Transcript Random Variable

Welcome to MM207 - Statistics!
Unit 4 Seminar
Good Evening Everyone!
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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. (500.12 pounds, 534.1534 pounds,
etc…
Probability Distribution
• 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
Score, x
Probability, P(x)
1
0.16
2
0.22
3
0.28
4
0.20
5
0.14
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.
Mean and Standard Deviation: Example 2
(page 199)
x
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
1
2.94
∑=
μ = ∑ x P(x) = 2.94
σ = √(x-μ)2 P(x) = √ 1.616 = 1.271219 ≈ 1.3
1.616
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) = ?
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
Using the Binomial Table
n = 10
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
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)
nC x
is the binomial coefficient give by nCx = n! / [x! (n-x)!]
Remember 4! = 4*3*2*1 = 24 and is called factorial notation.
Using the Binomial Formula
n = 10
p = 0.4
x=6
Find P(x = 6)
P(x) = nCx px qn-x
nCx = 10C6 = 210
px = 0.46 = 0.004096
qn-x = 0.610-6 = 0.64 =0.1296
210* 0.004096 * 0.1296 = 0.111476736 ≈ 0.111
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) = 0.111476736 ≈ 0.111
Finding Cumulative Probabilities
n = 10
p = 0.4
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)
= 0.006 + 0.040 + 0.121 + 0.215 + 0.251 + 0.201 + 0.111 = 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 – [0.042 + 0.011 + 0.002 + 0.000] = 1 – 0.055 = 0.945
Finding Cumulative Probabilities con’t
n = 10
p = 0.4
x≤6
Find P(x ≤ 6)
Use Excel
• Number_s: Enter 6
• Trials: Enter 10
• Probability: Enter .4
• Cumulative: True
P(x ≤ 6) = 0.945238118 ≈ 0.945
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)
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 ≈ 1.549
Variance
σ2 = npq = 2.4