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I was doing some problems on the topic of The Normal Curve and Standard Scores.

Using the same population parameters as in Problem 14, what percentage of the scores are between the following scores?

b. 5.4 and 8.0

$\mu = 8.2$

$\sigma = 2.4$

Calculations:

$z = \frac{5.4 - 8.2}{2.4} = -1.67 $

$z = \frac{8.0 - 8.2}{2.4} = -0.08 $

From the tables in the book:

z: 0.08  
Area between mean and z B: .0319
Area beyond z C: .4681

z: 1.67
Area between mean and z B: .4525
Area beyond z C: .0475

The final answer for this problem in the back side of the book is: 34.71%.

I do not understand how they derived to this answer. I got something different, but, I am probably doing something wrong.

Book:

 Understanding Statistics in the behavioral sciences
 Robert R. Pagano
 International Edition

EDIT: The book does not provide a table for negative Z-scores.

enter image description here

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    $\begingroup$ Seems like it should be $.4525-.0319=.4206$. $\endgroup$
    – Benjamin Wang
    Commented Aug 3 at 7:04
  • $\begingroup$ Final answer is: $.3471$, I did what you did too, but, it is incorrect per the answer in the book. Now, if someone knows this well, they can tell me if the book is mistaken. $\endgroup$
    – Alix Blaine
    Commented Aug 3 at 7:10
  • $\begingroup$ Not knowing what you got as an answer - when I ran the calculation myself (with no context, so I do not know if some rounding happened along the way), I go 0.345121 which would be 34.51% Since this differs by only one digit, maybe it is a typographical error? $\endgroup$
    – Red Five
    Commented Aug 3 at 7:17
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    $\begingroup$ Perhaps your book is wrong. Incidentally, I get $\Phi\left(\frac{8.0-8.2}{2.4}\right)- \Phi\left(\frac{5.4-8.2}{2.4}\right)$ $\approx \Phi\left(-0.0833\right)- \Phi\left(-1.1667\right)$ $\approx 0.4668 - 0.1217$ $\approx 0.3451$ though using tables there could be rounding errors $\endgroup$
    – Henry
    Commented Aug 3 at 7:20
  • $\begingroup$ Your error is looking at $-1.67$ rather than $-1.17$. The book's smaller error is poor precision and quoting too many decimal places. $\endgroup$
    – Henry
    Commented Aug 3 at 7:29

1 Answer 1

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Let $X \sim \operatorname{Normal}(\mu = 8.2, \sigma = 2.4)$. Then we are asked for $$\Pr[5.4 \le X \le 8.0].$$ Standardizing, we obtain $$\begin{align} \Pr[5.4 \le X \le 8.0] &= \Pr\left[\frac{5.4 - 8.2}{2.4} \le \frac{X - \mu}{\sigma} \le \frac{8.0 - 8.2}{2.4}\right] \\ &= \Pr[-1.16667 \le Z \le -0.083333], \end{align}$$ where $Z \sim \operatorname{Normal}(0,1)$ is standard normal. Consequently, $$\begin{align} \Pr[5.4 \le X \le 8.0] &= \Pr[Z \le -0.083333] - \Pr[Z \le -1.16667] \\ &= \Phi(-0.083333) - \Phi(-1.16667), \end{align}$$ where $\Phi(z) = \Pr[Z \le z]$ is the cumulative distribution function for the standard normal distribution. Since the standard normal is symmetric about $0$, we have $$\Phi(-z) = 1 - \Phi(z).$$ This is the reason why your table does not have values for negative $z$-scores. Looking up the values in a table*, $\Phi(0.083333) \approx 0.533207$, and $\Phi(1.16667) \approx 0.878327$. Therefore, the desired probability is $$\Pr[5.4 \le X \le 8.0] \approx (1 - 0.533207) - (1 - 0.878327) = 0.34512 \approx 34.5\%.$$ The reason why the book answer is inaccurate is because it presumably did not compute the cumulative distribution function with sufficient precision; i.e., no interpolation procedure was used for intermediate table values. For instance, if the table lookup were to round to the nearest hundredth, we have $\Phi(0.08) \approx 0.5319$ and $\Phi(1.17) \approx 0.8700$, and the result is $0.3471$, which matches the book's answer.

In the event that you would like to use (linear) interpolation to obtain a more precise result, the way you would do it is to look up the following values: $$\begin{array}{c|c} z & \Phi(z) \\ \hline 0.08 & 0.5319 \\ 0.09 & 0.5359 \\ 1.16 & 0.8770 \\ 1.17 & 0.8790 \\ \end{array}$$

Then, the interpolated values are given by $$\Phi(0.083333) \approx \frac{0.09 - 0.083333}{0.09 - 0.08} \Phi(0.08) + \frac{0.083333 - 0.08}{0.09 - 0.08} \Phi(0.09) \approx 0.533233,$$ and $$\Phi(1.166667) \approx \frac{1.17 - 1.166667}{1.17 - 1.16} \Phi(1.16) + \frac{1.166667 - 1.16}{1.17 - 1.16} \Phi(1.17) \approx 0.878333.$$ Thus the linear interpolation yields the result $0.34510$, which is much closer to the exact answer than the naive (uninterpolated) computation.

*Note. Now that I can see your table, I should point out that column $B$ corresponds to the calculation $$\Pr[0 \le Z \le z] = \Pr[Z \le z] - \Pr[Z \le 0] = \Phi(z) - \frac{1}{2}.$$ Column $C$ corresponds to the calculation $$\Pr[Z > z] = 1 - \Pr[Z \le z] = 1 - \Phi(z).$$

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  • $\begingroup$ Thanks for this detailed answer. I need to do some further reading and research to fully grasp all of this. $\endgroup$
    – Alix Blaine
    Commented Aug 3 at 7:53

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