Question

In which of the following situations would it be appropriate to use a Normal distribution to approximate probabilities for a binomial distribution with the given values of n and p ?

a. $n=10,p=0.5$

b. $n=40,p=0.88$

c. $n=100,p=0.2$

d. $n=100,p=0.99$

e.$n=1000,p=0.003$

Short answer

(a) both the conditions aren't satisfied, so normal approximation isn't appropriate.

(b)Here, the first condition is satisfied but the second isn't satisfied, so normal approximation isn't appropriate

(c) Here, both the conditions are satisfied, so normal approximation is appropriate

(d)first condition is satisfied but the second is unsatisfied, so normal approximation isn't appropriate

(e) the second condition is satisfied but the first isn't, so normal approximation isn't appropriate.

Step-by-step solution

Part (a) Step 1: Given Information

Given,

$$\begin{gathered} n=10 \\ p=0.5 \end{gathered}$$

Used concept:

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

Both of the following conditions must be met for the binomial distribution to be regularly approximated.

$$\begin{gathered} n\times p>5 \\ n\times (1-p)>5 \end{gathered}$$

Part (a) Step 2: Simplification

Lets Check the following condition

$$\begin{gathered} \mathrm{n}\times \mathrm{p}=10\times 0.5=5 \\ \mathrm{n}\times 1-\mathrm{p}=10\times 1-0.5=5 \end{gathered}$$

Because both assumptions are not met in this case, typical approximation is not acceptable.

Part (b) Step 1: Given Information

Given,

$$\begin{gathered} n=40 \\ p=0.88 \end{gathered}$$

Part (b) Step 2: Simplification

Lets Check the following condition

$$\begin{gathered} \mathrm{n}\times \mathrm{p}=40\times 0.88=35.2 \\ \mathrm{n}\times 1-\mathrm{p}=10\times 1-0.88=4.8 \end{gathered}$$

Because the first requirement is met but the second is not, normal approximation is not appropriate.

Part (c) Step 1: Given Information

Given,

$$\begin{gathered} n=100 \\ p=0.2 \end{gathered}$$

Part (c) Step 2: Simplification

Lets Check the following condition

$$\begin{gathered} \mathrm{n}\times \mathrm{p}=100\times 0.2=20 \\ \mathrm{n}\times 1-\mathrm{p}=100\times 1-0.2=80 \end{gathered}$$

Because both prerequisites are met in this case, conventional approximation is appropriate.

Part (d) Step 1: Given Information

Given,

$$\begin{gathered} n=100 \\ p=0.99 \end{gathered}$$

Part (d) Step 2: Simplification

Lets Check the following condition

$$\begin{gathered} \mathrm{n}\times \mathrm{p}=100\times 0.99=99 \\ \mathrm{n}\times 1-\mathrm{p}=100\times 1-0.99=1 \end{gathered}$$

Because the first criterion is satisfied but the second is not, normal approximation is not appropriate.

Part (e) Step 1: Given Information

Given,

$$\begin{gathered} n=1000 \\ p=0.003 \end{gathered}$$

Part (e) Step 2: Simplification

Lets Check the following condition

$$\begin{gathered} \mathrm{n}\times \mathrm{p}=1000\times 0.003=3 \\ \mathrm{n}\times 1-\mathrm{p}=1000\times 1-0.003=997 \end{gathered}$$

Because the second criteria is met but the first is not, normal approximation is not applicable.