CHAPTER 8

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CHAPTER 8

SOLUTIONS TO PROBLEMS

a.
b.
c.

d. To answer this question, you first have to determine the z score that corresponds to a raw IQ score of 116 (since it asks for those who score above 115: . Now, you go to the z table with a z score of 1.60. You know the area of the curve to the left of the mean is .50. From the table you can see that the area to the right of the mean to a z score of 1.60 is .4452. The area to the left of a z score of 1.60, therefore, is equal to .50 + .4452 = .9452. Since the total area under the curve is 1.0, the area to the right of a z score of 1.60 must be 1 - .9452 or .0548. We show this for you in the picture below. That means that .0548 or slightly more than 5% of the cases have an IQ score above 140.

e. In order to answer this question you have to find the corresponding z scores for the raw scores of 90 and 110: . Now, you have to go to the z table and find the area of the normal curve that lies between -1.0 and +1.0. Looking at the z table you can see that the area of the curve from the mean to a z score of 1.0 is equal to .3416. Since the nor mal curve is symmetric, the area of the curve from the mean to a z score of -1.0 is also .3416. Therefore, the area of the curve between -1.0 and +1.0z is equal to .3416 + .3416 or .6832. We show you this in the picture below.

You should recall that in the chapter we noted that approximately 68% of the normal curve is within ±1 standard deviation units. A z score of 1 is one standard deviation unit away from the mean.
a. a raw score of 95 corresponds to a z score of 1.50: . Now go to the z table. You know that .50 of the curve lies to the left of the mean. How much of the curve lies lies to the right of the mean to a z score of 1.50? Looking at the table you can see that it is .4332. A raw score of 95, then, is better than .50 + .4332 = .9332 or 93% of the scores. It is not in the top 5% of scores, therefore, and this candidate would not be accepted.

b. A raw score of 110 corresponds to a z score of 3.0: . How good of a score is this? Again, you know that .50 of the curve lies to the left of the mean. Looking at the z table you can see that the area of the curve to the right of the mean to a z score of 3.0 is .4986. A raw score of 110, therefo re, is better than .50 + .4986 or .9986 or 99% of the scores. It is in the top 5%, and this candidate would be accepted.

c. This question asks you to find the minimum raw score that would be good enough to be accepted. You know that in order to be accepted you have to score in the top 5% or the top .05 of the curve. Find the z score that corresponds to .05 of the normal curve. Go to the z table and find .4500, and look for the z score for this area. You will find .4495 and .4505. Let's use .4505. You will see that the z score is 1.65. This tells you that a z score of 1.65 or higher will place you in the top 5% of the class, giving you an acceptable score. Now you have to figure out what the raw score is for a z score of 1.65. To solve this, just take the normal z equation, and put in what you know (z of 1 .65, mean of 80, standard deviation of 10):
.
Now, simply solve for x. First, multiply both sides of the equation by 10:

then add 80 to both sides of the equation: .

Now you know that a z score of 1.65 corresponds to a raw score of 96.5, and that you must have a raw score of at least 96.5 on the test to be accepted as a correctional officer. If you do not believe it, just reverse the arithmetic an d find the z score that corresponds to a raw score of 95.6: .

Now, go to the z table with a z score of 1.65. You know that .50 of the curve lies to the left of the mean. Look in the table and find that .4505 of the curve lies to the right of the mean to a z score of 1.65. The area to the left of a z score of 1.65, then, is equal to .50 + .4505 or .9505. Therefore, z score of 1.65 or higher is better than 95% of the scores, and that is the minimum score you need to get accepted. We show this for you in the picture.
a. To find the proportion of cases between two raw scores, first convert them to z scores. .

Now we need to find the area of the curve between a z score of -.63 and 1.37. Go to the z table and you will find that the area of the curve from the mean to a z score of -.63 (remember, the normal curve is symmetric so the distance from the mean to a z score of .63 is the same as that from the mean to a z score of -.63) is .2367. The area of the curve from the mean to a z score of 1.37 is .4147. As we show you in the picture, the area of the curve b etween the two points is .2367 + .4147 = .6514.

b. First find the z scores for the two raw scores: . Now go to the z table. The area of the curve from the mean to a z score of -1.25 is .3944. The area of the curve from the mean to a z score of 2.50 is .4938. As you can see from the picture, the area in between the two scores is equal to .3944 + .4938 = .8882. In other words, almost 90% of the area of the standard normal curve lies between those two z scores.

c. Find the z scores for the two raw scores: . From the table you can see that the area of the curve from the mean to a z score of -1.50 is .4332, while the area of the curve from the mean to a z score of -.37 is .1433. As shown by the picture, we have to subtract the second area from the first in order to find the area of the curve between the two scores: .4332 - .1433 = .2899.

d. In this question, we want to know the raw score that is better than 90% of all the other scores. First, find the z score that is in the upper 10% or .10 of the tail (the right tail). You know that .50 of the curve lies to the left of the mean. Move to the right of the mean until you cover an area equal to .40. You will not find .40 exactly, but .3397 and .4015. We will use the latter number. If we add this to the area to the left of the mean we will have a total area of the curve e qual to .50 + .4015 = .9015. The area to the right of this point is in the upper 10% or .10 of the curve. Find the z score corresponding to this point. You see that it is equal to 1.29. This tells you that z scores of 1.29 or greater are in the upper 10% of the curve. But, the question really asks for the raw score that is in the upper 10%. We now have to transform our z score of 1.29 into a raw score. Knowing that we have a mean of 60 and a standard deviation of 8, we c an substitute what we know into our ususal z score formula, and then solve for x: .

