Question

A normal distribution has a standard deviation equal to 39. What is the mean of this normal distribution if the probability of scoring above x = 209 is 0.0228?

Answer 1

Answer:

The mean is $\mu = 131$

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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 problem, we have that:

$\sigma = 39$

What is the mean of this normal distribution if the probability of scoring above x = 209 is 0.0228?

This means that when X = 209, Z has a pvalue of 1-0.0228 = 0.9772. So when X = 209, Z = 2.

$Z = \frac{X - \mu}{\sigma}$

$2 = \frac{209 - \mu}{39}$

$209 - \mu = 2*39$

$\mu = 209 - 78$

$\mu = 131$

The mean is $\mu = 131$