1. Welcome to MM207 - Statistics!
Unit 4 Seminar
Good Evening Everyone!
To resize your pods:
Place your mouse here.
Left mouse click and hold.
Drag to the right to enlarge the pod.
To maximize chat, minimize roster by clicking here

2. Definitions
Statistical Experiment: Any process by which we obtain measurements or data.
In Unit 3 seminar we discussed roulette. Spinning the roulette wheel is a statistical experiment.
Random Variable: A random variable is the outcome of a statistical experiment. We don’t know that this outcome will be before conducting the experiment
Discrete random variable: The possible values of the experiment take on a countable number of results.
For the roulette case the there were 38 possible results.
Continuous random variable: The possible values of the experiment are infinite.
For example, measure the weight of 1 year old cows is continuous. The number of possibilities are uncountable. ( pounds, pounds, etc…

3. Probability Distribution
Score, x
Probability, P(x) 1
0.16 2
0.22 3
0.28 4
0.20 5
0.14
Probability distribution is the assignment of probabilities to specific values for a random variable or to a range of values for the random variable
Plain English: Each random variable (outcome) from a random experiment has a particular probability of occurring.
See page 196 Example 2

4. Probability distribution properties
Mean: This is the expected value of a probability distribution. This is the outcome about which the distribution is centered. μ = ∑ x P(x)
Standard deviation: This is the spread of the data around the expected value (mean) σ = √ ∑ (x – μ)2 P(x)
How do you use the equations? Let’s work an example.

5. Mean and Standard Deviation: Example 2 (page 199)
P(x)
x P(x)
x - μ
(x-μ)2
(x-μ)2 P(x) 1
0.16
1 * 0.16 = 0.16
1 – 2.94 = -1.94
(-1.94)2 = 3.764
3.764 * 0.16 = 0.602 2
0.22
0.44
-0.94
0.884
0.194 3
0.28
0.84
0.06
0.004
0.001 4
0.20
0.80
1.06
1.124
0.225 5
0.14
0.70
2.06
4.244
0.594
∑ =
2.94
1.616
μ = ∑ x P(x) = 2.94
σ = √(x-μ)2 P(x) = √ = ≈ 1.3

6. Features of the Binomial Experiment
Fixed number of trials denoted by n
n trials are independent and performed under identical conditions
Each trial has only two outcomes: success denoted by S and failure denoted by F
For each trial the probability of success is the same and denoted by p. The probability of failure is denote by q and q = 1 - p)
The central problem is to determine the probability of x successes out of n trials. P(x) = ?

7. Example n = 10 p = 0.4 x = 6 Find P(x = 6)
Using the binomial table (Table 2 A8-A10)
Using the binomial formula
Using Excel

8. Using the Binomial Table
p = 0.4
x = 6
Find P(x = 6)
Find the block for 10
Find the row for 6
Find the column for 0.4
P(x = 6) = 0.111

9. Using the Binomial Formula
P(x) = nCx px qn-x
x = number of successes
n = number of trials
p = probability of one success
q = probability of one failure (1 – p)
nCx is the binomial coefficient give by nCx = n! / [x! (n-x)!]
Remember 4! = 4*3*2*1 = 24 and is called factorial notation.

10. Using the Binomial Formula
p = 0.4
x = 6
Find P(x = 6)
P(x) = nCx px qn-x
nCx = 10C6 = 210
px = 0.46 =
qn-x = = 0.64 =0.1296
210* * = ≈ 0.111

11. Using Excel n = 10 p = 0.4 x = 6 Find P(x = 6)
Click on the cell where you want the answer.
Under fx, find BINOMDIST
Number_s: Enter 6
Trials: Enter 10
Probability: Enter 0.4
Cumulative: False
P(x = 6) = ≈ 0.111

12. Finding Cumulative Probabilities
x ≤ 6
Find P(x ≤ 6)
Find each probability using the binomial table or the formula.
P(x ≤ 6) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6)
= = 0.945
Use the complement
P(x ≤ 6) = 1 – P(x > 6) = 1 – [P(x = 7) + P(x = 8) + P(x = 9) + P(x = 10)]
= 1 – [ ] = 1 – = 0.945

13. Finding Cumulative Probabilities con’t
x ≤ 6
Find P(x ≤ 6)
Use Excel
Number_s: Enter 6
Trials: Enter 10
Probability: Enter .4
Cumulative: True
P(x ≤ 6) = ≈ 0.945

14. Mean and Standard Deviation of the binomial probability distribution
Mean or expected number of success μ = np
Standard deviation σ = √ npq
Where: n = number of trials p = probability of success q = probability of failure (q = 1 – p)

15. Computing the Mean, Standard Deviation, and Variance for a Binomial Distribution
n = 10 p = 0.4 Mean μ = np μ = 10 * 0.4 μ = 4 Standard deviation σ = √ npq σ = √ 10 * 0.4 * 0.6 σ = √ 2.4 ≈ Variance σ2 = npq = 2.4