Question

A football player throws an average of 168 yards per game. If the total yardage

that the football player throws is normally distributed with a standard deviation equal to 74 yards:

(a) What fraction of the games that the football player will throw more than 224 yards?

(b) What is the probability that the football player throws between 91 and 129 yards?

(c) If the football player keeps doing what he has been doing, how many 300+ yard games do we expect the football player to perform for the next 50 games?

Answer 1

Given: = 168, = 74

To find the probability, we need to find the Z scores first.

Z = (X - )/ [/Sqrt(n)]. Since n = 1, Z = (X - )/

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(a) P(X > 224) = 1 - P(X < 224) ), as the normal tables give us the left tailed probability only.

For P( X < 224), Z = (224 – 168)/74 = 0.76

The probability for P(X < 224) from the normal distribution tables is = 0.7764

Therefore the required probability = 1 – 0.7764 = 0.2236

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(b) P(91 < X < 129) = P(X < 129) - P(X < 91)

For P( X < 129), Z = (129 – 168)/74 = -0.53. The probability = 0.2981

For P( X < 91), Z = (91 – 168)/74 = -1.04. The probability = 0.1492

Therefore the required probability = 0.2981 - 0.1492 = 0.1489

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(c) P( X > 300),

For P(X < 300); Z = (300 - 168)/74 = 1.78

P(X < 300) = 0.9625

Therefore P(X > 300) = 1 - 0.9625 = 0.0375

Therefore in the next 50 games, the expected number of games where he will score a 300+

Expected value= n * p = 50 * 0.0375 = 1.875 2 games

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