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

Answer:

0.0066

Step-by-step explanation:

Given that:

Mean $\mu$ = 75

Variance $\sigma^2 =961$

The standard deviation $\sigma = \sqrt{961} = 31$

The sample size n = 100

$\mu = \mu_{\overline x} = 75$

$\sigma _{\overline x} = \dfrac{\sigma }{\sqrt{n}}$

$\sigma _{\overline x} = \dfrac{31 }{\sqrt{100}}$

$\sigma _{\overline x} = \dfrac{31 }{10}}$

$\sigma _{\overline x} = 3.1$

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

$P(\overline x > 83.7) = 1 -P\bigg ( \dfrac{\overline x - \mu_{\overline x} }{\sigma _{\overline x}} < \dfrac{83.7 - 76}{3.1} \bigg)$

$P(\overline x > 83.7) = 1 -P\bigg ( Z< \dfrac{7.7}{3.1} \bigg)$

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

Using the Z table;

P(x > 83.7) = 1 - 0.9934

P(x > 83.7) = 0.0066

Thus, probability = 0.0066