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## G. Measures of Dispersion and Asymmetry.

### 1. Range

Downing & Clark, problem 7 above (Use data to find IQR). Review solutions and terms on page 41 (36 in 3rd ed.) of Downing & Clark.

### 2. The Variance and Standard Deviation of Ungrouped Data.

Text exercises 3.1b, 3.2b, 3.6, 3.37, 3.24 [3.1b, 3.2b, 3.7, 3.37, 3.23] (3.1b, 3.2b, 3.7, 3.23, 3.33)

### 3. The Variance and Standard Deviation of Grouped Data.

Text exercises 3.28, 3.30 (3.68, 3.70) (work 3.30 in thousands), Downing & Clark pg 42 or 37, problems 6,7 (Find sample standard deviation – hint: run problem 6 in hundreds) (Note that you can use the Excel or Minitab techniques in the graded assignment to compute and sum the and columns in problems 6 and 7.), Problems G1, G2. **Graded Assignment 1**

### 4. Skewness and Kurtosis.

Find the standard deviation, coefficient of variation and measures of skewness in Text problem 3.1, 3.2. Problems G3A, G4 (See 251wrksht).

### 5. Review

### a. Grouped Data.

### b. Ungrouped Data.

Sections 1 and 2 are in this document.

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**Downing and Clark, pg. 37, Application 7(Use data IQR or midrange):**

Solution: From section F we know

.

.

So

Exercise 3.1b: The numbers are 7, 4, 9, 8, 2. Find the range, IQR, variance, standard deviation and coefficient of variation.

Solution: In order these numbers are 2, 4, 7, 8, 9.

Index / / / /1 / 2 / 4 / 2–6 = -4 / 16

2 / 4 / 16 / 4 – 6 = -2 / 4

3 / 7 / 49 / 7 – 6 = 1 / 1

4 / 8 / 64 / 8 – 6 = 2 / 4

5 / 9 / 81 / 9 - 6 = 3 / 9

30 / 214 / 0 / 34

i) Range: .

ii) Variance: First, find the sample mean. . The preferred method for finding the variance is the computational formula, which has the advantage that the andcolumns are not needed. . If we use the definitional formula instead, .

iii) IQR: To find the first quartile so . So and . Thus

. To find the third quartile so . So and . Thus

. Finally, .

iv) Standard deviation:

v) Coefficient of variation: or 48.59%.

Exercise 3.2b: The numbers are 7, 4, 9, 7, 3, 12. Find the range, IQR, variance, standard deviation and coefficient of variation.

Solution: In order these are 3, 4, 7, 7, 9, 12.

Index / / / /1 / 3 / 9 / 3 – 7 = -4 / 16

2 / 4 / 16 / 4 – 7 = -3 / 9

3 / 7 / 49 / 7 – 7 = 0 / 0

4 / 7 / 49 / 7 – 7 = 0 / 0

5 / 9 / 81 / 9 – 7 = 2 / 4

6 / 12 / 144 / 12 - 7 = 5 / 25

42 / 348 / 0 / 54

i) Range: .

ii) Variance: First, find the sample mean. . By the computational method . If we use the definitional formula instead, .

iii) IQR: To find the first quartile so . So and . Thus

. To find the third quartile so . So and . Thus

. Finally, . (A more accurate answer than the text answer)

iv) Standard deviation:

v) Coefficient of variation: or 46.95%.

**Exercise 3.6[3.7 in 9th]:** The numbers are 568, 570, 575, 578, 584 and the numbers are 573, 574, 575, 577, 578.These are the inner diameters of two grades of tires.a) Find the mean, median and standard deviation for each of the two grades. b) Which grade is providing better quality? c) How would your answer change if we change the 578 for grade y to 588?

Theseare, in order as below.

Index / / / /1 / 568 / 322624 / 573 / 328329

2 / 570 / 324900 / 574 / 329476

3 / 575 / 330625 / 575 / 330625

4 / 578 / 334084 / 577 / 332929

5 / 584 / 341056 / 578 / 334084

2875 / 1653289 / 2877 / 1655443

a) Grade x: First, find the sample mean. The median is the middle number, 575 and the variance is . .

Grade y: First, find the sample mean. The median is the middle number, 575 and the variance is ..

