Correcting for a Finite Population In a study of babies born w...

Question

Correcting for a Finite Population In a study of babies born with very low birth weights, 275 children were given IQ tests at age 8, and their scores approximated a normal distribution with μ = 95.5 and σ = 16.0 (based on data from “Neurobehavioral Outcomes of School-age Children Born Extremely Low Birth Weight or Very Preterm,” by Anderson et al., Journal of the American Medical Association, Vol. 289, No. 24). Fifty of those children are to be randomly selected without replacement for a follow-up study.

b. Find the probability that the mean IQ score of the follow-up sample is between 95 and 105.

b. Find the probability that the mean IQ score of the follow-up sample is between 95 and 105.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says in a survey of 210 college students, the number of hours spent studying per week is normally distributed with our mean mu equal to 13.2 hours, and our standard deviation sigma equal to 4.1 hours. If 35 students are randomly selected without replacement for a special academic program, what is the probability that their mean study time is between 12 hours and 15 hours per week? And we're given 4 possible choices as our answers. Your choice A, We have 0.937, for choice B, we have 0.948, for choice C, we have 0.969, and for choice D, we have 0.983. So to find this probability, the first thing we're going to do is compare our sample size with our population size. So we're going to look at the ratio of little N divided by capital N. where little N here is the sample size, and N is our population size. And so, based on the information problem, this is going to be equal to, or low end, our sample size was 35, we'll have 35, and this can be divided by our population size, which is 210. And when we evaluate this fraction, this is going to be equal to 0.119. And here we just round to three decimal places, and this is actually greater than 0.05. So since this value is greater than 0.05 and we're sampling without replacement, we need to adjust the standard deviation of the sample means, sigma sub X bar by multiplying it by the finite population correction factor, which is the square root of the quantity of capital N minus little N, and that quantity is going to be divided by the quantity of capital N minus 1. So that means that sigma of X bar is going to be equal to sigma, divided by square root of N. And that fraction is multiplied by the square root of the quantity of capital N minus little N, and that is divided by the quantity of capital N minus 1. And we actually values for all those quantities. So we'll substitute in those values. So this is going to be equal to. 4.1 Divided by root of 35. And the fraction is multiplied by the square root of the quantity of 210. Minus 35, and that quantity is divided by. The quantity of 210 minus 1. And we can evaluate this expression, and we will get 0.634. And so we're now going to use this quantity to calculate our Z scores. And so, recall your formula for calculating the Z score, that we will have Z as equal to the quantity of X bar minus mu. In quantity divided by Sigma sub x bar. So, we're gonna have to calculate 2 Z scores here, one when X bar is equal to 12 hours, and another when it's equal to 15 hours. So, we'll begin by calculating it for 12 hours, so we'll call that the score Z1. And so that's going to be equal to the quantity of 12 minus from you, we have 13.2. And that quantity is going to be divided by sigma sub X bar, which we just calculated, which is 0.634. And when we evaluate this expression, we see that Z1 is going to be equal to. -1.892. We'll do the same thing again for Z2, except this time we'll have X equal to 15. We have Z2 is equal to the quantity of 15 minus 13.2 in quantity, divided by 0.634. And evaluating this expression, this is going to be equal to. 2.838. So, now we need to find the probabilities associated with these Z scores. So, we're gonna be looking at the probability P, that Z is going to be less than, and for our first Z score, we'll use Z1, so less than minus 1.892. And so this is going to be equal to the value that we find in our tables. However, we're actually going to need to use interpolation in order to find this value since it falls in between -1.89 and -1.90. So, using interpolation, that means that our probability P here is going to be equal to the quantity, and here we're gonna have a fraction where in the numerator, we're going to have our value for the probability when Z is less than -1.89. And looking up that value in the table, we get 0.0294, and then we'll have minus. The value of P, when Z is less than -1.9, and again, looking up that in the table, we get. 0.0287. And that difference is going to be divided by the quantity of minus 1.89. 1.90. And we'll need to multiply this fraction here, which is essentially like a slope. By the quantity of the value that we're looking for, which is -1.892 minus. The minus 1.89 in quantity. Then we'll have to add to this. The probability when Z is less than -1.89, which we found earlier to be 0.0294. And evaluating this expression, we see that P or Z less than minus 1.892 is going to be equal to. 0.0292. And here we just round it to 4 decimal places. We're also going to need the probability P when D is less than 2.838. And we can actually look this up just directly on the table and not have to do any interpolation. In doing so, this is going to be equal to. 0.9977. That's because the value for the probability of Z less than 2.83 and the probability of P that Z is less than 2.84, give us the same value on the table, so there's no need to do interpolation. So now to find our probability that the question is asking for. Which is going to be the probability that. is when minus 1.892 is less than Z, which is less than 2.838, which would give us the probability that the number of our study is between 12 and 15. So this is just going to be equal to. The probability P Of the less than. 2.838. Minus P of Z less than minus 1.892. So this is going to be equal to 0.9977 minus 0.0292. And so this is going to be equal to. 0.969. And here we've just rounded to three decimal places. So that is the answer that we are looking for for this problem. And if we compare that to our answer choices, we see that the correct answer choice corresponds to answer choice C. So, I hope that this explanation has been useful, and I'll see you in the next video.

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Key Concepts

Normal Distribution

Sampling Distribution of the Mean

Central Limit Theorem

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