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Numerade Educator



Problem 49 Medium Difficulty

As shown in Example $6.8,$ IQ scores are considered normally distributed, with a mean of 100 and a standard deviation of 16.
a. Find the probability that a randomly selected person will have an IQ score between 100 and $120 .$
b. Find the probability that a randomly selected person will have an IQ score above $80 .$


a. .3944; b. .8943


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Video Transcript

okay for this problem, we're gonna talk about some likelihood or some probability of my i Q scores. So the biggest fact we're going normal applications in this chapter. So we're told that's normally distributed with a mean of 100 and a standard deviation of 16. Okay, I said no. That's to me. The 100. And if we got rid of 1 16 we have the next one over there. If we went down one standard deviation, uh, be down to before. So that's like some rough boundaries here you could schedule over now, but let's just look at plus or minus one standard deviation roughly. And then what I like to dio and I think your teachers want you to do is to show the direction with what you have. So let's restate what we want. It's always good to restate using probability notation what you want. So for part A, they specifically asked you the likelihood that you have scores between 120. So the way you write the probability rotation is the probability that you have scores between 101 20. It's always a good idea to show that direction so 100 is here went when he just passed there. So I left in space here just to do this. So not the most beautiful sketch, but the sketches for direction. You're not sketching a graph to be grated. So 101 20 is that chunk there? So you have a decent size number, but not a huge number for the probability. Now you look at table three and like a disease course reach of those and they get the probabilities or a more efficient way to do this is to use a calculator. Workers APP lets you need to use normal CEO cumulative density frequency. So calculators gonna want you to say worried. Look, So you're telling the calculator Well, I want to look from 100 to 101 120 for this problem. So 100 is my lower boundary and my upper boundary Is that 1 20? Um, and I take this in a minute ago. But if you don't have this year, 100 is my mean and my standard deviation. So everything matches what's on my calculator? I think much of my diagram teachers replied. Rather see the diagram. I don't even see a calculator, but let's get the probability of that 0.394 that seems to make sense. Based on the relative size of the distribution of the area. Under the curve is always one for everything. 10.394 39.43 percent is the likelihood of that score. Not for part B. Let's look at this party Wants to know, uh, the likelihood that your score over 80 Okay, it's we're going to say X is greater than he. This is a pretty big numbers of 80 Force here, once in a deviation below 80. Would be about right here, brought up here for reference. Um, so I'm gonna little overlapping sketching here, but it's all right. So over 1 80 So what really happens here? He's really looking at a relatively big number because it's the whole area under the curve. So everything above 80 so should have a pretty big number. So let's go to our calculator and see what we got. So we're gonna find the distribution of the scores. It's normally distributed him since we've been told that, um, human live Lee looking left to right. So after, right, we know 80 is the lowest point that we concerned with for this problem, and we want to go to infinity. And 99999 is perfectly fine for the relative size of the numbers. The mean is 100. This an innovation 16. So it's gonna give us the area to that curve with proportion, no scores of all in that area. So we have a number. It's pretty big, great ones, the whole thing. So that says about 89% 890.89 for three for four do around in there. So the probability that we get scores that a greater than 80 is pointing nine for four.