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F. Measures of Central Tendency.

1. The Arithmetic Mean of Ungrouped Data.

2. The Arithmetic Mean of Grouped Data.

3. The Weighted Arithmetic Mean.Problem F1 (In Syllabus Supplement)

4. The Median of Ungrouped Data.

5. The Median of Grouped Data.

6. The ModeExercises 3.1a, 3.2a (3.1a, 3.2a,) Downing and Clark pg 42 (pg 37 in 3rd ed) Problems 6, 7(means, medians and modes only).

7. Other Means.Problem F2.

8. Measures of Position.Downing & Clark, problem 7 above (Use data to find 1st decile, 99th percentile, 1st quintile, and 1st and 3rd quartile), Problems F3, F4.data to find 1st decile, 99th percentile, 1st quintile, and 1st and 3rd quartile), Problems F3, F4.-------------------------------------------------------------------------------------------------------------------------

Problem F1: (Weighted Average)In 1994 the four Scandinavian countries had per capita incomes and populations as follows:

Country Income Population (Millions)

Finland $18850 5.1Sweden $23530 8.8Norway $26390 4.3Denmark $27070 5.2

a. Compute the per capita income of Scandinavians, using population as weights.b. What percent of the population was in each of the four countries? Using these percents as three-place decimals, compute the average income of Scandinavians.

Solution:a.

Finland $18850 5.1 96135Sweden $23530 8.8 207064Norway $26390 4.3 113477Denmark $27070 5.2 140764Total 23.4 557440

We round to whole numbers here because the original

incomes had been rounded to the nearest dollar and there is no way our average could be more accurate.

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b. Finland $18850 .218 4109.30Sweden $23530 .376 8847.28Norway $26390 .184 4855.76Denmark $27070 .222 6009.54Total 1.000 23821.88We could use the same formula here, but, since the sum of the weights is 1, we need not divide.

Because of the accuracy in the original numbers our answer is still 23822.

Exercise 3.1a: The numbers are 7, 4, 9, 8 and 2. Compute the mean, median and mode.

Solution: In order these are 2, 4, 7, 8, 9. is the index of the median, so

According to the Instructor’s Solutions Manual 3.1 (a) Mean = 6 Median = 7 There is no mode.

Exercise 3.2a: The numbers are 7, 4, 9, 7, 3 and 12. Compute the mean, median and mode.

Solution: In order these are 3, 4, 7, 7, 9, 12. is the index of the median. So

According to the Instructor’s Solutions Manual 3.2 (a) Mean = 7 Median = 7 Mode = 7

See next page.

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Downing and Clark, pg. 37, Application 6: Compute the means, medians and modes only from the following data. Class 0 – 1000 12 1000 – 2000 15 2000 – 3000 19 3000 – 4000 22 4000 – 5000 30 5000 – 6000 56 6000 – 7000 48 7000 – 8000 40 8000 – 9000 30 9000 – 10000 1610000 – 11000 411000 - 12000 2

Solution:Class (midpoint) 0 – 1000 500 12 12 6000 1000 – 2000 1500 15 27 22500 2000 – 3000 2500 19 46 47500 3000 – 4000 3500 22 68 77000 4000 – 5000 4500 30 98 135000 5000 – 6000 5500 56 154 308000 6000 – 7000 6500 48 202 312000 7000 – 8000 7500 40 242 300000 8000 – 9000 8500 30 272 255000 9000 – 10000 9500 16 288 15200010000 – 11000 10500 4 292 4200011000 - 12000 11500 2 294 23000Total 294 1680000

(i) So , , so that .

(ii) For the median, , where is the proportion below the median. . If we compare this with the cumulative

frequency column, 147.5 is above 98 and below 154 , so the median group is 5000 - 6000. We then use

the formula to find the median. For the group that we

have chosen, and . So the median is

.

(iii) The mode is 5500, the midpoint of the largest group (the class with the highest frequency), 5000 to 6000.

