MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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![Example 2: The monthly demand for kamote and kamoteng kahoy is known to have the following probability
distribution.
1
2
3
4
5
6
Demand
Probability
0.10
0.15
0.20
0.25
0.20
0.10
Find the mean and variance for the demand of kamote and kamoteng kahay
Solution:
a. In computing for the mean µ, we will represent our solution in a tabular form
X
P(x)
xP(x)
1
0.10
0.10
2
0.15
0.30
3
0. 20
0.60
4
0.25
1.00
5
0. 20
1.00
6
0.10
0.60
xP(x) = 3.6
So, the mean μ = 3.6
b. In computing for the variance and standard deviation, we will represent our solution in a tabular form
X
P(x)
xP(x)
X-μ
(x-μ)²
(x-μ)² P(x)
1
0.10
0. 10
-2.6
6.76
0.676
2
0.15
0.30
-1.6
2.56
0.384
3
0.20
0.60
-0.6
0.36
0.072
4
0.25
1.00
0.4
0. 16
0.04
5
0.20
1.00
1.4
1.96
0.392
6
0.10
0.60
2.4
5. 76
0.576
xP(x) = 3.6
Σ[(x-μ) P(x)] = 2.14
*Using the formula in computing for the variance o², we get Σ[(x − µ)² P(x)] =2. 14
*Using the formula [(x − µ)² P(x)] in computing for the standard deviation o, we get
√2.14 = 1.4629
Practice Exercise
1. Let X represent the number of times a grade 10 student visits an Internet Café to submit his outputs
in their Google classroom in a one-week period. The probability distribution of X is shown below.
X
P(x)
0
0.06
1
0.18
2
0.40
3
0.28
4
0.08
Find the mean and the variance.
2. The weekly load allowance of a grade 11 STEM student in an online class in Calculus is shown
below together with the associated probabilities.
X
2
3
4
5
1
0.05
P(x)
0.15
0.20
0.40
0.20
Find the mean and the variance.](https://content.bartleby.com/qna-images/question/9478e1a7-af69-4fc6-9548-6390cd64c84b/c9bafc52-2f8b-45f6-bf03-00fb6991ba9d/0pp299_processed.jpeg)
Transcribed Image Text:Example 2: The monthly demand for kamote and kamoteng kahoy is known to have the following probability
distribution.
1
2
3
4
5
6
Demand
Probability
0.10
0.15
0.20
0.25
0.20
0.10
Find the mean and variance for the demand of kamote and kamoteng kahay
Solution:
a. In computing for the mean µ, we will represent our solution in a tabular form
X
P(x)
xP(x)
1
0.10
0.10
2
0.15
0.30
3
0. 20
0.60
4
0.25
1.00
5
0. 20
1.00
6
0.10
0.60
xP(x) = 3.6
So, the mean μ = 3.6
b. In computing for the variance and standard deviation, we will represent our solution in a tabular form
X
P(x)
xP(x)
X-μ
(x-μ)²
(x-μ)² P(x)
1
0.10
0. 10
-2.6
6.76
0.676
2
0.15
0.30
-1.6
2.56
0.384
3
0.20
0.60
-0.6
0.36
0.072
4
0.25
1.00
0.4
0. 16
0.04
5
0.20
1.00
1.4
1.96
0.392
6
0.10
0.60
2.4
5. 76
0.576
xP(x) = 3.6
Σ[(x-μ) P(x)] = 2.14
*Using the formula in computing for the variance o², we get Σ[(x − µ)² P(x)] =2. 14
*Using the formula [(x − µ)² P(x)] in computing for the standard deviation o, we get
√2.14 = 1.4629
Practice Exercise
1. Let X represent the number of times a grade 10 student visits an Internet Café to submit his outputs
in their Google classroom in a one-week period. The probability distribution of X is shown below.
X
P(x)
0
0.06
1
0.18
2
0.40
3
0.28
4
0.08
Find the mean and the variance.
2. The weekly load allowance of a grade 11 STEM student in an online class in Calculus is shown
below together with the associated probabilities.
X
2
3
4
5
1
0.05
P(x)
0.15
0.20
0.40
0.20
Find the mean and the variance.
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