We are given the following information
Population mean = μ = 83.3
Population standard deviation = σ = 5.8
Population size = N = 85
Sample size = n = 33
As you can see, the sample size is quite large (n≥30) so the sampling distribution of the sample mean will be approximately normal.
The sample mean will be the same as the population mean.
[tex]\mu_x=83.3[/tex]The standard error of the sample will be
[tex]\sigma_x=\frac{\sigma}{\sqrt[]{n}}=\frac{5.8}{\sqrt[]{33}}=1.01[/tex]Now let us multiply this standard error by the correction factor since the sample size (33) is greater than 5% of the population (85)
[tex]CF=\sqrt[]{\frac{N-n}{N-1}}=\sqrt[]{\frac{85-33}{85-1}}=0.787[/tex]So, the standard error after the correction is
[tex]\sigma_x=1.01\cdot0.787=0.79[/tex]b. Sample mean less than 85
The probability that the sample mean will be less than 85 is given by
[tex]P(\mu_x<85)=P(z<\frac{x-\mu_x}{\sigma_x})[/tex]The Excel function is given by
=NORM.DIST(x, mean, standard error, TRUE)
So, substituting the values, it becomes
=NORM.DIST(85, 83.3, 0.79, TRUE)
Excel returns the following probability
[tex]P(\mu_x<85)=0.9843[/tex]c. Sample mean more than 84
The probability that the sample mean will be more than 84 is given by
[tex]P(\mu_x>84)=1-P(z<\frac{x-\mu_x}{\sigma_x})[/tex]Using Excel,
=NORM.DIST(84, 83.3, 0.79, TRUE)
Now subtract the result from 1
1 - 0.8122 = 0.1878
[tex]P(\mu_x>84)=0.1878[/tex]d. sample mean between 82.5 and 84.5
The probability that the sample mean will be between 82.5 and 84.5 is given by
[tex]P(82.5<\mu_x<84.5)=P(z<84.5)-P(z<82.5)[/tex]Using Excel,
=NORM.DIST(84.5, 83.3, 0.79, TRUE)
0.9356
=NORM.DIST(82.5, 83.3, 0.79, TRUE)
0.1556
So, the probability is
0.9356 - 0.1556 = 0.78
[tex]P(82.5<\mu_x<84.5)=0.7800[/tex]