Question

1. A department store manager has monitored the number of complaints received per week about poor service. The probabilities for numbers of complaints in a week, established by this review, are shown in the table.

Number of complaints	0	1	2	3	4	5
Probability	0.16	0.26	0.35	0.10	0.08	0.05

What is the standard deviation of complaints received per week?

Please round your answer to the nearest hundredth.

2. A department store manager has monitored the number of complaints received per week about poor service. The probabilities for numbers of complaints in a week, established by this review, are shown in the table.

Number of complaints	0	1	2	3	4	5
Probability	0.10	0.29	0.33	0.13	0.07	0.08

What is the mean of complaints received per week?

Please round your answer to the nearest hundredth.

Note that the correct answer will be evaluated based on the full-precision result you would obtain using Excel.

Answer 1

Solution :

(1)

X	P(X)	X.P(X)	X ² .P(X)
0	0.16	0	0
1	0.26	0.26	0.26
2	0.35	0.7	1.4
3	0.10	0.3	0.9
4	0.08	0.32	1.28
5	0.05	0.25	1.25
Total	1	1.83	5.09

Mean μ =  Σ X.P(X) = 1.83

Standard deviation σ = √ ( Σ X ² .P(X) - μ ² )

= √ ( 5.09 - (1.83) ² )

= √ 1.7411

= 1.31950

= 1.32

standard deviation of complaints received per week = 1.32

(2)
E(X) = Σ X * P(X)
For discrete probability distribution,

= 0 * 0.10+ 1 * 0.29 + 2 * 0.33 + 3 * 0.13 + 4 * 0.07 + 5 * 0.08

= 2.02

The mean of the complaints received per week = 2.02

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