Question

hi,how do you use Microsoft Excel to find a z-score when you have a given point in the distribution, population size, a sample size, a mean, and a standard deviation?

Answer 1

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.

`\mu_x=83.3`

The standard error of the sample will be

`\sigma_x=\frac{\sigma}{\sqrt[]{n}}=\frac{5.8}{\sqrt[]{33}}=1.01`

Now let us multiply this standard error by the correction factor since the sample size (33) is greater than 5% of the population (85)

`CF=\sqrt[]{(N-n)/(N-1)}=\sqrt[]{(85-33)/(85-1)}=0.787`

So, the standard error after the correction is

`\sigma_x=1.01\cdot0.787=0.79`

b. Sample mean less than 85

The probability that the sample mean will be less than 85 is given by

`P(\mu_x<85)=P(z<(x-\mu_x)/(\sigma_x))`

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

`P(\mu_x<85)=0.9843`

c. Sample mean more than 84

The probability that the sample mean will be more than 84 is given by

`P(\mu_x>84)=1-P(z<(x-\mu_x)/(\sigma_x))`

Using Excel,

=NORM.DIST(84, 83.3, 0.79, TRUE)

Now subtract the result from 1

1 - 0.8122 = 0.1878

`P(\mu_x>84)=0.1878`

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

`P(82.5<\mu_x<84.5)=P(z<84.5)-P(z<82.5)`

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

`P(82.5<\mu_x<84.5)=0.7800`