Question

. A normally distributed population has mean 57,800 and standard deviation 750.

a. Find the probability that a single randomly selected element X of the population is between 57,000 and 58,000.

b. Find the mean and standard deviation of 𝑥̅for samples of size 100.

c. Find the probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000. 

Answer 1

We are given that a normally distributed population has mean 57,800 and standard deviation 75, i.e.; $\mu = 57800$  and  $\sigma=750$ .

Let X = randomly selected element of the population

The z probability is given by;

      Z = $\frac{X-\mu}{\sigma}$  ~ N(0,1) 

a. So, P(57,000 <= X <= 58,000) = P(X <= 58,000) - P(X < 57,000)

P(X <= 58,000) = P(  $\frac{X-\mu}{\sigma}$ <= $\frac{58000-57800}{750}$  ) = P(Z <= 0.27) = 0.60642

P(X < 57000) = P(  $\frac{X-\mu}{\sigma}$ < $\frac{57000-57800}{750}$  ) = P(Z < -1.07) = 1 - P(Z <= 1.07) = 1 - 0.85769 = 0.14231

Therefore, P(31 < X < 40) = 0.60642 - 0.14231 = 0.46411 .

b. Now, we are given sample of size, n = 100

So, Mean of X, $\overline{X}$ = 57,800 same as before

But standard deviation of X, s = $\frac{\sigma}{\sqrt{n}}$ = $\frac{750}{\sqrt{100}}$ = 75

c. The z probability is given by;

      Z = $\frac{\overline{X}-\mu}{s}$  ~ N(0,1) 

Now, probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = P(57,000 < $\overline{X}$ < 58,000)

P(57,000 <= $\overline{X}$ <= 58,000) = P($\overline{X}$ <= 58,000) - P($\overline{X}$ < 57,000)

P($\overline{X}$ <= 58,000) = P(  $\frac{\overline{X}-\mu}{s}$  <= $\frac{58000-57800}{75}$ ) = P(Z <= 2.67) = 0.99621

P(X < 57000) = P( $\frac{\overline{X}-\mu}{s}$ < $\frac{57000-57800}{75}$) = P(Z < -10.67) = P(Z > 10.67)

This probability is that much small that it is very close to 0

Therefore, P(57,000 < X bar < 58,000) = 0.99621 - 0 = 0.99621 .