91Ó°ÊÓ

Let X follows binomial distribution with parameter n and p. Also, Y follows binomial distribution with parameters n and p/k with k>1.

Let, \(Z = kY\)

Since, Y follows binomial distribution with parameters n and p/k with k>1.

Therefore,

\(E\left( Y \right) = n.\frac{p}{k}\)

Now,

\(\begin{array}{c}E\left( Z \right) = kE\left( Y \right)\\ = k.n.\frac{p}{k}\\ = np\end{array}\)

But, \(E\left( X \right) = np\)

Hence, X and Z have the same mean.

Since, X follows binomial distribution with parameter n and p.

Therefore,

\(V\left( X \right) = np\left( {1 - p} \right)\)

Also,

Y follows binomial distribution with parameter n and p/k.

Therefore,

\(V\left( Y \right) = n\frac{p}{k}\left( {1 - \frac{p}{k}} \right)\)

So,

\(\begin{array}{c}V\left( {kY} \right) = {k^2}V\left( Y \right)\\ = {k^2}n\frac{p}{k}\left( {1 - \frac{p}{k}} \right)\\ = knp\left( {1 - \frac{p}{k}} \right)\end{array}\)

If p is small, then both \(\left( {1 - p} \right)\) and \(\left( {1 - \frac{p}{k}} \right)\) will be close to 1, and \(V\left( Z \right)\) is approximately knpwhile the variance of X is approximately np.

In Fig. 6.1, each bar has height equal to 0.01 times a binomial random variable with parameters 100 and the probability that \({X_1}\) is in the interval under the bar.

In Fig. 6.2, each bar has height equal to 0.02 times a binomial random variable with parameters 100 and probability that\({X_1}\) is in the interval under the bar.

The bars in Fig.6.2 have approximately one-half of the probability of the bars in Fig.6.1, but their heights have been multiplied by 2. By part (b), we expect the heights in Fig.6.2 to have approximately twice the variance of the heights in Fig.6.1.