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 80% of the cases. Suppose the 13 cases reported today are representative of all complaints.
Required:
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. 10.4 problems are expected to be resolved today, with a standard deviation of 1.44 problems.

b. 0.2457 = 24.57% probability 10 of the problems can be resolved today

c. 0.5137 = 51.37% probability 10 or 11 of the problems can be resolved today

d. 0.5017 = 50.17% probability more than 10 of the problems can be resolved today

Step-by-step explanation:

For each case, there are only two possible outcomes. Either they are solved, or they are not. Cases are independent of each other. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

It can resolve customer problems the same day they are reported in 80% of the cases.

This means that $p = 0.8$

Sample of 13 cases:

This means that $n = 13$

a. How many of the problems would you expect to be resolved today? What is the standard deviation?

Expected:

$E(X) = np = 13*0.8 = 10.4$

Standard deviation:

$\sqrt{V(X)} = \sqrt{13*0.8*0.2} = 1.44$

10.4 problems are expected to be resolved today, with a standard deviation of 1.44 problems.

b. What is the probability 10 of the problems can be resolved today?

This is $P(X = 10)$. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{13,10}.(0.8)^{10}.(0.2)^{3} = 0.2457$

0.2457 = 24.57% probability 10 of the problems can be resolved today.

c. What is the probability 10 or 11 of the problems can be resolved today?

This is $p = P(X = 10) + P(X = 11)$. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{13,10}.(0.8)^{10}.(0.2)^{3} = 0.2457$

$P(X = 11) = C_{13,11}.(0.8)^{11}.(0.2)^{2} = 0.2680$

$p = P(X = 10) + P(X = 11) = 0.2457 + 0.2680 = 0.5137$

0.5137 = 51.37% 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?

This is

$P(X > 10) = P(X = 11) + P(X = 12) + P(X = 13)$. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 11) = C_{13,11}.(0.8)^{11}.(0.2)^{2} = 0.2680$

$P(X = 12) = C_{13,12}.(0.8)^{12}.(0.2)^{1} = 0.1787$

$P(X = 13) = C_{13,13}.(0.8)^{13}.(0.2)^{0} = 0.0550$

$P(X > 10) = P(X = 11) + P(X = 12) + P(X = 13) = 0.2680 + 0.1787 + 0.0550 = 0.5017$

0.5017 = 50.17% probability more than 10 of the problems can be resolved today