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Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls.

a. If Xthe average distance in feet for 49 fly balls, then X¯~

b. What is the probability that the 49 balls traveled an average of fewer than 240 feet? Sketch the graph. Scale the horizontal axis forX¯. Shade the region corresponding to the probability. Find the probability.

c. Find the 80thpercentile of the distribution of the average of 49 fly balls.

Short Answer

Expert verified
  1. X~N(250,7.143)
  2. The probability that the 49balls traveled an average of fewer than 240feet is 0.0808.

c. The 80thpercentile of the distribution of the average of 49fly balls is246

Step by step solution

01

Given information part (a)

Given in the question that, the distance of fly balls hit to the outfield is normally distributed with a mean of 250feet and a standard deviation of 50feet. We randomly sample 49fly balls.

02

Explanation (Part a)

Consider X¯ as the average distance in feet for 49fly balls, then

localid="1649344986876" X¯~N250,5049

03

Given Information (part b)

According to the information,

X¯~N250,5049=N(250,7.143)

04

Explanation (Part b)

Let's find the probability:

P(X¯<240)=PX¯2507.143<2402507.143

=Px¯2507.143<1.4

=ϕ(1.4)

=10.9192

=0.0808

05

Graphical Representation (Part b)

06

Given Information (Part c)

The distance of fly balls hit to the outfield is normally distributed with a mean of 250feet and a standard deviation of 50feet. We randomly sample 49fly balls.

07

Explanation (Part c)

Let's find the value of zfrom

P(Σx<z)=0.8

PΣx2407.143<z2407.143=0.8

If a variable zhas a normal distribution, then

P(Z<0.84)=0.8

Now we have

z2407.143=0.84

z-240=6

z=246

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Most popular questions from this chapter

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