Chapter 1, Problem 25SE

The following data give the lengths of time to failure for n = 88 radio transmitter-receivers:

1. a Use the range to approximate s for the n = 88 lengths of time to failure.
2. b Construct a frequency histogram for the data. [Notice the tendency of the distribution to tail outward (skew) to the right.]
3. c Use a calculator (or computer) to calculate $\bar{y}$ and s. (Hand calculation is much too tedious for this exercise.)
4. d Calculate the intervals $\bar{y}\pm ks,$ k = 1, 2, and 3, and count the number of measurements falling in each interval. Compare your results with the empirical rule results. Note that the empirical rule provides a rather good description of these data, even though the distribution is highly skewed.

To determine

Calculate the approximate value of s by using the range.

Answer to Problem 25SE

The approximate value of s by using the range is 177.

Explanation of Solution

An empirical rule suggests that the standard deviation is approximately one-fourth of the range.

The maximum observation is 716; the minimum observation is 8. Thus, the range is $708\left(=716-8\right)$.

Thus, the approximate value of s by using the range is $\underset{\_}{177}\left(=\frac{708}{4}\right)$.

To determine

Make a frequency histogram for the given data.

Answer to Problem 25SE

The frequency histogram for the give data is obtained as follows:

Explanation of Solution

Calculation:

The lowest and highest values of the give data are 8 and 716, respectively.

Consider the class width for each interval is 130 units.

For convenience, start the class interval of the lowest class at 8 and end the class interval of the highest class at 788. Although the class intervals should ideally be continuous, for convenience, the classes have been taken as disjoint, such as 8 – 137, 138 – 267, etc. Note that, in the first class, both 8 and 137 are included, in the second class, both 138 and 267 are included, and so on.

Hence, the frequency distribution table of the class intervals for the given data.

Time to Failure	Frequency
8 – 137	35
138 – 267	28
268 – 397	13
398 – 527	6
528 – 657	5
658 – 788	1
Total	25

Graphical procedure:

Step-by-step procedure to draw the relative frequency histogram is given below:

• Draw the horizontal axis to represent the class intervals; mark the points listed under the column of “Time to Failure” in the above table.
• Draw the vertical axis to represent the frequencies; mark the points from 0 to 40, at intervals of 5, and label them.
• Corresponding to each class interval, draw a box with the height that is indicated under the column of “Frequency” in the above table.
• Under each box, label with the class interval that it represents.

The frequency histogram for the give data is obtained.

By carefully observing the histogram, it can be seen that there is a longer tail towards the higher values, which indicates a distribution that is skewed to the right.

To determine

Calculate the values of $\overline{y}$ and s.

Answer to Problem 25SE

The value of $\overline{y}$ is 210.8.

The value of s is 162.1728.

Explanation of Solution

The formula for the mean is, $\overline{y}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{y}_i}$. The formula for the standard deviation is, $s=\sqrt{\frac{1}{n-1}\left[{\displaystyle \sum _{i=1}^n{y}_i^2}-\frac{1}{n}{\left({\displaystyle \sum _{i=1}^n{y}_i}\right)}^2\right]}$. The calculation is shown below:

Thus, the mean and standard deviation are obtained as follows:

$y_i$	$y_i^2$
16	256
392	153,664
358	128,164
304	92,416
108	11,664
156	24,336
438	191,844
60	3,600
360	129,600
56	3,136
168	28,224
224	50,176
576	331,776
384	147,456
16	256
194	37,636
216	46,656
120	14,400
208	43,264
232	53,824
72	5,184
168	28,224
16	256
128	16,384
256	65,536
72	5,184
136	18,496
168	28,224
308	94,864
340	115,600
40	1,600
64	4,096
114	12,996
80	6,400
56	3,136
246	60,516
8	64
224	50,176
184	33,856
32	1,024
104	10,816
112	12,544
40	1,600
280	78,400
96	9,216
656	430,336
328	107,584
80	6,400
80	6,400
552	304,704
272	73,984
72	5,184
112	12,544
184	33,856
152	23,104
536	287,296
224	50,176
464	215,296
72	5,184
16	256
72	5,184
152	23,104
168	28,224
288	82,944
264	69,696
208	43,264
400	160,000
40	1,600
448	200,704
56	3,136
424	179,776
184	33,856
328	107,584
40	1,600
168	28,224
96	9,216
160	25,600
80	6,400
32	1,024
716	512,656
608	369,664
264	69,696
240	57,600
480	230,400
152	23,104
352	123,904
224	50,176
176	30,976
${\displaystyle \sum _{i=1}^{88}{y}_i}=18,550$	${\displaystyle \sum _{i=1}^{88}{y}_i^2}=6,198,356$

