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.

  • 1
    $\begingroup$ 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 $n=5$ $\endgroup$ – 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-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.

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

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