Chapter 10.4, Problem 56E

Solution

To verify: if $np$ and $nq$ are both at least 10, then 0 and n are both more than 3 standard deviations from the mean of a binomial distribution.

If $np$ and $nq$ are both at least 10, then 0 and $n$ are both more than 3 standard deviations from the mean. This is because the inequalities $np\ge 10>9\left(1-p\right)$ and $nq\ge 10>9p$ hold true.

Given information:

In any binomial probability distribution, the number of successes must be at least 0 and at most n , the number of trials. Therefore, when we want to use a normal model as an approximation, we must cut off the curve’s unbounded tails. That will not cause much inaccuracy if what we cut off lies more than 3 standard deviations from the mean

Concept Used:

Expected Value for a Binomial Distribution:

If the binomial random variable $X$ counts the number of successes in $n$ independent trials with probability of success $p$ , then $\mu =E(x)=np$ .

Standard Deviation for a Binomial Distribution:

If the binomial random variable $X$ counts the number of successes in $n$ independent trials with probability of success $p$ , then $\sigma =\sqrt{npq}$ where $q$ represents the probability of failure.

Calculation:

For the model, 0 and n are both more than 3 standard deviations from the mean.

  $\begin{array}{l}\mu \pm 3\sigma \in \left(0,n\right)\\ np\pm 3\sqrt{npq}\in \left(0,n\right)\\ np\pm 3\sqrt{np(1-p)}\in \left(0,n\right)\\ \begin{array}{ccc}\begin{array}{l}np-3\sqrt{np(1-p)}>0\\ np>3\sqrt{np(1-p)}\\ n^2{p}^2>9np\left(1-p\right)\\ np\ge 10>9\left(1-p\right)\end{array}& & \begin{array}{l}np-3\sqrt{np(1-p)}<n\\ -3\sqrt{np(1-p)}<n-np\\ 9np\left(1-p\right)<{\left(np-n\right)}^2\\ 9p>n\left(1-p\right)\\ n\left(1-p\right)\ge 10>9p\end{array}\end{array}\end{array}$