Hours Actually Worked. Repeat Problem $10$, assuming that the number of hours worked by female marketing and advertising managers is normally distributed.

Female marketing and advertising managers work an average of $\mu =45$hours per week (on average).

A $\sigma =7$hour standard deviation

The formula used: Standard deviation${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

The number of hours spent by female marketing and advertising executives is typically distributed. As a result, the random variable $\overline{x}$has a normal distribution with a mean of ${\mu }_x$and a standard deviation of $\frac{\sigma }{\sqrt{n}}$

The standard deviation for sample means is, and ${\mu }_X=45$

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{7}{\sqrt{196}} \\ =\frac{7}{14} \\ =0.5 \end{gathered}$$

Thus,

$$\begin{gathered} P(31\le \overline{x}\le 59)=P\left(\frac{31-45}{0.5}\le \overline{x}\le \frac{59-45}{0.5}\right) \\ =P\left(\frac{-14}{0.5}\le z\le \frac{14}{0.5}\right) \\ =P(-28\le z\le 28) \\ \approx 1 \end{gathered}$$

Thus, it is around 95% incorrect that the sample's average number of hours worked will be between $31$ and $59$

The population distribution is typically distributed in this case. As a result of part (a), the likelihood is one. Thus, about $95\%$ of all conceivable observations of female marketing and advertising managers working between $31$ and $59$ hours are erroneous.

Here,

$$\begin{gathered} P(44\le \overline{x}\le 46)=P\left(\frac{44-45}{0.5}\le \overline{x}\le \frac{46-45}{0.5}\right) \\ =P\left(\frac{-1}{0.5}\le z\le \frac{1}{0.5}\right) \\ =P(-2\le z\le 2) \\ =P(z\le 2)-P(z\le -2) \\ =0.9772-0.0228\left[\begin{array}{l}\text{From TABLE II: Areas under}]\\ \text{the standard normal curve}\end{array}\right] \\ =0.9544 \end{gathered}$$

As a result, there is a $95\%$ chance that the sample's average number of hours worked will be between $44$ and $46$