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?

Respuesta :

Answer:

0.0066

Step-by-step explanation:

Given that:

Mean [tex]\mu[/tex] = 75

Variance [tex]\sigma^2 =961[/tex]

The standard deviation [tex]\sigma = \sqrt{961} = 31[/tex]

The sample size n = 100

[tex]\mu = \mu_{\overline x} = 75[/tex]

[tex]\sigma _{\overline x} = \dfrac{\sigma }{\sqrt{n}}[/tex]

[tex]\sigma _{\overline x} = \dfrac{31 }{\sqrt{100}}[/tex]

[tex]\sigma _{\overline x} = \dfrac{31 }{10}}[/tex]

[tex]\sigma _{\overline x} = 3.1[/tex]

[tex]P(\overline x > 83.7) = 1 - P( \overline x < 83.7)[/tex]

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

[tex]P(\overline x > 83.7) = 1 -P\bigg ( Z< \dfrac{7.7}{3.1} \bigg)[/tex]

[tex]P(\overline x > 83.7) = 1 -P ( Z< 2.48)[/tex]

Using the Z table;

P(x > 83.7) = 1 - 0.9934

P(x > 83.7) = 0.0066

Thus, probability = 0.0066