Respuesta :
Given that the mean and standard deviation of the population are $24,215 and $3712 respectively,
[tex]\begin{gathered} \mu=24215 \\ \sigma=3712 \end{gathered}[/tex]The sample size taken is 80,
[tex]n=80[/tex]Consider that the salary of students in the sample is assumed to follow Normal Distribution with mean and standard deviation as follows,
[tex]\begin{gathered} \mu_x=\mu\Rightarrow\mu_x=24215 \\ \sigma_x=\frac{\sigma}{\sqrt[]{n}}=\frac{3712}{\sqrt[]{80}}\approx415 \end{gathered}[/tex]So the probability that the mean salary (X) is $24250 or more, is calculated as,
[tex]\begin{gathered} P(X\ge24250)=P(z\ge\frac{24250-24215}{415}) \\ P(X\ge24250)=P(z\ge0.084) \\ P(X\ge24250)=P(z\ge0)-P(0From the Standard Normal Distribution Table,[tex]\begin{gathered} \emptyset(0.08)=0.0319 \\ \emptyset(0.09)=0.0359 \end{gathered}[/tex]So the approximate value for z=0.084 is,
[tex]\emptyset(0.084)=\frac{0.0319+0.0359}{2}=0.0339[/tex]Substitute the value in the expression,
[tex]\begin{gathered} P(X\ge24250)=0.5-0.0339 \\ P(X\ge24250)=0.4661 \end{gathered}[/tex]Thus, there is a 0.4661 probability that the mean salary offer for these 80 students is $24,250 or more.
