00:01
This problem says suppose a single random sample of size n equals 36 is obtained from a population with a mean of 83 and a standard deviation of 12.
00:08
And first we're asked to describe the sampling distribution of x and then find several probabilities.
00:12
And here our sampling distribution of x would be approximately normal with the mean of our sampling distribution of 83, that's equivalent to the mean of the population, and the standard deviation of our sample mean being equivalent to the standard deviation of the population divided by the square of the sample size, which in our case would be 12, divided by the square root of 36, which is 6, which gives us 2.
00:32
Now the reason we can treat our sampling distribution as approximately normal, which would give us choice d, is because our sample size is greater than or equal to 30, which means the central limit theorem applies and allows us to treat our sampling distribution as approximately normal, which means we can find our next three probabilities using normal cdf in our calculator.
00:49
And for normal cdf in our calculator to find a probability we need four values, starting with the lower bound and upper bound that we want the probability between.
00:55
And for b we want the probability that we're greater than 86 .9 for our sample mean, so that's our lower bound.
01:01
And then we want to make sure that we include everything greater than 86 .9 or to the right of 86 .9, so we use infinity for the upper bound...