Question

The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it can resolve customer problems the same day they are reported in 75% of the cases. Suppose the 15 cases reported today are representative of all complaints.
(a) How many of the problems would you expect to be resolved today? What is the standard deviation?
(b) What is the probability 10 of the problems can be resolved today?
(c) What is the probability 10 or 11 of the problems can be resolved today?
(d) What is the probability more than 10 of the problems can be resolved today?

Answer 1

Answer:

(a) 11.25 and 1.68  

(b) 0.1651

(c) 0.3903

(d) 0.6865

Step-by-step explanation:

We are given that GTC, the Georgetown Telephone Company, reports it can resolve customer problems the same day they are reported in 75% of the cases and suppose the 15 cases reported today are representative of all complaints.

This situation can be represented through Binomial distribution as;

$P(X=r)= \binom{n}{r}p^{r}(1-p)^{n-r} ; x = 0,1,2,3,....$

where,  n = number of trials (samples) taken = 15

             r = number of success

             p = probability of success which in our question is % of cases in

                  which customer problems are resolved on the same day, i.e.;75%

So, here X ~ $Binom(n=15,p=0.75)$

(a) Expected number of problems to be resolved today = E(X)

            E(X) = $\mu$ = n * p = 15 * 0.75 = 11.25

    Standard deviation = $\sigma$ = $\sqrt{n*p*(1-p)}$ = $\sqrt{15*0.75*(1-0.75)}$ = 1.68

(b) Probability that 10 of the problems can be resolved today = P(X = 10)

     P(X = 10) = $\binom{15}{10}0.75^{10}(1-0.75)^{15-10}$

                    = $3003*0.75^{10} *0.25^{5}$ = 0.1651

(c) Probability that 10 or 11 of the problems can be resolved today is given by = P(X = 10) + P(X = 11)

    = $\binom{15}{10}0.75^{10}(1-0.75)^{15-10}+\binom{15}{11}0.75^{11}(1-0.75)^{15-11}$

    = $3003*0.75^{10} *0.25^{5} + 1365*0.75^{11} *0.25^{4}$ = 0.3903

(d) Probability that more than 10 of the problems can be resolved today is

    given by = P(X > 10)

P(X > 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)  

= $\binom{15}{11}0.75^{11}(1-0.75)^{15-11}+\binom{15}{12}0.75^{12}(1-0.75)^{15-12} + \binom{15}{13}0.75^{13}(1-0.75)^{15-13}+\binom{15}{14}0.75^{14}(1-0.75)^{15-14} + \binom{15}{15}0.75^{15}(1-0.75)^{15-15}$

= $1365*0.75^{11} *0.25^{4} + 455*0.75^{12} *0.25^{3}+105*0.75^{13} *0.25^{2} + 15*0.75^{14} *0.25^{1}+1*0.75^{15} *0.25^{0}$

= 0.6865