Chapter 3, Problem 30CE

Below are batting averages of the New York Yankees players who were at bat five times or more in 2006. (a) Construct a frequency distribution. Explain how you chose the number of bins and the bin limits. (b) Make a histogram and describe its appearance. (c) Repeat, using a different number of bins and different bin limits. (d) Did your visual impression of the data change when you changed the number of bins? Explain.

Batting Averages for the 2006 New York Yankees

To determine

Construct a frequency distribution and explain how the number of bins and the bin limits are chosen.

Answer to Problem 30CE

Thefrequency distribution is tabulated below,

Bin limits		Mid point	Width	Frequency $f$	Percent	Cumulative
Lower	Upper					Frequency	Percent
0	< 0.07	0.035	0.07	1	4	1	4
0.07	< 0.14	0.105	0.07	0	0	1	4
0.14	< 0.21	0.175	0.07	4	16	5	20
0.21	< 0.28	0.245	0.07	9	36	14	56
0.28	< 0.35	0.315	0.07	10	40	24	96
0.35	< 0.42	0.385	0.07	1	4	25	100
Total

Explanation of Solution

Calculation:

The given information is that, the data representsthebatting averages of the New York Yankees players who were at bat five times or more in 2006.

Frequency distribution:

It is a tabulation of n data values which are divided into k classes called bins. The bin limits are the cutoff points which defines each bin. These generally have equal interval and the limits do not overlap.

Step-by-step procedure to construct frequency distribution table is as follows:

• • The smallest and largest data values are 0 and 0.417.
• • Here the sample size is 25. By Sturge’s Rule, $k=1+3.3\log \left(n\right)$

  $\begin{array}{c}k=1+3.3\log \left(25\right)\\ =1+4.6132\\ =5.61\\ \approx 6\end{array}$

• • Bin width is obtained by dividing the range by the number of bins.

  $\begin{array}{c}\text{Bin}\text{width}=\frac{x_{\max }-x_{\min }}{k}\\ =\frac{0.417-0}{6}\\ =0.0695\\ \approx 0.07\end{array}$

Hence, the bin width is 0.07.

• • The minimum value in the data is 0 hence the first bin should start at 0.

Tally mark:

• • Make a tally mark for each score in the corresponding class and continue for all reading times in the data.
• • The number of tally marks in each class represents the frequency, f of that class.

Thus, the frequency distribution table is as follows:

Bin limits		Tally	Frequency $f$	Percent
Lower	Upper
0	< 0.07	$|$	1	$\frac{1}{25}\times 100=4$
0.07	< 0.14		0	$\frac{0}{25}\times 100=0$
0.14	< 0.21	$\left|\right|\left|\right|$	4	$\frac{4}{25}\times 100=16$
0.21	< 0.28	$\cancel{\left|\right|\left|\right|}\left|\right|\left|\right|$	9	$\frac{9}{25}\times 100=36$
0.28	< 0.35	$\cancel{\left|\right|\left|\right|}\cancel{\left|\right|\left|\right|}$	10	$\frac{10}{25}\times 100=40$
0.35	< 0.42	$|$	1	$\frac{1}{25}\times 100=4$
Total			25

Mid point:

The midpoint is the average of the lower limit and upper limit of a particular class. It is also called as class mark.

$\text{Mid}\text{point}=\frac{\left(\text{Lower}\text{class limit+Upper}\text{class}\text{limit}\right)}{2}$

Thus, the mid points for each class is tabulated below:

Bin limits		Frequency $f$	Mid point
Lower	Upper
0	< 0.07	1	$\frac{0+0.07}{2}=0.035$
0.07	< 0.14	0	$\frac{0.07+0.14}{2}=0.105$
0.14	< 0.21	4	$\frac{0.14+0.21}{2}=0.175$
0.21	< 0.28	9	$\frac{0.21+0.28}{2}=0.245$
0.28	< 0.35	10	$\frac{0.28+0.35}{2}=0.315$
0.35	< 0.42	1	$\frac{0.42+0.49}{2}=0.455$
Total		25

Cumulative frequency:

Cumulative frequency is the sum of all frequencies up to that class. The last class’s cumulative frequency is equal to the sample size $n$.