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Downing and Clark, pg. 37, Application 7: Use the data in columns 1 and 3 below to compute means, medians and modes only.Solution:Class (midpoint) 0 – 10 5 122 122 610 10 – 20 15 180 302 2700 20 – 30 25 256 558 6400 30 – 40 35 350 908 12250 40 – 50 45 311 1219 13995 50 – 60 55 278 1497 15290 60 – 70 65 250 1747 16250 70 – 80 75 211 1958 15825 80 – 90 85 180 2138 15300 90 – 100 95 175 2313 16625100 – 110 105 143 2456 15015110 – 120 115 120 2576 13800120 – 130 125 106 2682 13250130 – 140 135 99 2781 13365140 – 150 145 97 2878 14065150 – 160 155 75 2953 11625Total 2953 196365

(i) So , , so that .

(ii) For the median, , where is the proportion below the median. . If we compare this with the cumulative

frequency column, 1477 is above 1219 and below 1497, so the median group is 50 - 60. We then use the

formula to find the median. For the group that we have

chosen, and . So the median is

.

(iii) The mode is 35, the midpoint of the largest group (the class with the highest frequency), 30 to 40.

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Problem F2: Take the numbers 100, 200 and 300, and compute four different kinds of means.Solution:

a) Arithmetic mean:

b) Geometric mean: (Method i)

(Method ii)

and .

(Method iii)

and

.

c) Harmonic mean:

and

d) Root mean square (RMS):

and .

Downing and Clark, pg. 37, Application 7: Use the first and third columns of the data below to find the 1st decile, 99th percentile, 1st quintile, and 1st and 3rd quartile): Solution:Class (midpoint) 0 – 10 5 122 122 610 10 – 20 15 180 302 2700 20 – 30 25 256 558 6400 30 – 40 35 350 908 12250 40 – 50 45 311 1219 13995 50 – 60 55 278 1497 15290 60 – 70 65 250 1747 16250 70 – 80 75 211 1958 15825 80 – 90 85 180 2138 15300 90 – 100 95 175 2313 16625100 – 110 105 143 2456 15015110 – 120 115 120 2576 13800120 – 130 125 106 2682 13250130 – 140 135 99 2781 13365140 – 150 145 97 2878 14065150 – 160 155 75 2953 11625Total 2953 196365

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251solnF 1/22/08 (Open this document in 'Page Layout' view!) 6


The general formula is Recall that and

For the first decile, and . Using the cumulative distribution column we pick the group 10 – 20 because 295.4 is above 122 and below 302 . (Alternately

. This gives the same result.) For this group,

and . So .

For the 99th percentile, and . (Alternately . This gives the same result.) Using the

cumulative distribution column we pick the group 150 – 160.. For this group,

and . So .

(Corrected by R Bliss)

For the first quintile, and . (Alternately . This gives the same result.) Using the cumulative

distribution column we pick the group 30 – 40. For this group, and .

So .

For the first quartile, and . (Alternately . This gives the same result.) Using the cumulative

distribution column because 738.5 is above 558 and below 908 we pick the group 30 – 40. For this group,

and . So .

For the third quartile, and . (Alternately . This gives the same result.) Using the cumulative

distribution column we pick the group 90 – 100. For this group, and .

So .

Problem F3: Using the numbers below, compute a) and b) . 1, 5, 7, 9, 9, 11, 13, 14, 17, 19

Solution: The basic formulas are and

a) so . So and . And

(Alternately . This gives the same result.)

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b) Since , so . So and . And

(Alternately . So and . And

.) Problem F4: Compute a) and b) from the data below.

Solution: The basic formulas are (Alternately )

and . Note that the cumulative distribution has been added in the column

to the data originally given.a) so . (Alternately

. This gives the same result.) If we look at the column, we find that 50.50 is above 50 and below 60, so that is in the 50 to 59.9 class. In the formula is the frequency of the chosen group, is the cumulative frequency up to but not including the chosen group, is the class interval (the interval between the lower limit of the chosen group and the lower limit of the next group) and is the lower limit of the chosen group.

So

.

b) Since , so . If we look at the column, we find that 99.99 is above 90, so that is in the 90 to 99.9 class. (Alternately

. This gives the same result.) In the formula is the frequency of the chosen group, is the cumulative frequency up to but not including the chosen group, is the class interval (the interval between the lower limit of the chosen group and the

Group 0- 9.9 10 1010-19.9 10 2020-29.9 10 3030-39.9 10 4040-49.9 10 5050-59.9 10 6060-69.9 10 7070-79.9 10 8080-89.9 10 9090-99.9 10 100

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lower limit of the next group) and is the lower limit of the chosen group. So

.

Parts not copied ©2003 Roger Even Bove

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