A company services home air conditioners. It is known that times for service calls follow a normal distribution with a mean of 60 minutes and a standard deviation of 10 minutes. A random sample of eight service calls is taken. What is the probability that exactly two of them take more than 68.4 minutes

Step-by-step explanation:

If the times for service calls follow a normal distribution with mean μ = 60 minutes and standar deviation σ = 10 minutes, then we can calculate the probability that one call take more than 68.4 minutes, thus:

P(x>68.4) = 1 - P(X<68.4)

We standardize as follow

Since the table of the distribution normal, we obtain:

1-  P(z <0.84) = 1 - 0.7995  = 0.2005

Therefore:

P(x>68.4) = 0.2005

Now if you select random eight service calls, the number of calls takes more than 68.4 minutes is a binomial variable with

p = 0.2005

n = 8

And we need calculate this probability

77

step-by-step explanation: