Chapter 6.3, Problem 117E

(a)

To determine

To check: Whether the given distribution can be normally approximated.

Answer to Problem 117E

Not appropriate

Explanation of Solution

Concept used:

If the sample size or the number of trials be n and the probability of success be p.

For the binomial distribution to be normally approximated, both the below conditions should be satisfied.

  $\begin{array}{l}\text{n }\times \text{ p > 5}\\ \text{n }\times \text{ (1 - p) > 5}\end{array}$

Given information:

  $\begin{array}{l}n=10\\ p=0.5\end{array}$

Calculation:

Checking condition

  $\begin{array}{l}\text{n }\times \text{ p = 10 }\times \text{ 0}\text{.5 = 5}\\ \text{n }\times \text{ 1 - p = 10 }\times \text{ 1 - 0}\text{.5 = 5}\end{array}$

Here, both the conditions aren’t satisfied, so normal approximation isn’t appropriate.

(b)

To determine

To check: Whether the given distribution can be normally approximated.

Answer to Problem 117E

Not appropriate

Explanation of Solution

Given information:

  $\begin{array}{l}n=40\\ p=0.88\end{array}$

Calculation:

Checking condition

  $\begin{array}{l}\text{n }\times \text{ p = 40 }\times \text{ 0}\text{.88 = 35}\text{.2}\\ \text{n }\times \text{ 1 - p = 10 }\times \text{ 1 - 0}\text{.88 = 4}\text{.8}\end{array}$

Here, the first condition is satisfied but the second isn’t satisfied, so normal approximation isn’t appropriate.

(c)

To determine

To check: Whether the given distribution can be normally approximated.

Answer to Problem 117E

Appropriate

Explanation of Solution

Given information:

  $\begin{array}{l}n=100\\ p=0.2\end{array}$

Calculation:

Checking condition

  $\begin{array}{l}\text{n }\times \text{ p = 100 }\times \text{ 0}\text{.2 = 20}\\ \text{n }\times \text{ 1 - p = 100 }\times \text{ 1 - 0}\text{.2 = 80}\end{array}$

Here, both the conditions are satisfied, so normal approximation is appropriate.

(d)

To determine

To check: Whether the given distribution can be normally approximated.

Answer to Problem 117E

Not appropriate

Explanation of Solution

Given information:

  $\begin{array}{l}n=100\\ p=0.99\end{array}$

Calculation:

Checking condition

  $\begin{array}{l}\text{n }\times \text{ p = 100 }\times \text{ 0}\text{.99 = 99}\\ \text{n }\times \text{ 1 - p = 100 }\times \text{ 1 - 0}\text{.99 = 1}\end{array}$

Here, first condition is satisfied but the second is unsatisfied, so normal approximation isn’t appropriate.

(e)

To determine

To check: Whether the given distribution can be normally approximated.

Answer to Problem 117E

Not appropriate

Explanation of Solution

Given information:

  $\begin{array}{l}n=1000\\ p=0.003\end{array}$

Calculation:

Checking condition

  $\begin{array}{l}\text{n }\times \text{ p = 1000 }\times \text{ 0}\text{.003 = 3}\\ \text{n }\times \text{ 1 - p = 1000 }\times \text{ 1 - 0}\text{.003 = 997}\end{array}$

Here, the second condition is satisfied but the first isn’t, so normal approximation isn’t appropriate.