Calculating probability for a normal distribution

The question I wish to solve is as follows

According to records of 520 employees, the weekly overtime work of an employee follows a Normal distribution with mean 135 minutes and standard deviation 30 minutes. Calculate the probability that a randomly selected employee has worked overtime more than 165 minutes in at least 4 of the previous 5 weeks.

My thinking for this is to use the following formula for a normal distribution with x = 165 and mu= 135 and sigma = 30. $P(z >\frac{(x-\mu)}{\sigma})$

If I find the corresponding probability using z tables for this would I be right in assuming that this is the probability that a randomly selected employee works overtime more than 165 minutes in any given week?

how do I account for at least 4 of the previous 5 weeks?

Is this normal approximation to the binomial? I'm not quite sure about that part.

• You are on the right track. Use the Normal distribution to find the probability of working more than $165$ minutes in any one week, then use this probability in a Binomial distribution with $n=5$ – David Quinn Mar 12 '18 at 19:50

You'll want to find the probability that someone works overtime more than $165$ minutes in a given week. You're on the right track with this. The probability of it occurring in a given week is

$$1-\Phi\left(\frac{165-135}{30}\right)\approx0.1587$$

> 1-pnorm((165-135)/30)
[1] 0.1586553


To find the probability of it happening in at least four of the five weeks, you'll want to make use of the binomial distribution:

$$P(X=k)={n \choose k}p^k(1-p)^{n-k}$$

We wish to find $P(X=4)+P(X=5)$ where $n=5$ and $p$ is the probability obtained above.

• Excellent that clears up some of my confusion. Thanks! – user8728111 Mar 13 '18 at 13:14