Thus, the cumulative frequency for each calss is tabulated below:

Bin limits		Frequency $f$	Cumulative frequency
Lower	Upper
0	< 0.07	1	1
0.07	< 0.14	0	$1+0=1$
0.14	< 0.21	4	$1+4=5$
0.21	< 0.28	9	$5+9=14$
0.28	< 0.35	10	$14+10=24$
0.35	< 0.42	1	$24+1=25$
Total		25

Cumulative Relative frequency:

Bin limits		Cumulative frequency	Cumulative percent
Lower	Upper
0	< 0.07	1	$\frac{1}{25}\times 100=4$
0.07	< 0.14	1	$\frac{1}{25}\times 100=4$
0.14	< 0.21	5	$\frac{5}{25}\times 100=20$
0.21	< 0.28	14	$\frac{14}{25}\times 100=56$
0.28	< 0.35	24	$\frac{24}{25}\times 100=96$
0.35	< 0.42	25	$\frac{25}{25}\times 100=100$
Total

To determine

Constuct a histogram and describe its appearance.

Answer to Problem 30CE

The histogram for the data is as follows,

Explanation of Solution

Calculation:

Software procedure:

Step-by-step procedure to obtain histogram using MINITAB software is as follows,

• • Choose Graph > Histogram.
• • Choose Simple, and then click OK.
• • In Graph variables, enter the corresponding column of Yankees.
• • Click Scale>Y-Scale Type > Percent
• • Click OK.
• • To modify the interval settings, double click on the horizontal axis of the graph. Then, select Binning>Cutpoint>Cutpoint Positions, in this box, enter the values for the cut points of the bin intervals (0, 0.07, 0.14, 0.21, 0.28, 0.35 and 0.42).

From the histogram it is observed that the tail on the left is more elongated than the tail on the right. Hence the distribution of the histogram seems to be skewed left.

To determine

Repeat the procedure using different number of bins and different bin limits.

Answer to Problem 30CE

Thefrequency distribution using 5 bins is as follows,

Bin limits		Mid point	Width	Frequency $f$	Percent	Cumulative
Lower	Upper					Frequency	Percent
0	< 0.1	0.05	0.1	1	4	1	4
0.1	< 0.2	0.15	0.1	3	12	4	16
0.2	< 0.3	0.25	0.1	15	60	19	76
0.3	< 0.4	0.35	0.1	5	20	24	96
0.4	< 0.5	0.45	0.1	1	4	25	100
Total

The histogram for 5 bins with 0.1 width is as follows,

Thefrequency distribution using 0.15 width and 9 bins is as follows,

Bin limits		Mid point	Width	Frequency $f$	Percent	Cumulative
Lower	Upper					Frequency	Percent
0	< 0.05	0.025	0.05	1	4	1	4
0.05	< 0.10	0.075	0.05	0	0	1	4
0.10	< 0.15	0.125	0.05	1	4	2	8
0.15	< 0.20	0.175	0.05	2	8	4	16
0.20	< 0.25	0.225	0.05	6	24	10	40
0.25	< 0.30	0.275	0.05	9	36	19	76
0.30	< 0.35	0.325	0.05	5	20	24	96
0.35	< 0.40	0.375	0.05	0	0	24	96
0.45	< 0.45	0.425	0.05	1	4	25	100
Total

The histogram for 9 bins with width 0.05 is as follows,

Explanation of Solution

Calculation:

For 5 bins:

Here the number of bins are considered as 5 with width 0.1.

Bin limits		Tally	Frequency $f$	Percent
Lower	Upper
0	< 0.1	$|$	1	$\frac{1}{25}\times 100=4$
0.1	< 0.2	$\left|\right||$	3	$\frac{3}{25}\times 100=12$
0.2	< 0.3	$\cancel{\left|\right|\left|\right|}\cancel{\left|\right|\left|\right|}\cancel{\left|\right|\left|\right|}$	15	$\frac{15}{25}\times 100=60$
0.3	< 0.4	$\cancel{\left|\right|\left|\right|}$	5	$\frac{5}{25}\times 100=20$
0.4	< 0.5	$|$	1	$\frac{1}{25}\times 100=4$
Total			25

Mid point:

The midpoint is the average of the lower limit and upper limit of a particular class. It is also called as class mark.

$\text{Mid}\text{point}=\frac{\left(\text{Lower}\text{class limit+Upper}\text{class}\text{limit}\right)}{2}$

Thus, the mid points for each class is tabulated below:

Bin limits		Frequency $f$	Mid point
Lower	Upper
0	< 0.1	1	$\frac{0+0.1}{2}=0.05$
0.1	< 0.2	3	$\frac{0.1+0.2}{2}=0.15$
0.2	< 0.3	15	$\frac{0.2+0.3}{2}=0.25$
0.3	< 0.4	5	$\frac{0.3+0.4}{2}=0.35$
0.4	< 0.5	1	$\frac{0.4+0.5}{2}=0.45$
Total		25

Cumulative frequency:

Cumulative frequency is the sum of all frequencies up to that class. The last class’s cumulative frequency is equal to the sample size $n$.

Thus, the cumulative frequency for each calss is tabulated below:

Bin limits		Frequency $f$	Cumulative frequency
Lower	Upper
0	< 0.1	1	1
0.1	< 0.2	3	$1+3=4$
0.2	< 0.3	15	$15+4=19$
0.3	< 0.4	5	$19+5=24$
0.4	< 0.5	1	$24+1=25$
Total		25

Cumulative Relative frequency:

Bin limits		Cumulative frequency	Cumulative percent
Lower	Upper
0	< 0.1	1	$\frac{1}{25}\times 100=4$
0.1	< 0.2	4	$\frac{4}{25}\times 100=16$
0.2	< 0.3	19	$\frac{19}{25}\times 100=76$
0.3	< 0.4	24	$\frac{24}{25}\times 100=96$
0.4	< 0.5	25	$\frac{25}{25}\times 100=100$
Total

Software procedure:

Step-by-step software procedure to obtain histogram using MINITAB software is as follows,

• • Choose Graph > Histogram.
• • Choose Simple, and then click OK.
• • In Graph variables, enter the corresponding column of Yankees.
• • Click Scale>Y-Scale Type> Percent
• • Click OK.
• • To modify the interval settings, double click on the horizontal axis of the graph. Then, select Binning>Cutpoint>CutpointPositions,in this box, enter the values for the cut points of the bin intervals (0, 0.1, 0.2, 0.3, 0.4 and 0.5).
• For 9 bins:

Here the number of bins are considered as 9 with width 0.05.

The frequeny table is as follows,

Bin limits		Tally	Frequency $f$	Percent
Lower	Upper
0	< 0.05	$|$	1	$\frac{1}{25}\times 100=4$
0.05	< 0.10		0	$\frac{0}{25}\times 100=0$
0.10	< 0.15	$|$	1	$\frac{1}{25}\times 100=4$
0.15	< 0.20	$\left|\right|$	2	$\frac{2}{25}\times 100=8$
0.20	< 0.25	$\cancel{\left|\right|\left|\right|}|$	6	$\frac{6}{25}\times 100=24$
0.25	< 0.30	$\cancel{\left|\right|\left|\right|}\left|\right|\left|\right|$	9	$\frac{9}{25}\times 100=36$
0.30	< 0.35	$\cancel{\left|\right|\left|\right|}$	5	$\frac{5}{25}\times 100=20$
0.35	< 0.40		0	$\frac{0}{25}\times 100=0$
0.45	< 0.45	$|$	1	$\frac{1}{25}\times 100=4$
Total			25

Mid point:

The midpoint is the average of the lower limit and upper limit of a particular class. It is also called as class mark.

