91Ó°ÊÓ

Teachers in (public) classrooms earn an average yearly income of $\$49.0$thousand dollars. Assume a $\$9.2$thousand standard deviation.

The "annual pay" of classroom teachers is the variable of concern here. We have $\mathrm{μ}=49$and $\mathrm{σ}=9.2$based on the above data.

Formula used:

For samples of size $64$, the sampling distribution of the sample mean is as follows: Size of the sample,$n=64$

Mean of $\stackrel{¯}{x}$

$$\begin{gathered} {\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ} \\ =\$49.0 \end{gathered}$$

Standard deviation of $\stackrel{¯}{x}$

$$\begin{gathered} {\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}} \\ =\frac{9.2}{\sqrt{64}} \\ =\frac{9.2}{8} \\ =\$1.15 \end{gathered}$$

The mean and standard deviation of all possible sample mean wages for samples of $64$class teachers are $\$49$and $\$1.15$, respectively, for samples of size$64$

For samples of size $256$, the sampling distribution of the sample mean is as follows: The sample size is$n=256$

Mean of $\stackrel{¯}{x},{\mathrm{μ}}_x=\mathrm{μ}$

$=\$49.0$

Standard deviation of $\stackrel{¯}{x}$,

$$\begin{gathered} {\mathrm{σ}}_{\stackrel{¯}{x}}==\frac{\mathrm{σ}}{\sqrt{n}} \\ =\frac{9.2}{\sqrt{256}} \\ =\frac{9.2}{16} \\ =\$0.575 \end{gathered}$$

The mean and standard deviation of all potential sample means salaries for samples of $64$class teachers are $\$49.0$thousand and $\$0.575$thousand, respectively, for samples of size$64$

(c) There's no need to presume that classroom instructor salaries are spread evenly throughout answer parts.

(a) and (b) are due to the huge size of the samples$\left(nâ‰\yen 30\right)$

The likelihood that the average classroom teacher wage will be at least $\$1000$,

$$\begin{gathered} P\left(\right|\stackrel{¯}{X}-\mathrm{μ}|≤1)=P\left(\frac{|\stackrel{¯}{X}-\mathrm{μ}|}{\mathrm{σ}/\sqrt{n}}≤\frac{1}{9.2/\sqrt{64}}\right) \\ =P\left(\left|Z\right|<\frac{1}{1.15}\right) \\ =P(-0.87<Z<0.87) \\ =P(Z<0.87)-P(Z<-0.87) \\ =0.8078-0.1922 \\ =0.6156 \end{gathered}$$

As a result, the risk of making a sampling mistake in predicting the population mean wage of a sample of $64$classroom teachers $\$1000$is$0.6156$

Suppose samples of size,$n=256$

The likelihood that the average classroom teacher wage will be at least $\$1000$,

$$\begin{gathered} P\left(\right|\stackrel{¯}{X}-\mathrm{μ}|≤1)=P\left(\frac{|\stackrel{¯}{X}-\mathrm{μ}|}{\mathrm{σ}/\sqrt{n}}≤\frac{1}{9.2/\sqrt{256}}\right) \\ =P\left(\left|Z\right|<\frac{1}{0.575}\right) \\ ≈P\left(\right|Z|<1.74) \\ =P(-1.74<Z<1.74) \end{gathered}$$

$$\begin{gathered} =P(Z<1.74)-P(Z<-1.74) \\ =0.9591-0.0409 \\ =0.9182 \end{gathered}$$

As a result, the probability of making a sampling error in estimating the population means a sample of $256$classroom instructors' income $0.9182$will be at most $\$1000$