After studying the chapter Statistics[Class 9] you will be able to calculate range and coefficient of range for given data; to find out quartile deviation for regrouped and grouped data; to calculate mean deviation for regrouped data and grouped data; to draw histogram of varying width for inclusive and exclusive class intervals; to draw commutative frequency curve and identity quartile and median n it; to construct frequency polygons for inclusive and exclusive class intervals; to identify random experiment and types of probability.
1.5.1 Introduction – Statistics[Class 9]
The collection of numerical facts with particular information during experiment is called data. These data can be grouped into a table that displays frequencies of scores corresponding to various class interval. While grouping the data, if the end points o the groups do not overlap we call it inclusive method if grouping the data and if the end points of consecutive groups overlap, we calling it exclusive method.
Range – Statistics[Class 9]:
The difference between the highest and the lowest scores in a given distribution is called range.
Mean – Statistics[Class 9]:
It is the average of the scores, which s equal to the one of the scores divided by the number of scores.
For ungrouped data, the mean is calculated using the formula,
Ẍ = ^{⅀}^{x}/_{N}
For grouped data, the mean is given by Ẍ = ^{⅀}^{fx}/_{N}
Median – Statistics[Class 9]:
Median is the middle most score in a given set of scores. In ungrouped data, median is the middle score( when the scores are odd) or the average of two middle scores (when the scores are even), after the scores being arranged in ascending or descending order.
Median for grouped data is calculated using the formula,
Mode – Statistics[Class 9]:
Mode is the score that occurs frequently in a given set of scores. Most repeated score in a ungrouped data is the mode. Mode of the value around which the other scores cluster around densely. In a grouped data, the scores corresponding to the maximum frequency is the mode.
A collection of data can have more than one mode. If the data has only one mode, we say it has Uni mode, if it has 2 modes, we say it has bi mode and it has more than 3 modes, we say it has multimode.
1.5.2 Measures of dispersion – Statistics[Class 9]:
There are four measures of dispersion viz.
 range(R)
 Quartile Deviation(QD)
 Mean Deviation(MD)
 Standard deviation(SD)
(a) Range – Statistics[Class 9]:
To understand the range, consider the following set of data:
24, 52, 35, 28, 49, 21
Find out the highest and the lowest scores. Have you observed the highest scores is 52 and the lowest is 21? Take the differenece of these two scores. It is 52 – 21 = 31. What is the difference called? This difference os called range.
Example : Calculate the range from the following data:
Marks  26  38  54  65  72  88 
No. of students  5  10  15  20  25  30 
Solution:
We observe that, highest scores H = 88 and lowest scores L = 26
Therefore, range H – L = 88 – 26 = 62 marks
Range is the simplest measure of dispersion. The difference between the highest and the lowest scores of distribution is called range.
Range = Highest Score (H) – Lowest Score(L)
(b) Coefficient of range[Class 9]:
Consider the following example:
The wages of six workers of a factory in rupees are:
1600, 1500, 1750, 1800, 1250, 1400
What is the highest and the lowest wages? The highest is 1800 and the lowest is 1250. Let us calculate the ratio of the difference of the highest and the lowest wages to its sum. It is
^{H }^{–}^{ L}/_{H + L} = ^{1800 }^{–}^{ 1250}/_{1800 + 1250} = ^{550}/_{3050} = 0.18 (approximately)
Coefficient of range is a relative measure of dispersion and it is based on the value of range. It is also called the range coefficient of dispersion.
Coefficient of range is given by = ^{H }^{–}^{ L}/_{H + L}
Example : Calculate the coefficient of range for the following data:
No. of wards  1  2  3  4  5  6  7  8 
No. of houses  32  57  28  96  138  90  66  58 
Solution:
Here, H = 8 ; L = 1
Hence, coefficient of range = ^{H }^{–}^{ L}/_{H + L} = ^{81}/_{8+1} = ^{7}/_{8}
Merits and demerits of range[Class 9]:
Merits:
 It is the simplest measure of dispersion and easy to calculate.
 It does not require special knowledge to understand.
 Its calculation takes less time.
Demerits:
 It does not take into account all the scores/items of distribution.
 It is affected by extreme scores.
