Question

The speed with which utility companies can resolve problems is very important. GTC, the Georgetown TelephoneCompany, reports they can resolve customer problems the same day they are reported in 70 percent of the cases.Suppose the 15 cases reported today are representative of all complaints. a) What is the probability 10 of the problems can be resolved today? b) What is the probability 9 or 12 of the problems can be resolved today? c) What is the probability more than 10 of the problems can be resolved today?

Answer 1

a)
`P(10)=0.2061`

b)
`P(9\ or\ 12)=0.3172`

c)
`P(X>10)=0.51545`

Explanation:

From the question we are told that:

Probability of Same day resolution
`P=70\%=>0.7`

Probability of Another day resolution
`Q=1-P=>30%=>0.3`

Sample size
`n=15`

a)

Generally the equation for Probability 10 of the problems can be resolved today is mathematically given by

Since

`P(X)=^(n)C_(x)*p^(x)*Q^(n-x)`

`P(10)=^(15)C_(10)*p^(10)*Q^(15-10)`

`P(10)=(15!)/(10!(15-10)!*0.70^(10)*0.30^5)`

`P(10)=0.2061`

b)

Generally the equation for Probability 9 or 12 of the problems can be resolved today is mathematically given by

`P(9)=(15!)/(9!(15-9)!*0.70^(9)*0.30^6)`

`P(9)=5005*2.942*10^(-5)`

`P(9)=0.1472`

OR

`P(12)=(15!)/(12!(15-12)!*0.70^(12)*0.30^3)`

`P(12)=455*3.737*10^(-4)`

`P(12)=0.17`

Therefore Probability 9 or 12 of the problems can be resolved today is

`P(9\ or\ 12)=P(9)+P(12)`

`P(9\ or\ 12)=0.1472+0.17`

`P(9\ or\ 12)=0.3172`

c)Generally the equation for probability more than 10 of the problems can be resolved today is mathematically given by

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

Where

`P(11)=(15!)/(11!(15-11)!*0.70^(11)*0.30^4)`

`P(11)=0.2186`

`P(13)=(15!)/(13!(15-13)!*0.70^(13)*0.30^2)`

`P(13)=0.0916`

`P(14)=(15!)/(14!(15-14)!*0.70^(14)*0.30^1)`

`P(14)=0.0305`

`P(15)=(15!)/(15!(15-15)!*0.70^(15)*0.30^0)`

`P(15)=0.00475`

Therefore

`P(X>10)=0.2186+0.0916+0.0305+0.00475+0.17`

`P(X>10)=0.51545`