Answer:
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 = <u><em>times for service calls.</em></u>
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 = <u>0.1075</u>
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 = <u><em>Number of service calls which takes more than 93.6 minutes</em></u>
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 .