Example 5.14

EXAMPLE 5.14 Using the Normal Approximation

CASE 5.1 As described in Case 5.1 (page 247), a quality engineer inspects an SRS of 150 switches from a large shipment of which 8% fail to meet specifications. The count $X$ of nonconforming switches in the sample were thus assumed to be the $B (150, 0.08)$ distribution. In Example 5.10 (page 254), we found $μ_{X} = 12$ and $σ_{X} = 3.3226$ .

The Normal approximation for the probability of no more than 10 nonconforming switches is the area to the left of $X = 10$ under the Normal curve. Using Table A,

$\begin{array}{l} P (X \leq 10) & = P (\frac{X - 12}{3.3226} \leq \frac{10 - 12}{3.3226}) \\ = P (Z \leq - 0.60) = 0.2743 \end{array}$

In Example 5.5 (pages 247–248), we found that software tells us that the actual binomial probability that there is no more than 10 nonconforming switches in the sample is $P (X \leq 10) = 0.3384$ . The Normal approximation is only roughly accurate. Because $n p = 12$ , this combination of $n$ and $p$ is close to the border of the values for which we are willing to use the approximation.