# 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
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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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