Suppose that the 33rd percentile of a normal distribution is equal to 656 and that the 97.5th percentile of this normal distribution is 896. Find the mean μ and the standard deviation a of the normal distribution.
We find
P(x ≤ 63.5) = P (z ≤ - 3.85)
Using the normal table, we find that the area under the standard normal curve to the left of -3.85 is .00006. This says that, if/; equals .10, then in only 6 in 100,000 of all possible random samples of 1.000 purchasers would 63 or fewer say they would stop buying the cheese spread if the new spout were used. Since it is very difficult to believe that such a small chance (a .00006 chance) has occurred, we have very strong evidence that p does not equal .10 and is, in fact, less than .10. Therefore, it seems that using the new spout will be profitable.