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?

0.0066

Step-by-step explanation:

Given that:

Mean = 75

Variance

The standard deviation

The sample size n = 100

Using the Z table;

P(x > 83.7) = 1 - 0.9934

P(x > 83.7) = 0.0066

Thus, probability = 0.0066

12m

step-by-step explanation:

square root means have of a number so have of 3 is 1.5 then multiply it by 8 and you get your answer!

24+x=y

step-by-step explanation: