Question

7. The J.O. Supplies Company buys calculators from a Korean supplier. The probability of a defective calculator is 30%. If 14 calculators are selected at random, what is the probability that more than 6 of the calculators will be defective?

8. An important part of the customer service responsibilities of a cable company relates to the speed with which trouble in service can be repaired. Historically, the data show that the likelihood is 0.70 that troubles in a residential service can be repaired on the same day. For the first six troubles reported on a given day, what is the probability that: Fewer than 3 troubles will be repaired on the same day?

Answer 1

Ans:

7)

Use binomial distribution with n=14 and p=0.30

P(x=k)=¹⁴C_k*0.30^k*(1-0.30)^14-k

x	P(x)
0	0.0068
1	0.0407
2	0.1134
3	0.1943
4	0.2290
5	0.1963
6	0.1262

P(x>6)=1-P(x<=6)

=1-(0.0068+0.0407+0.1134+0.1943+0.2290+0.1963+0.1262)

=0.0933

or using calculator:

P(x>6) =1-binomcdf(14,0.30,6)=0.0933

8)

Use binomial distribution with n=6  and p=0.70

P(x=k)=⁶C_k*0.70^k*(1-0.70)^6-k

P(fewer than 3)=P(x<3)

=P(x=0)+P(x=1)+P(x=2)

=(1-0.70)^6+⁶C₁*0.7*(1-0.7)^5+⁶C₂*0.7^2*(1-0.7)^4

=0.0705