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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?

User Ericb
by
6.7k points

2 Answers

1 vote

Answer:

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

User Jesper Blad Jensen
by
6.2k points
3 votes

Answer:

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.

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