A bank loan officer rates applicants for credit the ratings...

A bank loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50. a) If an applicant is randomly selected, find the probability of a rating between 170 and 220. b) If an applicant is randomly selected, find the probability of a rating above 178. c) Find P80, the score which separates the lower 80% from the top 20%. d) If 40 different applicants are randomly selected, find the probability that their mean is above 215. e) If the loan officer processes 300 applications, how may would be expected to have a rating between 170 and 220.

Transcript

00:01 There is given a normal distribution for this question.

00:03 The mean and the standard division were given here.

00:05 So the mean, this is denoted by mu, this is equal to 200, and the standard division that we have here.

00:12 So the standard division, which is denoted by sigma, and that was given as 50 here.

00:17 So i'm going to define the random variable x, which is normal and the distributed.

00:21 So the mean is 200, and the standard division, 50.

00:24 Let's take a look at the first part of the question.

00:27 Let's say this is a.

00:28 So we're going to select just one applicant here and we're going to find the probability that rating between 170 and 220.

00:35 That means the random variable x should be less than 220 and greater than 170.

00:43 In order to get this probability, i'm going to use the graphing this way calculator application, the normal cdf.

00:48 So the lower boundary 170, upper boundary is 220.

00:52 And the mean is 200 and the standard division is 50 here.

00:55 Press second variance and the second option, the lower boundary, 170, upper boundary is 220, and the mean is 200, and the standard division 50.

01:05 So the answer would be, which is 0 .38 and 12.

01:10 And what about for part b? and in this case, which is ebo 178, so we need to get the probability of x is greater than 178.

01:20 Again, i'm going to use the normal cdf function.

01:24 So the lower boundary 178.

01:27 There is no upper boundary put very big number.

01:29 So the mean is 200 and the standard division 50.

01:32 Let me get the answer.

01:34 Press second variance, the second option.

01:36 The lower boundary, 178, the upper boundary 1.

01:39 This is second 899.

01:40 And the mean is 200 and the standard division 50...

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