Question

Population has a mean of 180 and a standard deviation of 24. a sample of 64 observations will be taken. the probability that the mean from that sample will be between 183 and 186 is

Answer 1

We use so called Central Limit Theorem as follows.

If the sample size is large enough, the sample mean would be normally distributed with a mean of µ and a standard deviation of σ/√n where µ and σ are the mean and standard deviation the random variable.

In this problem, µ = 180, σ = 24 and n = 64. Based off the Central Limit Theorem, the sample mean is normally distributed with a mean of 180 and a standard deviation of 24/√64 = 24/8 = 3.

Now let's convert the 183 and 186 into z scores. Formula for a z-score is
$z= \frac{x-mean}{std}$
So the z-score for 183 is (183–180)/3=1.
The z score for 186 is (186 –180)/3 = 2.

Now, let's look up the z-table for the areas to the left of those two z scores. We can see that the area to the left of z = 1 is 0.8413 and the area to the left of z = 2 is 0.9772. The probability that the z is between 1 and 2 would be 0.9772 – 0.8413 = 0.1359.

That's also the probability that the sample mean is between 183 and 186. Thus the answer is 0.1359.