Like

Report

Given that $x$ is a normally distributed random variable with a mean of 60 and a standard deviation of 10 find the following probabilities.

a. $\quad P(x>60)$

b. $\quad P(60< x<72)$

c. $\quad P(57< x<83)$

d. $\quad P(65< x<82)$

e. $\quad P(38< x<78)$

f. $\quad P(x<38)$

a. 0.5000

b. 0.3849

c. 0.6072

d. 0.2946

e. 0.9502

f. 0.0139

No Related Subtopics

You must be signed in to discuss.

it's This problem starts off with X being normally distributed. Every time we see that we can draw our bell curve. It tells us that our average is 60 and our standard deviation is 10. So the 60 would be placed right at the peak of that bell. And this problem involved six parts. The first part being What's the probability that are Ex score would be greater than 60. And if we look at our picture, greater than 60 is half of the bell. So therefore the probability is 0.5 for part B. We're asking what is the probability that 60 is less than X, which is less than 72? So if we think in terms of the picture that goes with that, we're talking about 60 up to 72. So we're talking about this little area right here, and in order to do that, we would have to utilize disease scores. And you have a formula about the scores that says the equals X minus new over Sigma. And if we were to utilize that formula with the 60 we would end up with a Z score of zero, just by the nature of where it fell. And if I did the Z score associated with 72 it would read 72 minus 60 divided by 10 and you would end up with a Z score off 1.2. So instead of our problem in terms of the X score, we can rewrite this problem now in terms of the Z score and the 60 corresponds to a Z score of zero and the 72 corresponded to Izzy score of 1.2. In order to do the probability of between two values, we would turn it into a subtraction problem and we would talk about Z being less than 1.2. Subtract the probability of Z being less than zero, and at that point you would need to utilize the back of your book. Ah, there is the normal distribution table and associate it with a Z score of 1.2 is an area of 0.8849 in the area associated with zero would be 00.5000 and therefore the area, or the probability that X falls between 60 and 72 would end up being 720.38 for nine part C. we're gonna do the same thing. We're going to set up a picture. We're going to find disease scores. So for part C, it's asking us for the probability that 57 is less than X, which is less than 83. So we continue to use the same bell set up with 60 in the centre. So 57 would be slightly to the left, 83 would be to the right, and we're asking for this area under the curve. And in order to do that, we would have to apply the Z score formula twice once for 57. So we would do 57 minus 60. Divide by 10. And in doing that, we would get a Z score of negative 0.3 and the Z score associated with 83 would be 83 minus 60. Divide by 10 which is a Z score of 2.3. So taking it back to our picture that saying that the 57 is a Z score of negative 570.3 and the 83 is a Z score positive 2.3 so we can now rewrite instead of in terms of X. We can put it in terms of Z and we could say negative 0.3 is less than is the which is less than 2.3. And again we would have to change it into a subtraction problem. So we would then say the probability that Z is less than 2.3, minus the probability that Z is less than negative 0.3 utilizing the back of the book, we would find the areas associated with each to be 0.9893 minus 0.38 to 1. So therefore, the probability that X falls between 57 83 would be 830.607 to in part D. Same scenario. I'm going to start taking a couple short cuts as we go. So in part D, I would recommend again drawing the picture. And in part D. We are asking for the probability that 65 is less than X, which is less than 82. So in our picture again, 60 is in the center, 65 would be to the right and 82 would be to the right. So we're asking for that probability or that area under the curve, so we would utilize our Z score formula again. Z equals X minus mu over sigma and you're going to get an X value associated with 65 or sorry Z value associated with 65 to be 650.5 and the Z score associated with 82 would be a 2.2. So we rewrite our probability problem in terms of see scores as the probability that 0.5 is less than Z, which is less than 2.2, we would then switch it into a subtraction. Problem Z is less than 2.2, minus the probability that Z is less than 0.5. Again, we would use the table in the back of your book. The probability that C is less than 2.2 would be 0.9861 The Probability Z is less than 0.5 this 0.6915 So therefore, the probability that X falls between 65 and 82 is going to be 820.2946 in Part E. We want to talk about the probability that ex falls between 38 and 78. So we're still using the same information about the bell again sixties in the center, 38 would be to the left. 78 would be to the right, and we're talking about this area under the curve again. We're going to use our Z score formula is e equals X minus mu over Sigma. We're going to do it for the 38 end for the 78 the Z score associated with 38 is negative. 2.2. Does he score? Associate it with 78 is 1.8. Therefore, we can rewrite our problem in terms of see scores Probability that negative 2.2 is less than Z is less than 1.8. We convinced. Turn it into a subtraction problem. Probability that Z is less than 1.8, minus the probability that C is less than negative 2.2 Utilize your chart from the back of the book. The probability that C is less than 1.8 would be 0.9641 And the probability that Z is less than negative 2.2 would be 0.139 So the answer for part e the probability that X falls between 38 78 would be 780.9 five zero to and then finally, part f of this problem again continues to work the same way in this problem. We're doing the probability that X is less than 38. If we think about what the picture would look like 60 being in the center 38 would be over here. We're looking for the area over into that left tail. So again, we're going to use our Z score formula. So we're gonna do 38 minus 60 divided by 10 and we end up with a Z score, associate it with 38 of negative 2.2. So negative 2.2 corresponds with 38 so we can rewrite our problem. The probability that Z is less than negative 2.2 and we would just have to look that up in the back of the book in the chart. So the probability of ex being less than 38 would be 380.13

WAHS

No Related Subtopics