# CJ 3320 Statistics for Criminal Justice-One statistics question

Unit 5 AssignmentComplete Exercise 3 on p. 139. This question consists of two part—part a is about CI 95% and

part b is about the z score. You may solve this question with reference to In-Class Exercise 5.3

for part a, and In-Class Exercise 5.2 for part b. Discard (20 points), which is redundant.

Type your answers and submit to the instructor via the Blackboard Messages by noon, Friday,

October 14.

Unit 5 Probability and the Normal Curve

A. Introduction

1. Sampling distribution is a theoretical frequency distribution of the scores of a variable or

a statistic. It is the bridge between descriptive statistics and inferential statistics.

(1) We usually are uncertain about the population, but we may know a great deal about

the sample.

(2) Central Limit Theorem allows us to use the information about samples (i.e.,

statistic) to make inferences about the information of a population (i.e., parameter)

and to have confidence in these inferences.

2. Unless the sample is a mirror image of the population, the statistic (e.g., sample mean)

will likely not equal to the parameter (e.g., population mean). In this case, the difference

between the sample mean and the population mean is called sampling error. On average,

the sampling error produced by large samples will be less than the sampling error from

small samples.

3. Probability theory is the foundation of inferential statistics.

(1) Probability involves the calculation of the chance that an event will happen. It can be

used to guide decision making. In criminal justice, one might wish to calculate the

chance of victimization or the area of a city where a certain type of crime is likely to

occur.

(2) The chance of an event taking place ranges from zero (no chance) to one (certainty).

This concept refers to the bounding rule of probability (0 < P(A) > 1).

B. Probability and the Normal Curve

1. The normal curve is a key statistical tool in determining probability. As a function of

nature, all social and moral data were distributed under the normal curve (Mueller,

Schuessler, and Costner, 1977:169).

2. The normal curve has a

number of useful

characteristics as below:

(1) It is unimodal.

(2) It has a symmetrical

curve.

(3) The mean, median, and

mode are equal.

(4) A constant area under the

curve regardless of its

specific mean or standard

deviation (Table A-2, pp.

230-231 provides normal

probabilities for many

different z-values).

(5) Empirical Rule: If a

sample of measurements has a mound-shaped distribution (See above figure or Figure

7.1, p.128)

x-bar + s = 68.26%

x-bar + 2s = 95.46%

x-bar + 3s = 99.72%

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3. The Standard Normal Distribution

The standard normal distribution or the standard form of the normal distribution refers

to the creating standard scores from raw scores.

z = x – x-bar/s or

z = x – μ/σ (when population mean and standard deviation are known)

In-Class Exercise 5.1: With a mean (x-bar) of 60 and a standard deviation (s) of 9, we

can transform the raw score of 75 into a z score as follows:

Step 1: Calculate the z score

z = x – x-bar/s = (75 – 60)/9 = 1,666666667 = 1.67 (rounding to the nearest hundredth)

Step 2: Find the area of certain z score by using Table A-2 (p. 238)

Go to the first column (z) and find 1.6, then move to right to the column .07. You should

find that the area of a z score of 1.67 is .9525.

238

2

Step 3: Interpretation

It can be interpreted that the probability of drawing a score larger than 75 is 4.75% (1

– .9525 = .0475).

In-Class Exercise 5.2: What is the probability that one may draw an inmate with age

beyond 32 from a jail with a mean age of 25 and a standard deviation of 5?

Step 1: Calculate the z score

z = x – x-bar/s = (32 – 25)/5 = 1.40

Step 2: Find the area of certain z score by using Table A-2 (p. 238)

Go to the first column (z) and find 1.4, then move to right to the column .00. You should

find that the area of a z score of 1.4 is .9192.

238

Step 3: Interpretation

It can be interpreted that the probability that one may draw an inmate with age beyond

32 is 8.08% (1 – .9192 = .0808) from a jail with a mean age of 25 and a standard

deviation of 5.

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C. Central Limit Theorem

1. A statistical proposition to the effect that, the large a sample size, the more closely the

sampling distribution of the mean will approach a normal distribution.

2. This is true even if the population from which the sample is drawn is not normally

distributed. A sample size of 30 or more will usually result in a sampling distribution of

the mean that is very close to a normal distribution.

3. The central limit theorem explains why sampling error (the difference between a

population parameter and a sample statistic used to estimate that parameter) is smaller

with a large sample than it is with a small sample.

D. Establishing Confidence Interval

1. The Confidence Interval (CI) is an estimate of the population value based upon our

sample results. It is our first brush with inferential statistics.

2. Three CI level are commonly being used: 99%, 95%, and 90%. The formulas are listed as

below. In criminal justice research, 95% is the most common one. It indicates that we are

95 percent “confidence” about certain issue.

CI 99% = x-bar + 2.575 (σ /√n)

CI 95% = x-bar + 1.96 (σ /√n)

CI 90% = x-bar + 1.645 (σ /√n)

In-Class Exercise 5.3: In a random sample of 100 observations from a population whose

standard deviation (σ) is 10, a statistician calculated the sample mean as x-bar = 75. Find

the CI 95% of the population mean. Interpret what the interval estimate tells you.

The innermost parentheses are

calculated first, you then follow

the normal order of operations

as laid out by PEMDAS

Step 1: Calculate CI 95%

CI 95% = x-bar + 1.96 (σ /√n)

= 75 + 1.96 (10/√100)

= 75 + 1.96 (1)

= 75 + 1.96

= (73.04, 76.96)

75–1.96 = 73.04, 75+1.96 = 76.96

Step 2: Interpretation

It is with 95% of confidence that the population mean falls between 73.04 and 76.96

(Don’t say “there is a 95% probability that the population mean lies between 73.04 and

79.96,” because the population mean is a fixed but unknown quantity).

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