Question

The average number of passes in a football game is right skewed with mean 75 and variance 961. If we sample 100 random games, what is the probability our average would be more than 83.7?

Answer 1

0.0066

Explanation:

Given that:

Mean
`\mu` = 75

Variance
`\sigma^2 =961`

The standard deviation
`\sigma = √(961) = 31`

The sample size n = 100

`\mu = \mu_(\overline x) = 75`

`\sigma _(\overline x) = (\sigma )/(√(n))`

`\sigma _(\overline x) = (31 )/(√(100))`

`\sigma _(\overline x) = (31 )/(10)}`

`\sigma _(\overline x) = 3.1`

`P(\overline x > 83.7) = 1 - P( \overline x < 83.7)`

`P(\overline x > 83.7) = 1 -P\bigg ( (\overline x - \mu_(\overline x) )/(\sigma _(\overline x)) < (83.7 - 76)/(3.1) \bigg)`

`P(\overline x > 83.7) = 1 -P\bigg ( Z< (7.7)/(3.1) \bigg)`

`P(\overline x > 83.7) = 1 -P ( Z< 2.48)`

Using the Z table;

P(x > 83.7) = 1 - 0.9934

P(x > 83.7) = 0.0066

Thus, probability = 0.0066