3
$\begingroup$

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.

$\endgroup$
  • 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
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.