 It does not indicate the direction of variability.
(c) Quartile Deviation – Statistics[Class 9]:
The points that divide the distribution in to four equal parts are called quartiles. If we take the difference between the third quartile and the first quartile, it gives us a value called inter quartile range. It is equal to
Q_{3} – Q_{1}. Half of this is the SemiInter Quartile Range or quartile deviation which is (Q_{3} – Q_{1})/_{2} . Quartile deviation is also called the semiinter quartile range.
(d) Quartile Deviation for ungrouped data:
Example: The runs scored by a batsman in five innings are 28, 60, 85, 58, 74, 20, 90. Find Q_{1}, Q_{2}, Q_{3} and quartile deviation.
Solution:
Arranging the scores in ascending we get, 28, 60, 85, 58, 74, 20, 90.
There are 7 scores and n = 7
 First quartile(Q_{1}) = ^{n+1}/_{4} th score = ^{7+1}/_{4} = 2^{nd} score = 28
 Median(Q_{2}) = ^{n+1}/_{2} th score = ^{7+1}/_{2} = 4^{th} score = 60
 Third quartile(Q_{3}) = ^{3(n+1)}/_{4} = ^{3(7+1)}/_{4} th score = 6^{th} score = 85
 Quartile deviation = ^{Q3 }^{–}^{ Q1}/_{2} = ^{85 }^{–}^{ 28}/_{2} = 28.5 ≅ 29
(e) Quartile deviation for grouped data – Statistics[Class 9]:
Example: Find the median and quartile deviation for the following data:
x  3  4  5  6  7 
f  12  35  52  41  18 
Solution:
Let us find the commutative frequency for the data given:
x  f  Commutative frequency  Remarks 
3  1  12  12, the first 12 scores correspond to x = 3 
4  35  47  (12 + 35) from 13^{th} to 47^{th} scores correspond to x = 5 
5  52  99  (47+52) from 48^{th} to 99^{th} scores correspond to x = 5 
6  41  140  (99+41) from 100^{th} to 140^{th} scores correspond to x = 6 
7  18  158  (140+18) from 141th to 158^{th} scores correspond to x = 7 
Observe that, n = 158, the last commutative frequency.
Median = ^{n+1}/_{2} th score = 79.5^{th} score ≅ 80^{th} score
From the column f_{c}, from 48^{th} to 99^{th} score corresponds to x = 5. Hence 80^{th} position = 5. Therefore, median = 5
To find the Quartile Deviation:
Q_{1} = ^{n}/_{4} th score = ^{158}/_{4} th score = 39.5 i.e., 40^{th} score
From column f_{c}, 13^{th} to 47^{th} scores correspond to x = 4. Hence the 40^{th} position is 4. Therefore,
Q_{1} = 4
Similarly, ^{3n}/_{4} = ^{3×158}/_{4} = 118.5
Hence, Q_{3} is 119^{th} score. The column for f_{c} shows that 100^{th} to 140^{th} scores correspond to x = 6. Therefore, Q_{3} = 6.
Quartile Deviation = ^{Q3 }^{–}^{ Q1}/_{2} = ^{6 }^{–}^{ 4}/_{2} = 1 mark.
(e) Quartile deviation for grouped data with class intervals:
Example: The heights of 100 students in 9^{th} standard are given below:
Height(cm)  100 – 110  110 – 120  120 – 130  130 – 140  140 – 150  150 – 160 
No. of Students(f)  10  12  16  30  12  20 
Find quartile deviation.
Solution:
Step 1: First let us find the commutative frequency corresponding to the frequencies given.
Height(cm)  f_{1}  f_{c} 
100 – 110  10  10 
110 – 120  12  22 
120 – 130  16  38(Q_{1} Class) 
130 – 140  30  68 
140 – 150  12  80 
150 – 160  20  100(Q_{3} Class) 
N = 100 
Step 2: To find Q_{1} : Recall the formula to find the median
where LRL = lower classs limit, f_{c} = commutative frequency just above the median class f_{m} = frequency corresponding the median class and I = size of the class interval. Now, to find Q_{1} replace ^{N}/_{2} by ^{N}/_{4} in the formula for median. Thus we get,
Now, find out, ^{N}/_{4} :^{ N}/_{4} = ^{100}/_{4} = 25
Locate 25 in the commutative frequency column. This corresponds to the CI 120 – 130. This is Q_{1} class.