$\text{Mid}\text{point}=\frac{\left(\text{Lower}\text{class limit+Upper}\text{class}\text{limit}\right)}{2}$

Thus, the mid points for each class is tabulated below:

Bin limits		Frequency $f$	Mid point
Lower	Upper
0	< 0.05	1	$\frac{0+0.05}{2}=0.025$
0.05	< 0.10	0	$\frac{0.05+0.10}{2}=0.075$
0.10	< 0.15	1	$\frac{0.10+0.15}{2}=0.125$
0.15	< 0.20	2	$\frac{0.15+0.20}{2}=0.175$
0.20	< 0.25	6	$\frac{0.20+0.25}{2}=0.225$
0.25	< 0.30	9	$\frac{0.25+0.30}{2}=0.275$
0.30	< 0.35	5	$\frac{0.30+0.35}{2}=0.325$
0.35	< 0.40	0	$\frac{0.35+0.40}{2}=0.375$
0.40	< 0.45	1	$\frac{0.40+0.45}{2}=0.425$
Total		25

Cumulative frequency:

Cumulative frequency is the sum of all frequencies up to that class. The last class’s cumulative frequency is equal to the sample size $n$.

Thus, the cumulative frequency for each calss is tabulated below:

Bin limits		Frequency $f$	Cumulative frequency
Lower	Upper
0	< 0.05	1	1
0.05	< 0.10	0	$1+0=1$
0.10	< 0.15	1	$1+1=2$
0.15	< 0.20	2	$2+2=4$
0.20	< 0.25	6	$4+6=10$
0.25	< 0.30	9	$10+9=19$
0.30	< 0.35	5	$19+5=24$
0.35	< 0.40	0	$24+0=24$
0.40	< 0.45	1	$24+1=25$
Total		25

Cumulative Relative frequency:

Bin limits		Cumulative frequency	Cumulative percent
Lower	Upper
0	< 0.05	1	$\frac{1}{25}\times 100=4$
0.05	< 0.10	1	$\frac{1}{25}\times 100=4$
0.10	< 0.15	2	$\frac{2}{25}\times 100=8$
0.15	< 0.20	4	$\frac{4}{25}\times 100=16$
0.20	< 0.25	10	$\frac{10}{25}\times 100=40$
0.25	< 0.30	19	$\frac{19}{25}\times 100=76$
0.30	< 0.35	24	$\frac{24}{25}\times 100=96$
0.35	< 0.40	24	$\frac{24}{25}\times 100=96$
0.40	< 0.45	25	$\frac{25}{25}\times 100=100$
Total

Software procedure:

Step-by-step software procedure to obtain histogram using MINITAB software is as follows,

• • Choose Graph > Histogram.
• • Choose Simple, and then click OK.
• • In Graph variables, enter the corresponding column of Yankees.
• • Click Scale>Y-Scale Type> Percent
• • Click OK.
• • To modify the interval settings, double click on the horizontal axis of the graph. Then, select Binning>Cutpoint>CutpointPositions,in this box, enter the values for the cut points of the bin intervals (0, 0.05, 0.10, 0.15, 0.20, 0.25, 0.30, 0.35, 0.40 and 0.45).

To determine

Explain whether the visual impression of the data change when the number of bins increased or not

Answer to Problem 30CE

Yes, the visual impression of the data changes when the number of bins are varied.

Explanation of Solution

From the histograms in part (c), it is observed that when the bins are 5 the histogram appears to be more symmetrical. And when the bins are 9 then, the histogram appears to be skewed left. Hence there is change the appearance of the histogram when the number of bins vary.