Question

AMR is a computer-consulting firm. The number of new clients that they have obtained each month has ranged from 0 to 6. The number of new clients has the probability distribution that is shown below.

Number of

New Clients

Probability

0

0.05

1

0.10

2

0.15

3

0.35

4

0.20

5

0.10

6

0.05

#1a). Refer to Exhibit 5-3. Compute the expected number and variance of new clients per month are respectively

#1b). Ten percent of the items produced by a machine are defective. Out of 15 items chosen at random, what is the probability that less than 3 items will be defective?

Answer 1

1a)

X	P(X)	X*P(X)	X² * P(X)
0	0.0500	0	0.000
1	0.1000	0.1	0.100
2	0.1500	0.3	0.6000
3	0.3500	1.05	3.1500
4	0.2000	0.8000	3.2000
5	0.1000	0.5000	2.5000
6	0.05	0.3000	1.8000

the expected number=mean = E[X] = Σx*P(X) =            3.05000

E [ X² ] = ΣX² * P(X) =            11.3500
          
variance = E[ X² ] - (E[ X ])² =            2.0475

1b)

n=15

p=0.10

P ( X = 0) = C (15,0) * 0.1^0 * ( 1 - 0.1)^15=      0.2059
P ( X = 1) = C (15,1) * 0.1^1 * ( 1 - 0.1)^14=      0.3432
P ( X = 2) = C (15,2) * 0.1^2 * ( 1 - 0.1)^13=      0.2669

P(X<3) = P(X=0) + P(X=1) + P(X=2) = 0.8159