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.$n=1000,p=0.003$

Given,

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.

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

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.

Given,

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.

Given,

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.

Given,

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.

Given,

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.