Question

The mean number of touchdowns per football game is 3.7 with a population standard deviation of 0.9. A random sample of 31 football games was taken and the probability that the mean number of touchdowns for the sample will be more than 3.9 was found.

Was the probability a Left-tail, Right -tail or Interval Probability? Use Excel to find the probability that the mean number of touchdowns for the sample will be more than 3.9.

Select the two correct answers that apply below.

Select all that apply:

• Right-Tail

• Left- Tail

• Interval

• P(x¯¯¯>3.9)=0.108

• P(x¯¯¯>3.9)=0.892

• P(x¯¯¯<3.9)=0.108

Answer 1

Given, $\mu$ = 3.7 , $\sigma$ = 0.9

For sampling distribution of sample mean,

$\mu$$\bar{x}$ = $\mu$ = 3.7 , $\sigma$ $\bar{x}$ = $\sigma$ / sqrt(n) = 0.9 / sqrt(31) = 0.161645

We have to calculate P( $\bar{x}$ > 3.9) = ?

This is right tailed probability.

Using Excel,

P( $\bar{x}$ > 3.9) = 1 - P($\bar{x}$ < 3.9)

= 1 - NORM.DIST ( x , mean , SD , cumulative)

= 1 - NORM.DIST ( 3.7 , 3.9 , 0.161645 , TRUE)

= 1 - 0.892

= 0.108

P( $\bar{x}$ > 3.9) = 0.108