Transcript Document

COURSE: JUST 3900
INTRODUCTORY STATISTICS
FOR CRIMINAL JUSTICE
Chapter 6: Probability
Peer Tutor Slides
Instructor:
Mr. Ethan W. Cooper, Lead Tutor
© 2013 - - PLEASE DO NOT CITE, QUOTE, OR REPRODUCE WITHOUT THE
WRITTEN PERMISSION OF THE AUTHOR. FOR PERMISSION OR QUESTIONS,
PLEASE EMAIL MR. COOPER AT THE FOLLWING: [email protected]
Key Terms and Formulas: Don’t
Forget Notecards





Probability (p. 165)
Random Sample (p. 167)
Independent Random Sample (p. 167)
Binomial Distribution (p. 185)
Binomial Formulas:

Mean: 𝜇 = 𝑝𝑛
Standard Deviation: 𝜎 =

z-Score: 𝑧 =

𝑋−𝜇
𝜎
=
𝑋−𝑝𝑛
𝑛𝑝𝑞
𝑛𝑝𝑞
Random Sampling

Question 1: A survey of students in a criminal justice
class revealed that there are 17 males and 8 females.
Of the 17 males, only 5 had no brothers or sisters, and 4
of the females were also the only child in the household.
If a student is randomly selected from this class,
a)
b)
c)
What is the probability of obtaining a male?
What is the probability of selecting a student who has at least
one brother or sister?
What is the probability of selecting a female who has no
siblings?
Random Sampling

Question 1 Answer:
a)
b)
c)
p = 17/25 = 0.68
p = 16/25 = 0.64
p = 4/25 = 0.16
Random Sampling With and
Without Replacement

Question 2: A jar contains 25 red marbles and 15 blue
marbles.
a)
b)
c)
If you randomly select 1 marble from the jar, what is the
probability of obtaining a red marble?
If you take a random sample of n = 3 marbles from the jar and
the first two marbles are both blue, what is the probability that
the third marble will be red?
If you take a sample (without replacement) of n = 3 marbles
from the jar and the first two marbles are both red, what is the
probability that the third marble will be blue?
Random Sampling With and
Without Replacement

Question 2 Answer:
a)
b)
c)
p = 25/40 = 0.625
p = 25/40 = 0.625
p = 15/38 = 0.395
Remember
that random
sampling
requires
sampling with
replacement.
Here, we did not replace the first two red
marbles that were drawn.
Probability and Frequency
Distributions

Question 3: Consider the following frequency distribution
histogram for a population that consists of N = 8 scores.
Suppose you take a random sample of one score from
this set.
a)
b)
c)
The probability that this score is equal to 4 is p(X = 4) = ____
The probability that this score is less than 4 is p(X < 4) = ____
The probability that this score is greater than 4 is p(X > 4) = __
Probability and Frequency
Distributions

Question 3 Answer:
a)
b)
c)
p(X = 4) = 4/8 = 0.500
p(X < 4) = 3/8 = 0.375
p(X > 4) = 1/8 = 0.125
Properties of the Normal Curve

Question 4: The scores for students on Dr. Anderson’s
research methods test had a mean of µ = 80 and a
standard deviation of σ = 5. Use the figure on the next
slide to answer the following questions.
a)
b)
c)
A score of 65 is ___ standard deviations below the mean, while
a score of 95 is ___ standard deviations above the mean. This
means that the percentage of students with scores between 65
and 95 is ___.
A score of 90 is ___ standard deviations above the mean. As a
result, the percentage of students with scores below 90 is ___.
You can infer that 84.13% of students have scores above ___.
Properties of the Normal Curve
Properties of the Normal Curve

Question 4 Answer:
a)
A score of 65 is _3_ standard deviations below the mean, while
a score of 95 is _3_ standard deviations above the mean. This
means that the percentage of students with scores between 65
and 95 is _99.74%.
Add the percentages
between -3σ and +3σ.
65
2.15 + 13.59 + 34.13 +
34.13 + 13.59 + 2.15 =
99.74%
70
75
80
85
90
95
Properties of the Normal Curve

