91Ó°ÊÓ

The standard deviation of the lengths of time women with one job worked over the last eight years is 92 weeks.

population S.D $\mathrm{σ}=92$week

Let the population mean is $\mathrm{μ}$ weeks

$\text{population mean and standard deviation:}{\mathrm{μ}}_{\stackrel{¯}{\tilde{x}}}=\mathrm{μ}\text{and}{\mathrm{σ}}_{\stackrel{¯}{\tilde{x}}}=\mathrm{σ}/\sqrt{n}\text{.}$

Sample size $n=50$

We can consider the sample size to be large because it is larger than $30$

As a result of using the C1.T sample, the mean $\stackrel{¯}{x}$follows a normal distribution with a mean$\mathrm{μ}$and S.D.

$$\begin{gathered} {\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}} \\ =\frac{92}{\sqrt{50}}\text{Days} \\ =13.01\text{Days} \end{gathered}$$

We have to find $P(\mathrm{μ}-20≤\stackrel{¯}{X}≤\mathrm{μ}+20)$

Where $\stackrel{¯}{X}~N\left(\mathrm{μ},{\mathrm{σ}}_{\stackrel{¯}{X}}^2\right)$

Where ${\mathrm{σ}}_{\stackrel{¯}{X}}=\frac{\mathrm{σ}}{\sqrt{n}}=13.01$days

$$\begin{gathered} P[\mathrm{μ}-20≤\stackrel{¯}{X}≤\mathrm{μ}+20] \\ =P\left[\frac{\mathrm{μ}-20-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}≤\frac{\stackrel{¯}{X}-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}≤\frac{\mathrm{μ}+20-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}\right] \\ =P\left[\frac{-20}{13.01}≤z≤\frac{20}{13.61}\right],z=\frac{\stackrel{¯}{X}-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}~N(0,1) \\ =P[-1.5372≤z≤1.5372] \\ =\mathrm{Φ}(1.5372)-\mathrm{Φ}(-1.5372) \\ =2\mathrm{Φ}(1.5372)-1\left[\mathrm{Φ}\right(-x)=1-\mathrm{Φ}(x\left)\right] \\ =0.8757556 \end{gathered}$$

As a result, the sample mean of samples of size $50$ has a probability of sampling error of at most $20$ weeks in $0.8757556$ the population mean length of time employed by all women with one job.