Question

The 400-meter race times recorded in the boys track meet was normally distributed with a standard deviation of 3 seconds.If David finished the race in 51.2 seconds with a z-score of -2.6, what is the mean time?

Answer 1

The z score is defined as:

`z =(X- \mu)/(\sigma)`

If we solve for the mean
`\mu` we got:

`\mu = X -z\sigma`

Replacing we got:

`\mu = 51.2 - (-2.6)*3 = 59`

Explanation:

For this case we have the following info given:

`z = -2.6` represent the z score

`\sigma=3` represent the population deviation

`x= 51.2` represent the time for David

The z score is defined as:

`z =(X- \mu)/(\sigma)`

If we solve for the mean
`\mu` we got:

`\mu = X -z\sigma`

Replacing we got:

`\mu = 51.2 - (-2.6)*3 = 59`

Answer 2

The mean time is 59 seconds.

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

`\sigma = 3`

If David finished the race in 51.2 seconds with a z-score of -2.6, what is the mean time?

This means that when X = 51.2, Z = -2.6. So

`Z = (X - \mu)/(\sigma)`

`-2.6 = (51.2 - \mu)/(3)`

`51.2 - \mu = -2.6*3`

`\mu = 51.2 + 2.6*3`

`\mu = 59`

The mean time is 59 seconds.