Multiply both sides of the equation by 8:
.
Now simply add 60 to each side:
.
Our raw score that is better than at least 90% of the other scores is 70.32.
The area under the standard normal (z) curve between: a. the mean and a z score of .20 is equal to .0793.

b. the mean and a z score of -2.34 is .4904.

c. a z score of -1.23 and a z score of .67 is .3907 + .2486 = .6393.
a. First, transform the two raw scores into z scores. The mean has a z score of 0. A raw score of 30 has a z score of:
.
The area of the curve between the mean and a z score of .5, then, is .1915.

b. Transform both raw scores into z scores:
.
The area of the curve between the mean and a z score of -.75 is equal to .2734. The area of the curve between the mean and a z score of 2.0 is equal to .4773. The area of the curve between the two z scores, then, is .2734 + .4773 = .7 507.

c. To find the raw score that is greater than 95% of all the other raw scores, first find what z score this would be. Go to the z table. You know that the area to the left of the mean is equal to .50 move to the right of the me an until you find .45 more of the area. You will not find .45 exactly, but .4495 and .4505. We will use the latter number. If you add these two areas we have a area equal to .50 + .4505 or .9505. The z score corresponding to this point on the curve is 1.65. Any z score this larger or larger, then, will fall in the top 5% of the curve, or will be greater than 95% of all the other scores. Now all we have to do is find the correct raw score that corresponds to a z score of 1.65 . Again, we know the mean is equal to 28 and the standard deviation is equal to 4. We can put what we now know into the formula for the z score and solve for the unknown:
.
Raw scores that are 34.6 or higher, then, are in the top 5% of all scores.

d. Here we have to find the area of the curve where the lowest 1% of the scores are. This means that we are interested in the left tail of the standard normal distribution. To find this area, keep in mind that the total area to the r ight of the mean is .50. We need to go to the z table and find .49 of the remaining area. We can find an area equal to .4901. When we add this to the .50 that is to the right of the mean, we have found .9901 of the curve. Find the z score th at corresponds to this point. You will see that it is -2.33. We show this for you in the picture below.

Now if have to find the raw score that corresponds to a z score of -2.33. This should be familar to you by now:
.
Scores that are 41.36 or lower, then, are less than 99% of all the other raw scores.
a. The area to the right of a z score of 1.65 is equal to .0495

b. The area to the left of a z score of -1.65 is equal to .0495

c. The area either to the left of a z score of -1.65 or to the right of a z score of 1.65 is equal to .099.
a. To see how unusual 9 prior arrests are in this population, let's transform the raw score into a z score: . Taking a z score of 1.50 to the z table we can see that the area to the right of this score comprises approximately 7% of the area of the normal curve. Those who have nine prior arrests then, are in the top 7% of this population. Since they are not in the top 5%, we would not consider them unusual.

b. A raw score of 11 prior arrests corresponds to a z score of: . Looking at the z table, a z score of 2.50 is way at the right or upper end of the distribution. Z scores of 2.50 or greater are greater than approximately 99% of all the other scores. This person does h ave an unusually large number of prior arrests since they are in the top 5%.

c. A raw score of two prior arrests corresponds to a z score of: . A z score of -2.0 falls in the lower end of the tail. Scores this lower are lower than almost 98% of all the other scores. The person with only two prior arrests, then, does have an unusually low number for this popu lation since they are in the bottom 5%.

8. The Central Limit Theorem is a statistical proposition which holds that if an infinite number of random samples of size N are drawn from any population with mean m and standard deviation s, then as the sample size becomes large, the sampling distribution of sample means will become normal with mean m and standard deviation equal to . The central limit theorem allows us to make three assumptions about sampling distributions when the sample size is large: a) We can assume that the mean of the sampling distribution is equal to the population mean, m, b) We can assume that the standard deviation of the sampling distribution is equal to s/ , and c) We can assume the sampling distribution is normally distributed even if the population from which the sample was drawn is not.
The first distribution is a distribution of sample scores. This distribution is empirical, in that it comes from our observed data. There is no assumption that the distribution of sample scores is normal. The second distribution is a di stribution of population values. This distribution is also empirical, in that it exists but we usually do not know much about its characteristics. What connects these two distributions is the third distribution, which is a distribution of sample means. T his distribution is theoretical, and based upon probability theory. It is the theoretical distribution of an infinite number of sample statistics of sample size N. The sampling distribution is what connects our known sample mean to our unknown popu lation mean. We can make inferences from our sample mean to the population mean by considering it as one sample mean from a theoretical distribution of all possible sample means. With our sampling distribution we can determine the probability of obtaining our sample statistic. Since the sampling distribution of means is normal and the population is assumed to be normal, we can make use of the standard normal distribution to determine this probability.

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