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

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
[tex]z= \frac{x-mean}{std} [/tex]
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.