Using Normal Approximation. In Exercises 5–8, do the following: If the requirements of $np⩾5$and$nq⩾5$are both satisfied, estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution; if$np<5$or$nq<5$, then state that the normal approximation should not be used.

Guessing on Standard TestsWith n= 50 guesses and p= 0.2 for a correct answer, findP(exactly 12 correct answers).

A sample of 50 guesses of answers on a test is considered. The probability of a correct answer is equal to 0.2.

It is required that$np⩾5$ and $nq⩾5$.

The values are computed below:

$$\begin{gathered} np=\left(50\right)\left(0.2\right) \\ =10 \\ ⩾5 \end{gathered}$$

$$\begin{gathered} nq=\left(50\right)\left(1-0.2\right) \\ =40 \\ ⩾5 \end{gathered}$$

As the requirement is fulfilled, the normal approximation can be applied to compute the probability value.

The mean value is equal to:

$$\begin{gathered} \mu =np \\ =50\times 0.2 \\ =10 \end{gathered}$$

The standard deviation is equal to:

$$\begin{gathered} \sigma =\sqrt{npq} \\ =\sqrt{50\times 0.2\times 0.8} \\ =2.83 \end{gathered}$$

Let x represent the number of correct answers.

Here, x is equal to 12.

The value of x is transformed as follows:

$$\begin{gathered} \left(x-0.5,x+0.5\right)=\left(12-0.5,12+0.5\right) \\ =\left(11.5,12.5\right) \end{gathered}$$

It is required to compute the probability of exactly 212 correct answers. Thus, the probability between the values 11.5 and 12.5 needs to be computed.

The required probability value is equal to:

$$\begin{gathered} P\left(11.5<x<12.5\right)=P\left(\frac{11.5-\mu }{\sigma }<\frac{x-\mu }{\sigma }<\frac{12.5-\mu }{\sigma }\right) \\ =P\left(\frac{11.5-10}{2.83}<z<\frac{12.5-10}{2.83}\right) \\ =P\left(0.53<z<0.88\right) \\ =P\left(z<0.88\right)-P\left(z<0.53\right) \\ =0.8106-0.7019 \\ =0.1087 \end{gathered}$$

Therefore, the probability of exactly 12 correct answers is equal to 0.1087.