Name: Bualan, Eliza Jiann B. and Aplicador, Dipper Ore
Filename: Last Name_ASS3 (BY-PAIR)
ASSESSMENT 3
Instruction: Solve the mean, variance, and standard deviation of the following probability
distributions.
1.
𝑿
𝑷(𝑿)
𝑿 ∙ 𝑷(𝑿)
𝑿𝟐 ∙ 𝑷(𝑿)
1
1
1
1
or 0.14
1 ∙ = 0.14
12 ∙ = 0.14
7
7
7
2
2
2
6
or 0.29
2
6 ∙ = 1.71
6 ∙ = 10.29
7
7
7
2
2
2
11
or 0.29
11 ∙ = 3.14
112 ∙ = 34.57
7
7
7
1
1
1
16
or 0.14
16 ∙ = 2.28
162 ∙ = 36.57
7
7
7
1
1
1
21
or 0.14
2
21 ∙ = 3
21 ∙ = 63
7
7
7
Show your Solution:
Solving for the Mean of the Probability Distribution:
𝜇 = ∑ 𝑋 ∙ 𝑃(𝑋) = 0.14 + 1.71 + 3.14 + 2.28 + 3 = 10.27
Therefore, the mean of the probability distribution is 9.25 or 9.
Solving for Sum of 𝑿𝟐 ∙ 𝑷(𝑿) :
∑ 𝑋 2 ∙ 𝑃(𝑋) = 0.14 + 10.29 + 34.57 + 36.57 + 63 = 144.57
Solving for Variance:
𝜎 2 = ∑ 𝑋 2 ∙ 𝑃(𝑋) − 𝜇2
𝜎 2 = 144.57 − (10.27)2
𝜎 2 = 144.57 − 105.47
𝜎 2 = 39.10
Solving for Standard Deviation:
𝜎 = √39.10
𝜎 = 6.25
2.
The number of smartphones sold per day at a local store, along with its corresponding
probabilities, is shown in the table. Find the mean, variance, and standard deviation of the
distribution.
Number of Smartphones Sold
X
0
2
4
6
8
Probability
𝑃(𝑋)
0.10
0.20
0.40
0.10
0.20
Show your solution:
Number of
Smartphones Sold
X
0
2
4
6
8
Probability
𝑃(𝑋)
𝑿 ∙ 𝑷(𝑿)
𝑿𝟐 ∙ 𝑷(𝑿)
0.10
0.20
0.40
0.10
0.20
0 ∙ 0.10 = 0
2 ∙ 0.20 = 0.40
4 ∙ 0.40 = 1.60
6 ∙ 0.10 = 0.60
8 ∙ 0.20 = 1.60
02 ∙ 0.10 = 0
22 ∙ 0.20 = 0.80
42 ∙ 0.40 = 6.40
62 ∙ 0.10 = 3.60
82 ∙ 0.20 = 12.80
Solving for the Mean of the Probability Distribution:
𝜇 = ∑ 𝑋 ∙ 𝑃(𝑋) = 0 + 0.40 + 1.60 + 0.60 + 1.60 = 4.20
Therefore, the mean of the probability distribution is 4.20 or 4.
This implies that the average number of smartphones that will be sold per day is 4.20 or
4 smartphones.
Solving for Sum of 𝑿𝟐 ∙ 𝑷(𝑿) :
∑ 𝑋 2 ∙ 𝑃(𝑋) = 0 + 0.80 + 6.40 + 3.60 + 12.80 = 23.6
Solving for Variance:
𝜎 2 = ∑ 𝑋 2 ∙ 𝑃(𝑋) − 𝜇2
𝜎 2 = 23.60 − (4.20)2
𝜎 2 = 23.60 − 17.64
𝜎 2 = 5.96
Solving for Standard Deviation:
𝜎 = √5.96
𝜎 = 2.44