 266.44`

The 70th percentile for the distribution of fly ball is 266.44 feet.