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) =
^{n}C_{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) =
^{n}C_{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) =
^{n}C_{x}*p^{x}*(1-p)^{(n-x)}

**P(x >
2) = 0.1517**

13. A customer service center receives a wide variety of calls
for a manufacturer, but 0.09 of these calls are warranty claims.
Assume that all calls are independent and that the probability of
each call being a warranty claim is 0.09. Let X denote the number
of warranty claims received in the first 16 calls. Find the
expected value of X.
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