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An hourly wage is normally distributed with a mean of $6.75 and a standard deviation of $0.55. What is the probability that an e

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Answer:

Option A. 81.5%

Explanation:

Population mean, $\mu = 6.75$

Standard deviation, $\sigma = 0.55$

The question is to find $p(5.65 < \bar{x} < 7.30)$

The z value is given by the formula:

$z = \frac{\bar{x} - \mu}{\sigma}$

z - value corresponding to the mean, $\bar{x} = 5.65$ :

$z_1 = \frac{5.65 - 6.75}{0.55}\\z_1 =-2.0$

z - value corresponding to the mean, $\bar{x} = 7.30$ :

$z_1 = \frac{7.30 - 6.75}{0.55}\\z_1 =1.0$

$p(5.65\leq \bar{x} \leq 7.30) = p(-2.0 \leq z \leq 1.0)\\p(-2.0 \leq z \leq 1.0) = p(z\leq 1) - p(z\leq -2)\\ p(-2.0 \leq z \leq 1.0) = 0.84134 - 0.02275\\p(-2.0 \leq z \leq 1.0) = 0.81859 \\p(-2.0 \leq z \leq 1.0) = 0.82$

Probability that an employee’s hourly wage is between $5.65 and $7.30 = 81.5%

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