Question

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 probability of between 3 and 5 (inclusive) complaints received per week?

Please specify your answer in decimal terms and round your answer to the nearest hundredth (e.g., enter 12 percent as 0.12).

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

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	0.20	0.25	0.26	0.14	0.10	0.05

What is the mode of complaints received per week?

Please round your answer to the nearest integer.

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

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	0.20	0.25	0.26	0.14	0.10	0.05

What is the variance 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

Number of Complaints (x)	Probability(p)
0	0.16
1	0.26
2	0.35
3	0.10
4	0.08
5	0.05
Total	1

The probability of between 3 and 5 (inclusive) complaints received per week = 0.10+ 0.08+0.05 = 0.23

-----------------------------------------------------------------------------------

For next two question distribution is same so we will use same table.

Number of Complaints (x)	Probability(P(x)	X*P(x)	X^2*P(x)
0	0.20	0	0
1	0.25	0.25	0.25
2	0.26	0.52	1.04
3	0.14	0.42	1.26
4	0.10	0.40	1.6
5	0.05	0.25	1.25
Total		1.84	5.4

Mode

Here Highest probability is 0.26 which for x=2

Therefore   Mode = 2

------------------------------------------------------------------------------------

Formula for variance is

$\sigma^{2}=\sum \left ( x^{2}*P\left ( X \right ) \right )-\mu^{2}$

Here

µ= Σx*P(x) = 1.84

Variance = 5.4 - (1.84)^2

             = 5.4 - 3.3856

Variance = 2.0144