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A multiple-choice test is considered where each question has 5 options, and only one of the options is correct. A sample of 100 questions is answered.

The following requirements must be fulfilled to use the normal distribution to approximate binomial distribution:

Here, the sample size is given to be equal to n=100.

Success is defined as getting a correct answer.

The probability of success is given below:

$$\begin{gathered} p=\frac{\text{Number}\text{of}\text{correct}\text{answers}}{\text{Number}\text{of}\text{options}} \\ =\frac{1}{5} \end{gathered}$$

First requirement:

Here, the sample given is a result of 100 independent trials, and the probability of success for each trial is equal to 0.2.

Thus, the first requirement is fulfilled.

Second requirement:

$$\begin{gathered} np=100\left(0.2\right) \\ =20 \\ >5 \end{gathered}$$

$$\begin{gathered} nq=100\left(1-0.2\right) \\ =100\left(0.8\right) \\ =80 \\ >5 \end{gathered}$$

Thus, the second requirement is also satisfied.

Since the procedure satisfies the necessary requirements, the method of normal approximation of binomial distribution can be applied.