Question

See attached image.

Answer 1

Using the normal distribution, it is found that the mean is of $\mu = 37.27$ and the standard deviation is of $\sigma = 5.68$.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, we have that the p-value of Z when X = 30 is of 0.1, hence, when X = 30, Z = -1.28, so:

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

$-1.28 = \frac{30 - \mu}{\sigma}$

$30 - \mu = -1.28\sigma$

$\mu = 30 + 1.28\sigma$

The p-value of Z when X = 32.5 is of 0.2, hence when X = 32.5, Z = -0.84, hence:

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

$-0.84 = \frac{32.5 - \mu}{\sigma}$

$32.5 - \mu = -0.84\sigma$

$\mu = 32.5 + 0.84\sigma$

Hence:

$30 + 1.28\sigma = 32.5 + 0.84\sigma$

$0.44\sigma = 2.5$

$\sigma = \frac{2.5}{0.44}$

$\sigma = 5.68$

$\mu = 32.5 + 0.84(5.68) = 37.27$

More can be learned about the normal distribution at brainly.com/question/27643290

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