Question 4 Answer:
A score of 90 is _2_ standard deviations above the mean. As a
result, the percentage of students with scores below 90 is
97.72%.
b)
13.59 + 34.13 + 34.13 +
13.59 + 2.15 + 0.13 =
97.72%
Score of 90.
or
100 – 2.15 – 0.13 = 97.72%
65
70
75
80
85
90
95
Properties of the Normal Curve

Question 4 Answer:
c)
You can infer that 84.13% of students have scores above
_75_.
Start from 100 and subtract
until you reach 84.13%.
84.13 % of students
scored above a 75.
100 – 0.13 – 2.15 – 13.59 –
34.13 - 34.13 = 84.13%
65
70
75
80
85
90
95
The Unit Normal Table

Question 5: Use the unit normal table (p. 699) to find the
proportion of a normal distribution that corresponds to
each of the following sections: (Hint: Make a sketch)
a)
b)
c)
d)
z < 0.28
z > 0.84
z > -1.25
z < -1.85
The Unit Normal Table

Question 5 Answer:
a)
b)
c)
d)
p = 0.6103
p = 0.2005
p = 0.8944
p = 0.0322
z < 0.28
z > 0.84
z > -1.25
z < -1.85
Binomial Data

Question 6: In the game Rock-Paper-Scissors, the
probability that both players will select the same
response and tie is p = 1/3, and the probability that they
will pick different responses is q = 2/3. If two people play
72 rounds of the game and choose there responses
randomly, what is the probability that they will choose the
same response (tie) more than 28 times?
Binomial Data
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Question 6 Answer:
1)
2)
Find µ and σ.
1)
𝜇 = 𝑝𝑛 =
2)
𝜎=
𝑛𝑝𝑞 =
∗ (72) = 24
1
3
72 ∗
2
∗ ( 3) =
16 = 4
Find z.
1)
3)
1
3
𝑧=
𝑋−𝜇
𝜎
=
𝑋−𝑝𝑛
𝑛𝑝𝑞
=
28.5−24
4
=
4.5
4
= 1.13
Use unit normal table.
1)
p(X > 28.5) = p(z > 1.13) = 0.1292.
Don’t forget real limits.
We’re looking for the probability
Of MORE than 28. Hence, we
Use the upper real limit of 28.5.
Binomial Data

Question 7: If you toss a balanced coin 36 times, you
would expect, on the average, to get 18 heads and 18
tails. What is the probability of obtaining exactly 18
heads in 36 tosses?
Binomial Data

Question 7 Answer:
1)
2)
3)
Find µ and σ.
1)
𝜇 = 𝑝𝑛 = 0.5 ∗ (36) = 18
2)
𝜎=
𝑛𝑝𝑞 =
36 ∗ 0.5 ∗ (0.5) =
9=3
Find z.
1)
𝑧=
𝑋−𝜇
𝜎
=
𝑋−𝑝𝑛
𝑛𝑝𝑞
=
17.5−18
3
=
−0.50
3
2)
𝑧=
𝑋−𝜇
𝜎
=
𝑋−𝑝𝑛
𝑛𝑝𝑞
=
18.5−18
3
=
0.50
3
= −0.17
= 0.17
Don’t forget to use real limits.
X = 18 spans the interval from
17.5 to 18.5. Therefore, we
have to find the z-score for
both the upper and lower real
limits.
Use the unit normal table to find the proportion between z and
the mean for each z-value.
1)
p(X = 18) = p(z = ±0.17) = 0.0675 + 0.0675 = 0.1350
Frequently Asked Questions
FAQs

How does one know if a question is asking for random
sampling with replacement or random sampling without
replacement?


Unless the question specifically states that the sample was taken
without replacement, always assume that the sample took place
with replacement.
Remember the requirements for random samples:
1)
2)
Every individual in the population must have an equal chance of
being selected.
The probability of being selected must stay constant from one
selection to the next if more than one individual is being selected.
Frequently Asked Questions
FAQs

A few things to keep in mind about binomial distributions:
A score of 1 spans
From 0.5 to 1.5.

A score of 10 spans
from 9.5 to 10.5.
Binomial distributions work with discrete variables, but the
normal distribution is continuous. However, binomial distributions
approximate the normal distribution when pn and qn are both
greater than or equal to 10. But keep in mind that each X value
actually corresponds to bar in the histogram. Therefore, a
score of 10 is bounded by the real limits of 9.5 and 10.5.