Chapter 7.5, Problem 67E

a.

To determine

Find the exact and approximate values of $P\left(Y\le \mu +3\right)$ for $n=5,10,15\text{and}20$.

Answer to Problem 67E

The exact and approximate values of $P\left(Y\le \mu +3\right)$ for $n=5$ are 0.9997 and 0.9532, respectively.

The exact and approximate values of $P\left(Y\le \mu +3\right)$ for $n=10$ are 0.9936 and 0.9731, respectively.

The exact and approximate values of $P\left(Y\le \mu +3\right)$ for $n=15$ are 0.9819 and 0.9761, respectively.

The exact and approximate values of $P\left(Y\le \mu +3\right)$ for $n=20$ are 0.9679 and 0.9689, respectively.

Explanation of Solution

From the given information, Y follows binomial distribution and $p=0.20$.

For $n=5$

$\begin{array}{c}P\left(Y\le \mu +3\right)=P\left(Y\le np+3\right)\\ =P\left(Y\le 5\left(0.20\right)+3\right)\\ =P\left(Y\le 4\right)\end{array}$

Step-by-step procedure to obtain the probability using APPLET:

• Choose Normal approximation to binomial distribution under Applets.
• Set the n value as 5 and p value as 0.20.
• Select the bars 0 and 4.

Output using APPLET is given below:

From the above output, it can be observed that exact and approximate values are 0.9997 and 0.9532, respectively.

Thus, the exact and approximate values of $P\left(Y\le \mu +3\right)$ for $n=5$ are 0.9997 and 0.9532, respectively.

For $n=10$

$\begin{array}{c}P\left(Y\le \mu +3\right)=P\left(Y\le np+3\right)\\ =P\left(Y\le 10\left(0.20\right)+3\right)\\ =P\left(Y\le 5\right)\end{array}$

Step-by-step procedure to obtain the probability using APPLET:

• Choose Normal approximation to binomial distribution under Applets.
• Set the n value as 10 and p value as 0.20.
• Select the bars 0 and 5.

Output using APPLET is given below:

From the above output, it can be observed that exact and approximate values are 0.9936 and 0.9731, respectively.

Thus, the exact and approximate values of $P\left(Y\le \mu +3\right)$ for $n=10$ are 0.9936 and 0.9731, respectively.

For $n=15$

$\begin{array}{c}P\left(Y\le \mu +3\right)=P\left(Y\le np+3\right)\\ =P\left(Y\le 15\left(0.20\right)+3\right)\\ =P\left(Y\le 6\right)\end{array}$

Step-by-step procedure to obtain the probability using APPLET:

• Choose Normal approximation to binomial distribution under Applets.
• Set the n value as 15 and p value as 0.20.
• Select the bars 0 and 6.

Output using APPLET is given below:

From the above output, it can be observed that exact and approximate values are 0.9819 and 0.9761, respectively.

Thus, the exact and approximate values of $P\left(Y\le \mu +3\right)$ for $n=15$ are 0.9819 and 0.9761, respectivley.

For $n=20$

$\begin{array}{c}P\left(Y\le \mu +3\right)=P\left(Y\le np+3\right)\\ =P\left(Y\le 20\left(0.20\right)+3\right)\\ =P\left(Y\le 7\right)\end{array}$

Step-by-step procedure to obtain the probability using APPLET:

• Choose Normal approximation to binomial distribution under Applets.
• Set the n value as 20 and p value as 0.20.
• Select the bars 0 and 7.

Output using APPLET is given below:

From the above output, it can be observed that exact and approximate values are 0.9679 and 0.9689, respectively.

Thus, the exact and approximate values of $P\left(Y\le \mu +3\right)$ for $n=20$ are 0.9679 and 0.9689, respectively.

b.

To determine

Give the observation about the shapes of the binomial histograms as the sample size increased by referring part (a).

Give the observation about the differences between the exact and approximate values of $P\left(Y\le \mu +3\right)$ as the sample size increased by referring part (a).

Answer to Problem 67E

The observation is that the shapes of the binomial histograms tend to bell shaped by increasing the sample size.

The difference between the exact and approximate values of $P\left(Y\le \mu +3\right)$ decreases as the sample size increases.

Explanation of Solution

From the histograms in part a, it can be observed that as the sample size increases the binomial histograms become bell shaped.

Thus, the observation is that the shapes of the binomial histograms tend to bell shaped by increasing the sample size.

From the part a, it can be observed that

The exact and approximate values of $P\left(Y\le \mu +3\right)$ for $n=5$ are 0.9997 and 0.9532, respectively. The difference is $0.9997-0.9532=0.0465$.

The exact and approximate values of $P\left(Y\le \mu +3\right)$ for $n=10$ are 0.9936 and 0.9731, repectively. The difference is $0.9936-0.9731=0.0205$.

The exact and approximate values of $P\left(Y\le \mu +3\right)$ for $n=15$ are 0.9819 and 0.9761, respectively. The difference is $0.9819-0.9761=0.0058$.

The exact and approximate values of $P\left(Y\le \mu +3\right)$ for $n=20$ are 0.9679 and 0.9689, respectively. The difference is $0.9679-0.9689=-0.001$.

From the above results it can be observed that as the sample size increases the difference between the exact and approximate values of $P\left(Y\le \mu +3\right)$ decreases.

Thus, the difference between the exact and approximate values of $P\left(Y\le \mu +3\right)$ decreases as the sample size increases.

c.

To determine

Find the sample size n for the approximation to be adequate.

Check whether this is consistent with that observed in parts (a) and (b).

Answer to Problem 67E

The sample size n for the approximation to be adequate is 36.

The sample size is consistent with that observed in parts (a) and (b).

Explanation of Solution

It is known that, the normal approximate to the binomial approximation is adequate if

$n>9\left(\frac{\text{larger}\text{of}p\text{and}q}{\text{smaller}\text{of}p\text{and}q}\right)$

From the given information, $p=0.2$. Then, $q=1-p=1-0.2=0.8$.

Then,

$\begin{array}{l}n>9\left(\frac{\text{larger}\text{of}0.2\text{and}0.8}{\text{smaller}\text{of}0.2\text{and}0.8}\right)\\ n>9\left(\frac{0.8}{0.2}\right)\\ n>9\left(4\right)\\ n>36\end{array}$

Thus, the sample size n for the approximation to be adequate is 36.

From the parts a and b it can be observed that as the sample size increases the exact and approximate values of $P\left(Y\le \mu +3\right)$ are nearly equal. Here, the sample size is large.

Thus, the sample size is consistent with that observed in parts (a) and (b).