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At a made-up help desk, waiting times have a mean of 6 minutes and a standard deviation of 6…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

At a made-up help desk, waiting times have a mean of $6$ minutes and a standard deviation of $6$ minutes. The share of waits longer than $t$ minutes is $e^{-t/6}$ . First, find the share of waits that a normal model with the same mean and standard deviation would put below $0$ minutes. Then find the real share of waits within one standard deviation of the mean, from $0$ to $12$ minutes. Give four decimal places.

Need a reminder? The key ideaShow

$B(n, p)$ is close to $N\!\left(np, \sqrt{np(1 - p)}\right)$ when $np \ge 10$ and $n(1 - p) \ge 10$ . Skewed or bounded data are often not normal.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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