Question

A​ toll-free phone number is available from 9 a.m. to 9 p.m. for your customers to register complaints about a product purchased from your company. Past history indicates that an average of 0.3 calls are received per minute. Complete parts​ (a) through​ (d). a. What properties must be true about the situation described here in order to use the Poisson distribution to calculate probabilities concerning the number of phone calls received in a? 1-minute period? Select all the assumptions for a Poisson distribution. A.The probability that two or more phone calls received in a time period approaches zero as the length of the time period becomes smaller. B. At least 30 calls are received. C. The probability of a phone call is the same in any given? 1-minute period. D. The number of phone calls received in a given? 1-minute period is independent of the number of phone calls received in any other? 1-minute period. b. What is the probability that during a? 1-minute period zero phone calls will be? received? The probability that during a? 1-minute period zero phone calls will be received is.... . ?(Round to four decimal places as? needed.) c. What is the probability that during a? 1-minute period three or more phone calls will be? received? The probability that during a? 1-minute period three or more phone calls will be received is ?(Round to four decimal places as? needed.) d. What is the maximum number of phone calls that will be received in a? 1-minute period? 99.99% of the? time? The maximum number of phone calls that will be received in a? 1-minute period? 99.99% of the time is..... phone calls.

Answer 1

Ans:

a)

The number of phone calls received in a given 1-minute period is independent of the number of phone calls received in any other 1-minute period.

b)mean rate=0.3 calls per minute

P(0 calls)=P(x=0)=exp(-0.3)=0.7408

c)

P(x>=3)=1-P(x<=2)

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

=1-exp(-0.3)*(0.3⁰/0!+0.3¹/1!+0.3²/2!)

=0.0036

d)

P(x<=a)=0.9999

P(x<=2)=exp(-0.3)*(0.3⁰/0!+0.3¹/1!+0.3²/2!)=0.9964

P(x<=3)=exp(-0.3)*(0.3⁰/0!+0.3¹/1!+0.3²/2!+0.3³/3!)=0.9997

P(x<=4)=exp(-0.3)*(0.3⁰/0!+0.3¹/1!+0.3²/2!+0.3³/3!+0.3⁴/4!)=0.99998

So,maximum number of calls with probability of 99.99% is=4

a	P(X<=a)
0	0.74082
1	0.96306
2	0.99640
3	0.99973
4	0.99998