5.2 - Triola textbook 6.2 - Sullivan textbook

1. Elementary Statistics
Chapter 5:
Discrete Probability
Distribution
5.2 Binomial
Probability
Distributions
1

2. Chapter 5: Discrete Probability Distribution
5.1 Probability Distributions
5.2 Binomial Probability Distributions
2
Objectives:
• Construct a probability distribution for a random variable.
• Find the mean, variance, standard deviation, and expected value for a discrete
random variable.
• Find the exact probability for X successes in n trials of a binomial experiment.
• Find the mean, variance, and standard deviation for the variable of a binomial
distribution.

3. Key Concept: Random variable & Probability Distribution.
A probability histogram is a graph that visually depicts a probability distribution.
Find the important parameters of mean, standard deviation, and variance for a probability distribution.
Determine whether outcomes are significant (significantly low or significantly high).
Random Variable: A random variable is a variable (typically represented by x) that has a single
numerical value, determined by chance, for each outcome of a procedure.
Probability Distribution: A probability distribution is a description that gives the probability for each
value of the random variable. It is often expressed in the format of a table, formula, or graph. A discrete
probability distribution consists of the values a random variable can assume and the corresponding
probabilities of the values.
Discrete Random Variable (DRV): A discrete random variable has a collection of values that is finite
or countable. (If there are infinitely many values, the number of values is countable if it is possible to
count them individually, such as the number of tosses of a coin before getting heads.)
Continuous Random Variable: A continuous random variable has infinitely many values, and the
collection of values is not countable. (That is, it is impossible to count the individual items because at
least some of them are on a continuous scale, such as body temperatures.) 3
Recall: 5.1 Probability Distributions

4. Probability Distribution Requirements (3):
1. There is a numerical (not categorical) random variable (RV) x, and its values are
associated with corresponding probabilities.
2. ∑P(x) = 1 where x assumes all possible values. (The sum of all probabilities must be 1,
but sums such as 0.999 or 1.001 are acceptable because they result from rounding errors.)
3. 0 ≤ P(x) ≤ 1 for every individual value of the random variable x. (That is, each
probability value must be between 0 and 1 inclusive.)
4
Recall: 5.1 Probability Distributions
Mean: ( ) ( )E x x p x   
Parameters of a Probability Distribution
2 2 2
SD: ( ) ( ) ( )x p x x p x      
2 2 2 2
Variance: ( ) ( ) ( )x p x x p x         

5. Key Concept:
The binomial probability distribution, its mean and standard deviation &
interpreting probability values to determine whether events are significantly low or
significantly high.
Binomial Experiment and P. D. (properties)
1. n identical ( fixed # of) trials (Each repetition of the experiment)
2. Each has only 2 categories of outcomes
3. Probability stays constant
4. Independent trials
If a procedure satisfies these 4 requirements, the distribution of the random
variable x is called a Binomial Probability Distribution.
(It is the Probability Distribution for the number of successes in a sequence of
Bernoulli trials.)
5.2 Binomial Probability Distributions
5

6. p = probability of success, q = 1 − p = probability of failure
Make sure that the values of p and x refer to the same category called a success (x and p are consistent).
x: A specific number of successes in n trials: 0 ≤ 𝑥(𝑎 𝑤ℎ𝑜𝑙𝑒 #) ≤ 𝑛
P(x): probability of getting exactly x successes among the n trials
The word success as used here is arbitrary and does not necessarily represent something good. Either of the two
possible categories may be called the success S as long as its probability is identified as p.
Parameters: n & p
Binomial Coefficient (# of outcomes containing exactly x successes):
Each trial is known as a Bernoulli trial (named after Swiss mathematician Jacob Bernoulli).
Notes: If the probability of x success in n trials is less than 0.05, it is considered unusual.
5% Guideline for Cumbersome Calculations: When sampling without replacement and the sample size is no
more than 5% of the size of the population, treat the selections as being independent (even though they are
actually dependent).
Warning: If we toss a coin 1000 times, P (H=501), or any specific number of heads is very small, however, we
expect P (at least 501 heads) to be high.
5.2 Binomial Probability Distributions ( ) n x n x
xp x C p q 

n
xC
6

7. The Probability Histogram is Symmetric if: p = 0.5
The Probability Histogram is Right Skewed: p < 0.5
The Probability Histogram is Left Skewed: p > 0.5
In a binomial experiment, the probability of exactly X successes in n trials is
5.2 Binomial Probability Distributions
( )
(1 )
n x n x
xp x C p q
np
npq
np p






 
!
( )
( )! !
x n xn
P x p q
n x x

  

number of possible probability of a
desired outcomes desired outcome
or
( ) X n X
n xP x C p q 
  
7
2
Variance:  npq
Mean:  np
Standard Deviation:  npq

8. When an adult is randomly selected (with replacement), there is a
0.85 probability that this person knows what Twitter is. Suppose that
we want to find the probability that exactly three of five randomly
selected adults know what Twitter is.
a. Does this procedure result in a binomial distribution?
b. If this procedure does result in a binomial distribution, identify
the values of n, x, p, and q.
Example 1
a. Solution: Yes
1. Number of trials: n = 5 (fixed)
3. 2 categories: The selected person knows what Twitter is or does not.
4. Constant probability: For each randomly selected adult: x = 3, p =
0.85 & q= 0.15
5. Independent: Different adults (All 5 trials are independent because
the probability of any adult knowing Twitter is not affected by
results from other selected adults.)
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Binomial Experiment and P. D.
(properties)
1. n identical ( fixed # of) trials (Each
repetition of the experiment)
2. Each has only 2 categories of
outcomes
3. Probability stays constant
4. Independent trials
!
( ) , ,
( )! !
x n xn
p x p q np npq
n x x
 
