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As shown in Example $6.12,$ incomes of junior executives are normally distributed with a standard deviation of $\$ 3828.$a. What is the mean for the salaries of junior executives, if a salary of $\$ 62,900$ is at the top end of the middle $80 \%$ of incomes?b. With the additional information learned in part a, what is the probability that a randomly selected junior executive earns less than $\$ 50,000 ?$

a $59,614.36 b .0068

Intro Stats / AP Statistics

Chapter 6

Normal Probability Distributions

Section 3

Applications of Normal Distributions

The Normal Distribution

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okay for this problem. We have some more normally distributed data, but this problems a little bit different because we don't know what the means. War is so with the most normal curve problems. You want a sketch? Uh, the normal distribution and write down what we know. So first I'm going to do is write down. We know, um, that standard deviation equals two ah, $3898 for junior executives at a company. So what's different about this problem in some of the earlier problems is that we want for this problem is we want to know for party. At least we want to know what, Thea, what the mean value is. So our plan is going to be to really use ah, the formula for the Z score to get the to get the mean Okay, so we're gonna have to find the Z score. So let's figure out how to get a Z score here. So first things first, we're gonna get ah, go to our calculator. And we were told in the problem that there's an 80% middle 80% of incomes have it a certain we have the high end of the salary for the 80% of the incomes. We know what that ISS. So let's do something here. We call the Universal Normal. We've been doing forward what I call forward problems we put in the the Z scores that we know are put in the parameters that we know and to get the probabilities. This case we know the amount of probability of the area is 80%. Guess so. Um, so it's 80. And so we're gonna get work backwards that to get the Z score that goes with 80%. You figure out that, IHS, I'm gonna leave the mean at zero on the stand, aviation at one. Because the score and we just mark this up here, too. Um and so the number that we know is the high end of the 80%. So it's the porn 80 number of kind of from their back is, um, 62,900. We note that case 62,900 of the high end of the middle. 80% of the owners we want to know is the is the meeting here. So, um, from our inverse, normal calculation work backwards on the seat. See table Technically, so we know that we use a Z score, so let's go back to our plan here. So now we know the Z score that goes with the number that we have. The 62,900 is 0.84. So that's the Z score that goes with it. And so we would know using the Z 62,900 minus the mean divided by the standard deviation which we stayed over there is 30 98 is the relationship. So we can work backwards to find at with that with that mean number is for this overall distribution. When party, we're gonna do a little bit more work to figure out a different parameter. So when he's my calculator, since it's up here, so see 0.84 I'm gonna solve for the for the mean value someone multiply both sides of the equation by 38 98 so we can see that thing. It was 3280 0.63 um, equals just 62,900 minus the means were you okay? So now I'm just gonna solve Really What I would do showing all my work. There's at immutable sides. And then what happened to do is take the 60 2900. It's attracted 30 to 80.63. So what type of my calculator over here? 62,900 my ass. The answer. That's my stack here. So that tells us the mean number is 52,000. 59,000 619 0.36 around. There's a penny there. Okay, so that's your answer for perch. What? I'm gonna really? That's what we wanted for up here. Okay, so we have the 59,000 619.3 sheets in really with that is that goes with better there. So I get that part done. Um, And now for part B. What? They asked us who's they asked us. Well, now use your information that you got a part. A to find an executive with an executive. The probability, um, that executive earns less than $50,000 so we wanted the probability the X is less than it 2000 giving this distribution. Okay, well, so now we know this 59,000 is here um, so 50,000. I'd like to make a sketch of the estimate there, so we want to basically find the probability to left in this line. So I liked Always sketch out what we see here. Okay, I'm just gonna keep keep it short arms and about 50 k All right, So when that Rooney's our calculator now, we're going to go forward in our calculator and use our second distribution function to do rz scoring and are, um, probability under understanding. So we're gonna do second distribution. Well, miss click their second distribution. Um, normal CDF So now asking our calculator. Well, let's go from negative infinity to the upper boundary of 50,000. And again, he used an apple it to do this or look from your table, give us the scores. Um, And for this problem, what we looked used in part A. We now know that the mean is 59,000 619 0.36. And standard deviation for this data is 38 98. So, um, that tells us that overall yeah, way have 0.67 the other 60 feet. So that's the probability, given the information we have. But there in that much

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