Question

The average number of hours worked per week for college students is 27, and the standard deviation is 6. Assume the data is normally distributed. Determine the probability of someone working between 10 and 15 hours.

Answer 1

Answer:

2.05% probability of someone working between 10 and 15 hours.

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:

$\mu = 27, \sigma = 6$

Determine the probability of someone working between 10 and 15 hours.

This is the pvalue of Z when X = 15 subtracted by the pvalue of Z when X = 10.

So

X = 15

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

$Z = \frac{15 - 27}{6}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228.

X = 10

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

$Z = \frac{10 - 27}{6}$

$Z = -2.83$

$Z = -2.83$ has a pvalue of 0.0023.

0.0228 - 0.0023 = 0.0205

2.05% probability of someone working between 10 and 15 hours.