Answer :
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;
[tex]P(X=r)= \binom{n}{r}p^{r}(1-p)^{n-r} ; x = 0,1,2,3,....[/tex]
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 ~ [tex]Binom(n=15,p=0.75)[/tex]
(a) Expected number of problems to be resolved today = E(X)
E(X) = [tex]\mu[/tex] = n * p = 15 * 0.75 = 11.25
Standard deviation = [tex]\sigma[/tex] = [tex]\sqrt{n*p*(1-p)}[/tex] = [tex]\sqrt{15*0.75*(1-0.75)}[/tex] = 1.68
(b) Probability that 10 of the problems can be resolved today = P(X = 10)
P(X = 10) = [tex]\binom{15}{10}0.75^{10}(1-0.75)^{15-10}[/tex]
= [tex]3003*0.75^{10} *0.25^{5}[/tex] = 0.1651
(c) Probability that 10 or 11 of the problems can be resolved today is given by = P(X = 10) + P(X = 11)
= [tex]\binom{15}{10}0.75^{10}(1-0.75)^{15-10}+\binom{15}{11}0.75^{11}(1-0.75)^{15-11}[/tex]
= [tex]3003*0.75^{10} *0.25^{5} + 1365*0.75^{11} *0.25^{4}[/tex] = 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)
= [tex]\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}[/tex]
= [tex]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}[/tex]
= 0.6865