$\begin{array}{c}\overline{y}=\frac{1}{88}{\displaystyle \sum _{i=1}^{88}{y}_i}\\ =\frac{18,550}{88}\\ \approx \mathrm{210.80.}\end{array}$

$\begin{array}{c}s=\sqrt{\frac{1}{88-1}\left[{\displaystyle \sum _{i=1}^{88}{y}_i^2}-\frac{1}{88}{\left({\displaystyle \sum _{i=1}^{88}{y}_i}\right)}^2\right]}\\ =\sqrt{\frac{1}{87}\left[6,198,356-\frac{1}{88}{\left(18,550\right)}^2\right]}\\ =\sqrt{\frac{2,288,100}{87}}\\ \approx \sqrt{26,300}\\ \approx \mathrm{162.1728.}\end{array}$

Hence, the value of $\overline{y}$ is 210.80; the value of s is 162.1728.

To determine

Find the intervals $\overline{y}\pm ks$, when $k=1,2,3$.

Find the number of measurements in each interval.

Make a comparison of the observed counts with the predicted counts, according to the empirical rule.

Answer to Problem 25SE

The intervals are, $\left(\overline{y}\pm s\right)=\underset{\_}{\left(48.6272,372.9728\right)}$, $\left(\overline{y}\pm 2s\right)=\underset{\_}{\left(-113.5456,535.1456\right)}$, and $\left(\overline{y}\pm 3s\right)=\underset{\_}{\left(-275.7184,697.3184\right)}$.

There are 63 measurements in the interval $\left(\overline{y}\pm s\right)$, 82 measurements in the interval $\left(\overline{y}\pm 2s\right)$, and 87 measurements in the interval $\left(\overline{y}\pm 3s\right)$.

The observed counts are very close to the predicted counts, according to the empirical rule.

Explanation of Solution

Calculation:

The intervals, $\overline{y}\pm ks$, when $k=1,2,3$ are calculated below:

For $k=1$:

 $\begin{array}{c}\left(\overline{y}\pm s\right)=\left(210.8\pm 162.1728\right)\\ =\underset{\_}{\left(48.6272,372.9728\right)}.\end{array}$

For $k=2$:

$\begin{array}{c}\left(\overline{y}\pm 2s\right)=\left(210.8\pm \left(2\right)\left(162.1728\right)\right)\\ =\underset{\_}{\left(-113.5456,535.1456\right)}.\end{array}$

For $k=3$:

$\begin{array}{c}\left(\overline{y}\pm 3s\right)=\left(210.8\pm \left(3\right)\left(162.1728\right)\right)\\ =\underset{\_}{\left(-275.7184,697.3184\right)}.\end{array}$

In order to find the number of measurements in each interval, first arrange the data in an ascending order.

Now, count the number of observations within each interval specified above.

It can be observed that, there are 63 measurements in the interval $\left(\overline{y}\pm s\right)$, 82 measurements in the interval $\left(\overline{y}\pm 2s\right)$, and 87 measurements in the interval $\left(\overline{y}\pm 3s\right)$.

According to the empirical rule, about 68% of the observations must lie within the interval $\left(\overline{y}\pm s\right)$, about 95% of the observations must lie within the interval $\left(\overline{y}\pm 2s\right)$, and approximately all the observations must lie within the interval $\left(\overline{y}\pm 3s\right)$.

Here, total number of observations is, $n=88$. The following percentages of 88 can be obtained:

$\begin{array}{c}68\%\text{ of }88=0.68\times 88\\ =\mathrm{59.84.}\\ 95\%\text{ of }88=0.95\times 88\\ =83.6\end{array}$

According to the empirical rule, about 59 or 60 observations should be in the interval $\left(\overline{y}\pm s\right)$; for the given data set, 63 observations are in this interval.

About 83 or 84 observations should be in the interval $\left(\overline{y}\pm 2s\right)$; for the given data set, 82 observations are in this interval.

Nearly all the observations should be in the interval $\left(\overline{y}\pm 3s\right)$; for the given data set, 87 out of the 88 observations are in this interval.

Thus, it can be said that the observed counts are very close to the predicted counts, according to the empirical rule.

Now, the histogram shows a distribution that is very prominently skewed to the right, that is, a distribution that is certainly not normal. However, the observations conform to the empirical rule, which is applicable for approximately normal distributions. An explanation for such a situation is possibly the large sample size of 88 observations; irrespective of the original distribution of the population, a large sample usually behaves like a normal distribution, at least approximately.