4.12 ([1], 4.68, 4.69, 4.70) The grade point averages (GPAs)...

4.12 ([1], 4.68, 4.69, 4.70) The grade point averages (GPAs) of a large population of college students are approximately normally distributed with mean 2.4 and standard deviation 0.8. a) What fraction of the students will possess a GPA in excess of 3.0? b) If students possessing a GPA less than 1.9 are dropped from college, what percentage of the students will be dropped? c) Suppose that three students are randomly selected from the student body. What is the probability that all three will possess a GPA in excess of 3.0?

Transcript

00:01 There is a normal distribution in this question and the mean value is given.

00:04 Let me just take this note.

00:06 The mean is 2 .4.

00:08 Let me just show the mean with this notation that is mu, 2 .4, and the standard deviation, which is denoted by sigma, that is equal to 0 .8.

00:21 Okay, the first part of the question, what fraction of the students will possess gpa in an excess of 3 .02? so since this is a normal distribution, i'm going to just define the random variable, x here, and normal a distributed, that is 2 .4 and 0 .8.

00:40 So in part a, we have to get the probability of x is greater than, which is 3 .02.

00:50 Okay.

00:51 In order to just get this one, i'm going to use the normal cdf function of the ti calculator.

00:57 So in this case, if i just graph the situation here, this is the normal distribution.

01:02 The mean value here is 2 .4 and 3 .02, which is here.

01:08 So we need to get the area of this region.

01:11 So for this shaded region, the minimum value is 3 .02.

01:15 And the upper boundary, that goes to positive infancy.

01:17 I'm going to put a very big number.

01:19 And the mean value is 2 .4.

01:22 And the standard division is 0 .8.

01:24 Let's get the answer.

01:25 This is second distribution, normal cdf.

01:29 Lower boundary 3 .02 comma.

01:34 The upper boundary here is, this is 1 second a99.

01:38 The mean value is 2 .4 and the standard deviation 0 .8.

01:43 Great.

01:45 We got the probability 0 .292.

01:49 I just give the answer with four decimal places.

01:54 Okay.

01:55 Let's keep solving with the number b.

01:59 Here and the part b suppose that the duropout rate of a college is approximately um two point um i think this is 0 .259 as far as i understand so the probability here is there should be some x1 value here x is greater than x1 which is equal to 0 .259 that's it so we have to get this x1 value here and so what am i supposed to do i'm going to use first of all, the inverse norm function in order to just get the standard z score.

02:36 But before inverse norm, i have to just express the probability of x is less than x1.

02:41 The total area under the curve is 1 minus the probability of x is greater than x1, which is 1 minus 0 .259, which is 1 minus 0 .259, that is equal to 0 .741.

02:57 So let's use the inverse norm function in order to get the standard z score 0 .741 because we have to take from left to right.

03:07 Because of that, we need to just get this expression here...

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