Math 1600 Basic Probability and Statistics
In this word document you have one example of what is expected when you have to compute
the mean and the variance of a discrete probability distribution
In page number 3 you have the modified exercise 4 for quiz No 8
Example
1.
Defective Transistors From past experience, a company has found that in cartons of
transistors, 92% contain no defective transistors, 3% contain one defective transistor, 3%
contain two defective transistors, and 2% contain three defective transistors.
Find the mean, variance, and standard deviation for the defective transistors. Construct
the probability histogram
Solution
Fill in the values in the table below
X
P(X)
X*P(X)
(X - µ)2
(1)
(2)
(3)
(4)
(X - µ)2 * P(X)
(5)
X2*P(X)
(6)
0
0.92
0.00
(0 – 0.15)2
0.000207
0.00
1
0.03
0.03
(1 – 0.15)2
0.029107
0.03
2
0.03
0.06
2 – 0.15)2
0.118207
0.12
3
0.02
0.06
(3 – 0.15)2
0.178205
0.18
1.00
0.15
(0 – 0.15)2
0.325725
0.33
Mean µ = 0.15  0.2 (rounded to one decimal)
Variance 2 = 0.325725  0.3 (rounded to one decimal)
Standard Deviation  = 0.570723  0.6 (rounded to one decimal)
The variance can be computed by using the formula
Variance 2 =  X2 * P(X) - µ2
Therefore 2 = 0.33 – 0.152
= 0.33 – 0.0225 = 0.3075  0.3 (rounded to one decimal)
1
Note: Create a table similar to the one above when you have to compute the mean,
variance and standard deviation for a discrete probability distribution
P(X)
0
1
2
3
X
2
Modification of Quiz8 Section 5.1
Consider the experiment of rolling two dice. Let X be the sum of the number of dots on the
upper faces.
(a) Construct the probability distribution for variable X
(b) Construct the probability histogram.
(c) Find the mean, variance, and standard deviation, for the variance and the standard deviation
use both formulas.
X
(1)
2
P(X)
(2)
1/36
X*P(X)
(3)
(X - µ)2
(4)
(X - µ)2 * P(X)
(5)
X2*P(X)
(6)
3
4
5
6
7
8
9
10
11
12
3