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Histograms

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Histograms

I. Summary



A histogram represents percents by area. It consists of a set of blocks, and the area of

each block represents the percent of cases in the corresponding class interval. With a

density scale, the height of each block equals the percent of cases in the corresponding

class interval divided by the length of that interval. With a density scale, the total area

of the histogram is 100%.



II. Examples



1. The following are 25 final averages in a math class:



46 64 72 79 89

49 66 74 79 91

53 66 75 80 94

60 67 76 83 95

61 71 79 88 98



Draw the histogram using the following class intervals:



A : 92  x  100

B : 80  x  92

C : 64  x  80

D : 52  x  64

F : 44  x  52



Class Number in % in Width of Height of

interval interval interval interval block



A 3 3  25  12% 100-92=8 12%  8  1.5% / Pt

B 5 5  25  20% 92-80=12 20% 12  1.7% / Pt

C 12 12  25  48% 80-64=16 48% 16  3% / Pt

D 3 3  25  12% 64-52=12 12% 12  1% / Pt

F 2 2  25  8% 52-44=8 8%  8  1% / Pt

25 100%









1

%/Pt

(3.0)

3.0



2.5



2.0 (1.7)

(1.5)

1.5

(1.0) (1.0)

1.0



.5



0

44 52 64 80 92 100

Final average (pts)





2. The following histogram was plotted showing the number of cigarettes per

day smoked by each subject. The class intervals include the right endpoint,

not the left. The block over the class interval 20 to 40 is missing.



%/cig



4

(3.0)

3



2

(1.0)

1 (.5)



0

0 10 20 40 80

Number of cigarettes



What is the height of the missing block?









2

Class Width of Height of % in

interval interval block interval

0  x  10 10  0  10cig 1% / cig (10cig )(1% / cig )  10%

10  x  20 20  10  10cig 3% / cig (10cig)(3% / cig)  30%

20  x  40 40  20  20cig ? ??

40  x  80 80  40  40cig 0.5% / cig (40cig )(0.5% / cig )  20%



Since the total percent in all the blocks is 100%, the percent in the class interval

20  x  40 must be 40% [40%  100%  (10%  30%  20%)]. Thus,

(20cig )(?)  40%  height of the missing block  40 % 20 cig  2% / cig.







3. In a small town, the number of children in each family was tabulated with the

following results:



Number of children Percent of families having

in family indicated number of children



0 20%

1 25%

2,3 40%

4,5,6 15%



Draw the histogram.



Class interval % Width of interval Height of block



 .5  x  .5 20% .5  (.5)  1 20% 1child  20% / child

.5  x  1.5 25% 1.5  .5  1 25% 1child  25% / child

1.5  x  3.5 40% 3.5  1.5  2 40%  2children  20% / child

3.5  x  6.5 15% 6.5  3.5  3 15%  3children  5% / child





%/child (25)

25 (20) (20)

20

15

10 (5)

5

0

0 1 2 3 4 5 6

Number of children



3

Practice Sheet – Histograms

I. The following table gives the distribution of final averages in a math class:



Class interval Number in interval

A : 93  x  100 5

B : 85  x  93 12

C : 78  x  85 31

D : 70  x  78 18

F : 50  x  70 4



Draw the histogram for this data.



II. The following histogram was plotted, using the results from a final exam. The

class interval include the left endpoint. The block over the class interval 65 to 80

is missing.



%/Pt

4



3



2 (1.5)

(1.0) (1.0)

1 (.5)



0

20 50 65 80 90 100

Points on final exam



What is the height of the missing block?



III. Four fair coins are tossed and the number of heads is recorded. This process is done

50 times with the following results:



Number of heads Frequency

0 2

1 13

2 22

3 10

4 3



Draw the histogram for this data.



4

Solution Key for Histograms

I. %/Pt (6.3)



6



5



4

(3.3)

3

(2.1)

2



1 (1)

(.3)

0



50 70 78 85 93 100

Final average (pts)



II. Height = 3.0%/Pt



III. %/Head



50 (44)



40



30 (26)

(20)

20



10 (6)

(4)

0

0 1 2 3 4

Number of heads









5



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