Question

URGENT PLEASE

A customer service center receives a wide variety of calls for a manufacturer, but 7% of these calls are warranty claims. Assume that all calls are independent and that the probability of each call being a warranty claim is 0.07.

A. What is the probability that 2 of the first 10 calls will be warranty claims?

B. On average, how many of the first 10 calls will be warranty claims?

C. What is the standard deviation of the number of claim calls in the first 10 calls?

D. What is the probability that 2 or less of the first 10 calls will be warranty claims?

E. What is the probability at least 2 of the first 10 calls will be warranty claims?

Answer 1

Solution:-

A) The probability that 2 of the first 10 calls will be warranty claims is 0.1234.

p = 0.07, n = 10

x = 2

By applying binomial distribution

P(x,n) = ⁿC_x*p^x*(1-p)^(n-x)

P(x = 2) = 0.1234

B) The expected number of calls in first 10 calls will be warranty claims is 0.70.

E(x) = n*p

E(x) = 10*0.07

E(x) = 0.70

C) The standard deviation of the number of claim calls in the first 10 calls is 0.807.

D) The probability that 2 or less of the first 10 calls will be warranty claims is 0.9717.

p = 0.07, n = 10

x = 2

By applying binomial distribution

P(x,n) = ⁿC_x*p^x*(1-p)^(n-x)

P(x < 2) = 0.9717

E) The probability at least 2 of the first 10 calls will be warranty claims is 0.1517.

p = 0.07, n = 10

x = 2

By applying binomial distribution

P(x,n) = ⁿC_x*p^x*(1-p)^(n-x)

P(x > 2) = 0.1517