Question

The annual commissions per salesperson employed by a retailer of mobile communication devices are normally distributed, and averaged $40,000, with a standard deviation of $5,000. what percent of the salespersons earn between $32,000 and $42,000?

Answer 1

Using the normal distribution, it is found that 60.06% of the salespersons earn between $32,000 and $42,000.

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.

The mean and the standard deviation are given, respectively, by:

$\mu = 40000, \sigma = 5000$

The proportion of salespersons that earn between $32,000 and $42,000 is the p-value of Z when X = 42000 subtracted by the p-value of Z when X = 32000, hence:

X = 42000:

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

$Z = \frac{42000 - 40000}{5000}$

Z = 0.4

Z = 0.4 has a p-value of 0.6554.

X = 32000:

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

$Z = \frac{32000 - 40000}{5000}$

Z = -1.6

Z = -1.6 has a p-value of 0.0548.

0.6554 - 0.0548 = 0.6006 = 60.06%.

60.06% of the salespersons earn between $32,000 and $42,000.

More can be learned about  the normal distribution at https://brainly.com/question/24663213

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