Question

The average number of passes in a football game is right skewed with mean 75 and variance 729. If we sample 121 random games, what is the probability our average would be more than 80.8? Use 4 decimal places

Answer 1

SOLUTION:

From given data,

The average number of passes in a football game is right skewed with mean 75 and variance 729. If we sample 121 random games, what is the probability our average would be more than 80.8?

Mean = = 75

Variance = = 729

Standard deviation = = sqrt(Variance ) = sqrt(729 ) = 27

Sample = n = 121

We have to find P( > 80.8)

To find this we convert this to standard normal using

z = ( - ) / = ( - ) / (/sqrt(n) )​​​​​​​

z = (80.8 - 75 ) / (27/sqrt(121) )​​​​​​​ = 2.36

P( > 80.8) =  P(z > 2.36)

P( > 80.8) = 1 - P(z < 2.36)

P( > 80.8) = 1 - 0.99086

P( > 80.8) = 0.0091

P(z < 2.36) in a z-table having area to the left of z, locate 2.3 in the left most column. Move across the row to the right under column 0.06 and get value 0.99086.

P( > 80.8) = 0.0091 (4 decimal places)