91Ó°ÊÓ

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

A $\mathrm{σ}=7$hour standard deviation

The formula used: Standard deviation${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}$

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

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

$$\begin{gathered} {\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}} \\ =\frac{7}{\sqrt{196}} \\ =\frac{7}{14} \\ =0.5 \end{gathered}$$

Thus,

$$\begin{gathered} P(31≤\stackrel{¯}{x}≤59)=P\left(\frac{31-45}{0.5}≤\stackrel{¯}{x}≤\frac{59-45}{0.5}\right) \\ =P\left(\frac{-14}{0.5}≤z≤\frac{14}{0.5}\right) \\ =P(-28≤z≤28) \\ ≈1 \end{gathered}$$

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,

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

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