00:01
In this problem, we are given that there are two z scores minus z and plus z.
00:07
And the area between these two z scores, which we will represent with a, this is equal to 0 .58.
00:16
And we have to determine the value of the z score that's represented by the letter z.
00:21
So we will consider this as z star instead, so as to distinguish this from the regular standard normal random variable z.
00:29
And the area between these two z scores, which is the probability that the z score is more than minus z star, but less than z star, this is equal to 0 .58.
00:40
So we can write it as twice the probability that the z score is greater than 0, but less than z star, this is equal to 0 .58.
00:50
And that implies the probability that the z score is less than z star minus probability that the z score is less than 0, it is equal to 0 .8...