Respuesta :
Answer:
The probability that exactly eight of them take more than 93.6 minutes is 5.6015 [tex]\times 10^{-6}[/tex] .
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([tex]\mu=75,\sigma^{2} =15^{2}[/tex])
The z-score probability distribution for the normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean time = 75 minutes
[tex]\sigma[/tex] = 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( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{93.6-75}{15}[/tex] ) = P(Z > 1.24) = 1 - P(Z [tex]\leq[/tex] 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;
[tex]P(Y = y) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ; y = 0,1,2,3,.........[/tex]
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) = [tex]\binom{12}{8}\times 0.1075^{8} \times (1-0.1075)^{12-8}[/tex]
= [tex]495 \times 0.1075^{8} \times 0.8925^{4}[/tex]
= 5.6015 [tex]\times 10^{-6}[/tex] .