From this class, LRL = 120, f_{c} = 22 , f_{m} = 16 and I = 10. Substituting these values in Q_{1} , we get,
= 120 + (0.1875 x 10)
= 120 + 1.875
= 121.88
Thus, Q_{1} ≅ 122
Step 3: To find Q_{3}:
In the median formula, replace, ^{N}/_{2} by ^{3N}/_{4} and follow the same steps as in step 2.
Here, ^{3N}/_{4} = ^{3×100}/_{4} = 75. In the commutative frequency column 75 corresponds to class interval 140 – 150. Therefore,
LRL = 140, f_{c} = 68 , f_{m} = 12 and I = 10
= 140 + (0.58 x 10)
= 120 + 5.8
= 145.8
Thus, Q_{3} ≅ 146
Step 4: Now, we can find quartile deviation using the formula:
Quartile deviation = ^{Q3 – Q1}/_{2} = ^{146 – 122}/_{2} = 12.
Statistics Exercise 1.5.3
 Calculate the range and coefficient of range from the following data.
a) The heights of 10 children in cm: 122, 144, 154, 101, 168, 118, 155, 133, 160, 140
Solution:
Heights of 10 children in cm: 122, 144, 154, 101, 168, 118, 155,133, 160,140.
Range: H – L = 168 – 101 = 67
Coefficient of Range: ^{H – L}/_{H + L}
= ^{67}/_{168 + 101} = ^{67}/_{269} = 0.249
b) Marks scored by 12 students in a test: 31, 18, 27, 19, 25, 28, 49, 14, 41, 22, 33, 13.
Solution:
Marks scored by 12 students
31, 18, 27, 19, 25, 28, 49, 14, 41, 22, 33, 13.
H = 49; L = 13
Range: H – L = 49 – 13 = 36
Coefficient of Range: ^{H – L}/_{H + L} = ^{36}/_{49 + 13} = ^{36}/_{62} = 0.58.
c) Number of trees planted in 6 months: 186, 234, 465, 361, 290, 142.
Solution:
No .of trees planted in 6 months:
186, 234, 465, 361, 290, 142.
H = 465; L = 142
Range: H – L = 465 – 142 = 323
Coefficient of Range: ^{H – L}/_{H + L} = ^{323}/_{465+142} = ^{323}/_{607} = 0.532
 State quartile deviation for the following data:
a) 30, 18, 23, 15, 11, 29, 37, 42, 10, 21.
Solution:
Arrange the scores in ascending order:
n=10,
10, 11, 15, 18, 21, 23, 29, 30, 37, 42.
(a) First Q1 = ^{n+1}/_{4}
= ^{10+1}/_{4} = ^{11}/_{4} = 2.75 = 3rd score = 15
(ii) Third Quartile
Q3 = ^{3(n+1)}/_{4} = ^{3×11}/_{4} = ^{33}/_{4} = 8.25 = 9th = 37
(iii) Quartile Deviation:
= (Q3−Q1)/_{2} = ^{37−15}/_{2} = ^{22}/_{2} = 11
b) 3, 5, 8, 10, 12, 7, 5.
