The following table uses $1992$data concerning the percentages of male and female full-time workers whose

annual salaries fall into different ranges:

Suppose that random samples of 200 male and 200 female full-time workers are chosen. Approximate the probability

that

(a) at least $70$of the women earn $\backslash (25,000$or more;

(b) at most $60$percent of the men earn $\backslash )25,000$or more;

(c) at least three-fourths of the men and at least half the women earn $\$20,000$or more.

Construct the poisson distribution $N$to symbolize the number of women earning $\$25,000$or more. $N$sim operator name in $0m(200,p)$, where $p$is the chance that a random woman earns $\$25,000$or more. As can be seen from the table, $p=0.34$. We'll use the Normal approximation method. Obtain

$$\begin{gathered} E\left(N\right)=200\times 0.34 \\ =68 \\ \mathrm{Var}\left(N\right)=200\times 0.34\times 0.66 \\ =44.88 \end{gathered}$$

$$\begin{gathered} P(N\ge 70)=1-P(N<69) \\ =1-P\left(\frac{N-68}{\sqrt{44.88}}<\frac{69-68}{\sqrt{44.88}}\right) \end{gathered}$$

$$\begin{gathered} \approx 1-\mathrm{\Phi }\left(0.1493\right) \\ P(N\ge 70)=0.4407 \end{gathered}$$

Create a dependent vector. The quantity of males earning $\$25,000$or more is denoted by the letter $M$. We have $M~\mathrm{Binom}(200,p)$, where $p$is the chance of random men earning $p=0.587$$\$25,000$equals $N\le 120$. We'll use the Average interpolation technique. Obtain

role="math" localid="1649529329060" $$\begin{gathered} E\left(N\right)=200\times 0.587 \\ =117.4 \\ \mathrm{Var}\left(N\right)=200\times 0.587\times 0.413 \\ =48.49 \end{gathered}$$

which implies,

$$\begin{gathered} P(M\le 120) \\ =P\left(\frac{M-117.4}{\sqrt{48.49}}<\frac{120-117.4}{\sqrt{48.49}}\right) \end{gathered}$$

$$\begin{gathered} \approx \mathrm{\Phi }\left(0.3734\right) \\ P(M\le 120)=0.6456 \end{gathered}$$

We're going to assume that men's and women's salaries are equal. Establish the random variables $X$and $Y$, which represent the number of women and men in the sample who have a paycheck minus sign.

$$\begin{gathered} 20,000\text{ormore. Wehavethat} \\ X~\mathrm{Binom}(200,0.534), \\ Y~\mathrm{Binom}(200,0.745) \end{gathered}$$

$\begin{array}{r}P(X\ge 100,Y\ge 150)=\left(1-P\left(\frac{X-106.8}{\sqrt{49.77}}\le \frac{99.5-106.8}{\sqrt{49.77}}\right)\right)\\ \left(1-P\left(\frac{Y-149}{\sqrt{38}}\le \frac{149.5-149}{\sqrt{38}}\right)\right)\end{array}$

$=(1-\mathrm{\Phi }(-1.035\left)\right)(1-\mathrm{\Phi }(0.0811\left)\right)$

$$\begin{gathered} =0.8497\times 0.4677 \\ =0.3974 \end{gathered}$$