  

b. n = 5, x = 3, p = 0.85, q = 1 − 𝑝 = 0.15
3 5 35!
(3) 0.85 0.15
(5 3)!3!
P 
  

10(0.614125)(0.0225)
0.1382
TI Calculator:
Binomial Distribution
1. 2nd + VARS
2. binompdf(
3. Enter: n, p, x
4. Enter
5. If you enter n, p only
6. Gives all
probabilities from 0
to n
7. If using Binomcdf(
8. Gives sum of the
probabilities from 0
to x.

9. Assume one out of five people has visited a doctor in any given month. If 10
people are selected at random,
a. find the probability that exactly 3 have visited a doctor last month.
b. find the probability that at most 2 have visited a doctor last month. (at Home)
c. find the probability that at least 3 have visited a doctor last month. (at Home)
9
Example 2
Solution
1
5: 10, , 3BD n p x  
3 7
10! 1 4
(3)
7!3! 5 5
P
   
    
   
0.20133
!
( )
( )! !
x n xn
P x p q
n x x

  

1010!
( ) (1/5) (4/5)
!(10 )!
x x
P x
x x

  

q = 1 − 𝑝 =
4
5
!
( ) , ,
( )! !
x n xn
p x p q np npq
n x x
 
  

TI Calculator:
Binomial Distribution
1. 2nd + VARS
2. binompdf(
3. Enter: n, p, x
4. Enter
5. If you enter n, p only
6. Gives all
probabilities from 0
to n
7. If using Binomcdf(
8. Gives sum of the
probabilities from 0
to x.

10. A survey found that 30% of teenage consumers receive their spending
money from part-time jobs. If 5 teenagers are selected at random,
a. find the probability that at least 3 of them will have part-time jobs.
b. find the probability that at most 2 of them will have part-time jobs.
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Example 3
Solution : 5, 0.30,"at least 3" 3,4,5BD n p X   
3 25!
(3) 0.3 0.7
2!3!
P   
4 15!
(4) 0.3 0.7
1!4!
P   
0.1323 ( 3)
0.16308
P x
Add
 

0.02835
5 05!
(5) 0.03 0.7
0!5!
P    0.00243
(at least 3) ). ( 3a P P x 
( 2)
1 ( )
.
3
b P x
P x
 
 
1 0.16308
0.83692
 

!
( ) , ,
( )! !
x n xn
p x p q np npq
n x x
 
  

TI Calculator:
Binomial Distribution
1. 2nd + VARS
2. binompdf(
3. Enter: n, p, x
4. Enter
5. If you enter n, p only
6. Gives all
probabilities from 0
to n
7. If using Binomcdf(
8. Gives sum of the
probabilities from 0
to x.

11. It is reported that 2% of all American births result in twins. If a random sample of 8000 births is taken,
a. find the mean, variance, and standard deviation of the number of births that would result in twins.
b. Use the range rule of thumb to find the values separating the numbers of twins that are
significantly low or significantly high.
c. Is the result of 200 twins in 8000 births significantly high?
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Example 4
BD, n = 8000, p = 0.02Solution: a.
np 
2
npq 
npq 
8000(0.02) 160 
8000(0.02)(0.98)
156.8 12.522 
156.8
b. 𝜇 ± 2𝜎 →
c. The result of 200 twins is
significantly high because it is
greater than the value of 185.044
twins found in part (b).
!
( ) , ,
( )! !
x n xn
p x p q np npq
n x x
 
  

µ − 2σ = 160 − 2(12.522)
= 134.956 Twins
µ + 2σ = 160 + 2(12.522)
= 185.044 Twins

12. McDonald’s has a 95% recognition rate. A special focus group consists of 12 randomly selected adults.
a. find the mean, variance, and standard deviation.
b. Use the range rule of thumb to find the minimum and maximum usual number of people who
would recognize McDonald’s.
c. Is the result of 12 people in a group of 12 recognizing the brand name of McDonald’s unusual?
12
Example 5
Solution: a.
np 
2
npq 
npq 
12(0.95) 11.4 
1290.95)(0.05)
0.57 0.754983
0.57
µ − 2σ = 11.4 − 2(0.755)
= 9.89 people
µ + 2σ = 11.4 + 2(0.755)
= 12.91 people
c. If a particular group of 12
people had all 12 recognize the
brand name of McDonald’s, that
would not be unusual.
BD, n = 12, p = 0.95
!
( ) , ,
( )! !
x n xn
p x p q np npq
n x x
 
  

b. 𝜇 ± 2𝜎 →

13. Range Rule of Thumb
Significantly low values ≤ (µ − 2σ)
Significantly high values ≥ (µ + 2σ)
Values not significant: Between (µ − 2σ) and (µ + 2σ)
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