A company services home air conditioners. It is known that times for service calls follow a normal distribution with a mean of 75 minutes and a standard deviation of 15 minutes. A random sample of twelve service calls is taken. What is the probability that exactly eight of them take more than 93.6 minutes

The probability that exactly eight of them take more than 93.6 minutes is 5.6015 .

Step-by-step explanation:

We are given that it is known that times for service calls follow a normal distribution with a mean of 75 minutes and a standard deviation of 15 minutes.

A random sample of twelve service calls is taken.

So, firstly we will find the probability that service calls take more than 93.6 minutes.

Let X = times for service calls.

So, X ~ Normal()

The z-score probability distribution for the normal distribution is given by;

                              Z  =    ~ N(0,1)

where, = mean time = 75 minutes

            = standard deviation = 15 minutes

Now, the probability that service calls take more than 93.6 minutes is given by = P(X > 93.6 minutes)

       P(X > 93.6 min) = P( > ) = P(Z > 1.24) = 1 - P(Z 1.24)

                                                                = 1 - 0.8925 = 0.1075

The above probability is calculated by looking at the value of x = 1.24 in the z table which has an area of 0.8925.

Now, we will use the binomial distribution to find the probability that exactly eight of them take more than 93.6 minutes, that is;

where, n = number of trials (samples) taken = 12 service calls

            r = number of success = exactly 8

            p = probability of success which in our question is probability that

                   it takes more than 93.6 minutes, i.e. p = 0.1075.

Let Y = Number of service calls which takes more than 93.6 minutes

So, Y ~ Binom(n = 12, p = 0.1075)

Now, the probability that exactly eight of them take more than 93.6 minutes is given by = P(Y = 8)

               P(Y = 8)  =  

                             =  

                             =  5.6015 .