normal distribution approximating beginners question

Question

woking out:

My textbook uses the same values as this source https://d2vlcm61l7u1fs.cloudfront.net/media/830/830f67c9-4684-47c0-89f6-fb355ca81bf5/phpRG5AJv.png And I pretty much get the wrong-by-decimal-places answer for other similar questions so it must be fundamentally my working out that has issue.

Answer 1

The issue is you have rounded your intermediate calculations and therefore have lost precision.

The lower limit $z_L$ is actually $$z_L = \frac{8.5 - 6}{\sqrt{4.2}} = 1.219875091185666\ldots,$$ and the upper limit is $$z_U = \frac{11.5 - 6}{\sqrt{4.2}} = 2.683725200608466\ldots.$$ So if you use a computer for computing $\Pr[z_L \le Z \le z_U]$, you would get $$\begin{align} \Pr[z_L \le Z \le z_U] &= \Phi(z_U) - \Phi(z_L) \\ &\approx 0.996359653878851 - 0.888743885137045 \\ &= 0.1076157687418\ldots. \end{align}$$

So in fact, the book's answer of $0.1075$ is actually slightly off, but reasonably accurate.

Now, if you did not have a computer and used a table, how would you proceed? One option is to round to two decimal places and compute $$\Pr[1.2\color{red}{2} \le Z \le 2.68] \approx 0.107551.$$ Note the correct digit in red. This is actually close enough for your purposes. However, an even better approach is through linear interpolation. From the standard normal distribution table, we can read off $$\Phi(1.21) \approx 0.8869, \\ \Phi(1.22) \approx 0.8888, \\ \Phi(2.68) \approx 0.9963, \\ \Phi(2.69) \approx 0.9964.$$

So because $1.219875 = (1 - 0.9875)(1.21) + (0.9875)(1.22)$, we have $$\Phi(1.219875) \approx (1 - 0.9875)\Phi(1.21) + (0.9875)\Phi(1.22) \approx 0.888776,$$ and similarly, $2.683725 = (1 - 0.3725)(2.68) + (0.3725)(2.69)$ gives the linear interpolation $$\Phi(2.683725) \approx (1 - 0.3725)\Phi(2.68) + (0.3725)\Phi(2.69) \approx 0.996359.$$ Then the desired probability is approximately $$\Pr[z_L \le Z \le z_U] \approx 0.996359 - 0.888776 = 0.107583.$$

This is closer to the exact probability calculated by computer, but still not perfect because the table doesn't have a lot of decimal digits--most have only four digits of precision.