According to the Instructor’s Solutions Manual

Grade X Grade Y

Mean575 575.4

Median575 575

Standard deviation 6.402.07

b) According to the Instructor’s Solutions Manual

(b) If quality is measured by the average inner diameter, Grade X tires provide slightly better quality because X’s mean and median are both equal to the expected value, 575 mm. If, however, quality is measured by consistency, Grade Y provides better quality because, even though Y’s mean is only slightly larger than the mean for Grade X, Y’s standard deviation is much smaller. The range in values for Grade Y is 5 mm compared to the range in values for Grade X which is 16 mm.

c) According to the Instructor’s Solutions Manual

(c)Grade X Grade Y, Altered

Mean575 577.4

Median575 575

Standard deviation6.40 6.11

In the event the fifth Y tire measures 588 mm rather than 578 mm, Y’s average inner diameter becomes 577.4 mm, which is larger than X’s average inner diameter, and Y’s standard deviation swells from 2.07 mm to 6.11 mm. In this case, X’s tires are providing better quality in terms of the average inner diameter with only slightly more variation among the tires than Y’s.

**Exercise 3.37 (3.23 in 8th edition):**The data set ‘batteries’ consists of the numbers 342, 426, 317, 545, 264, 451, 1049, 631, 512, 266, 492, 562 and 298. a) Produce a 5-number summary. b) Construct a box-and-whiskers plot and describe the shape.

Solution: The numbers in order are 264 266 298 317 342 426 451 492 512 545 562 631 1049

1 2 3 4 5 6 7 8 9 10 11 12 13

a) The five number summary consists of the lowest value, the first quartile, the median, the third quartile and the highest value.

(i) The lowest value is 264.

(ii) To find the first quartile so . So and . Thus

(iii) For the median and

(iv) To find the third quartile so . So and . Thus .

(v) The highest value is 1049.

According to the Instructor’s Solutions Manual

3.23(a) Five-number summary: 264 307.5 451 553.5 1,049

b) According to the Instructor’s Solutions Manual

(b)

The distribution is right-skewed.

c) In the 9th edition, you are asked to compare the answer to b) with the results of 313(h). According to the Instructor’s Solutions Manual the answer to 3.13h (3.12h) is

3.13(a) Mean = 473.46Median = 451There is no mode.

The median seems to be better descriptive measures of the data, since they are closer to the observed values than is the mean.

(h) The shape of the distribution of the original data is right-skewed, since the mean is larger than the median.

So, looking at the box-and-whisker plot, we can say

(c) Because the data set is small, one very large value (1,049) skews the distribution to the right.

**Exercise 3.24 [3.23 in 9th edition]:**You have data on mutual funds. For 1-year total returns you have a population mean of 8.20, a population standard deviation of 2.75, a range from -2.0 to +17.1, a first quartile of 5.5 and a third quartile of 10.5. a) According to the empirical rule, what percentage of the returns is within 1 standard deviation of the mean? b) within 2 standard deviations of the mean? c) According to the Tchebyschev rule what percentage of the returns should be within 1, 2 or 3 standard deviations of the mean? d) According to the Tchebyschev rule, between what two amounts should there be 93.75% of the returns?

Solution:Note that and . The Empirical rule states that about 67% will be within 1 standard deviations of the mean and about 95% are within 2 standard deviations of the mean. The Tchebyschev rule states . What this means is that for any value of above or equal to 1, we can divide a distribution into two types of areas. The center is the area between and . The tails are the areas below and above. The fraction of the data in the tails cannot exceed and the fraction in the center will be at least . In part d) the fraction within standard deviations of the mean is in part c) They ask where at least 93.75% of the data must be. Since 93.75% is 15/16, we can say , so according to the Instructor’s Solutions Manual the answer to 3.23 is:

3.23(a) 68%(b) 95%(c)not calculable(d) 75%(e) 88.89%

(f) or -2.8 to 19.2

**Exercise 3.33 (in 8th edition):**Note that and . The Empirical rule states that about 67% will be within 1 standard deviations of the mean and about 95% are within 2 standard deviations of the mean. The Chebyschef rule states. What this means is that for any value of above or equal to 1, we can divide a distribution into two types of areas. The center is the area between and . The tails are the areas below and above. The fraction of the data in the tails cannot exceed and the fraction in the center will be at least . In part d) the fraction within standard deviations of the mean is in part c) They ask where at least 93.75% of the data must be. Since 93.75% is 15/16, we can say , so and So According to the Instructor’s Solutions Manual the answer to 3.33 is:

3.33(a) 67%

(b) 90%-95%

(c) not calculable

(d) 75%

(e) 88.89%

(f) or 1.2 to 55.2

Parts not copied ©2003 Roger Even Bove