EXAMPLE 5.27

The amount of a particular impurity in a batch of a certain chemical product is a random variable with mean value $4.0\mathrm{g}$ and standard deviation $1.5\mathrm{g}$ . If 50 batches are independently prepared, what is the (approximate) probability that the sample average amount of impurity $\bar{X}$ is between 3.5 and $3.8\mathrm{g}$ ? According to the rule of thumb to be stated shortly, $n=50$ is large enough for the CLT to be applicable. $\bar{X}$ then has approximately a normal distribution with mean value ${\mu }_{\bar{X}}=4.0$ and ${\sigma }_{\bar{X}}=1.5/\sqrt{50}=.2121$ , so

$$\begin{array}{rl}P\left(3.5\le \bar{X}\le 3.8\right)& \approx P\left(\frac{3.5-4.0}{0.2121}\le Z\le \frac{3.8-4.0}{0.2121}\right)\\ & =\Phi \left(-0.94\right)-\Phi \left(-2.36\right)\\ & \approx \mathrm{0.1645.}\end{array}$$

Now consider randomly selecting 100 batches, and let $T_o$ represent the total amount of impurity in these batches. Then the mean value and standard deviation of $T_o$ are $100\left(4\right)=400$ and $\sqrt{100}\left(1.5\right)=15$ , respectively, and the CLT implies that $T_0$ has approximately a normal distribution. The probability that this total is at most $425\mathrm{g}$ is

$$\begin{array}{rl}P\left(T_0\le 425\right)& \approx P\left(Z\le \frac{425-400}{15}\right)\\ & =P\left(Z\le 1.67\right)\\ & =\Phi \left(1.67\right)=\mathrm{0.9525.}\end{array}$$