Solution:
3,5,5,7,8,10,12
n = 7
(i) Quartile Q1 = ^{n+1}/_{4} = ^{7+1}/_{4} = ^{8}/_{4} = 2nd score
Q1 = 5
(ii) Quartile Q3 = 3^{(}^{n + 1)}/_{4} = 3[(8)/_{4}] = 6th score
Q3 = 10
(iii) Quartile Deviation:
= [Q3 − Q1]/_{2} = ^{10 – 5}/_{2} = ^{5}/_{2} = 2.5
(c)
Age  3  6  9  12  15 
No. of children  4  8  11  7  12 
Solution:
x  f  commutative frequency 
3  4  4 
6  8  12 
9  11  23 
12  7  30 
15  12  42 
n = 42
Q1 = ^{n}/_{4} ; score = ^{42}/_{4} = 10.5;11th score
From fc ∴Q1 = 6
Q3 = ^{3n}/_{4} = ^{3 ×42}/_{4} = 31.5; 32nd score ∴Q3 = 15
Q.D = [Q3−Q1]/_{2} = ^{15−6}/_{2} = ^{9}/_{2} = 4.5
d)
Marks scored  10  20  30  40  50  60 
No. of students  12  7  16  08  18  22 
Solution:
x  f  fc 
10  12  12 
20  07  19 
30  16  35 
40  08  43 
50  18  61 
60  22  83 
n = 83
Q1= ^{n}/_{4} = ^{83}/_{4} = 20.75; 21st score
∴Q1 = 30
Q3 = ^{3n}/_{4} = ^{3 ×83}/_{4} = 3X20.75 = 62.25; 62nd score
∴Q3 = 60
Q.D = [Q3−Q1]/2 = ^{60 – 30}/_{2} = ^{30}/_{2} = 15
∴QD = 15
 Compute quartile deviation for each of the following tables.
a)
Class interval  Frequency  fc 
5 – 15  11  11 
15 – 25  5  16 
25 – 35  15  31 
35 – 45  9  40 
45 – 55  22  62 
55 – 65  8  70 
65 – 75  17  87 
Solution:
n = 87
Q1= ^{n}/_{4} = ^{87}/_{4} = 21.75
22nd score CI = 25 – 35
∴LRC = 25
fc = 31; i = 10
Q2 = LRL +( [^{N}/_{4}−fc]/_{fm})i
= 25 + [(^{87}/_{4}−16)/_{15}] 10
= 25 + [^{21.75 −16}/_{15}] ×10
Q2 = 28.83
Q3 = LRL + [(^{3N}/_{4} – fc)/_{fm}]*i
^{3N}/_{4} = ^{3 × 87}/_{4} = 65.25 class interval 55 – 65
L = 55, fc = 62, fm = 8, CI = 10
LRL = 55 +(^{65.25 – 62})/_{8} ×10
= 55 + 4.06
LRL = 59.06
Quartile Deviation = [Q3 − Q1]/_{2}
= ^{59.06 – 28.83}/_{2}
= ^{30.23}/_{2}
Q.D = 15.11
(b)
class interval  frequency  fc 
1 – 9  4  4 
10 – 19  3  7 
20 – 29  20  27 
30 – 39  12  39 
40 – 49  5  44 
50 – 59  8  52 
60 – 69  14  66 
70 – 79  27  93 
80 – 89  2  95 
90 – 99  5  100 
H = 100
Solution:
n = 100
^{100}/_{4} = 25th Score 20 – 29; LRL = 19.5
Fc = 7; fm = 20
Q1 = LRL+[ (^{N}/_{4}−fc)/_{fm}]*i
= 19.5 + [^{25 – 7}/_{20}]× 10
= 19.5 + ^{18}/_{20}*10
Q1 = 28.5
^{3N}/_{4} = ^{3 }^{× }^{100}/_{4} = 3 × 25 = 75th score cl 70 – 79
LRL = 69.5; fc = 66; fm = 14
Q3 = 69.5 + [^{(}^{75 – 66)}/_{14}] × 10
= 69.5 + 3.33
Q3 = 72.83
Quartile Deviation = (Q3 − Q1)/_{2}
= ^{72.83 – 28.5} /_{2}
= ^{44.33}/_{2}
Q.D = 22.16
1.5.3 Mean Deviation – Statistics[Class 9]:
Calculation of mean deviation for ungrouped data about median[Class 9]:
Consider the following set of scores:
15, 11, 13,20, 26, 18, 21
arranging scores in ascending order:
11, 13, 15, 18, 20, 21, 26.
Here we have 7 scores which is odd.
Therefore, median = (^{N+1}/_{2}) th score = (^{7+1}/_{2})th score = (^{8}/_{2}) th score = 4^{th} score = 18
Now let us find the deviation of each score from the median. the deviation D = score(X) – median.
Let us take only positive value of D. The positive value of D is called absolute value and denoted by D.
Scores(X)  Deviations from median
D = X – median 
D 
11  11 – 18 = 7  7 
13  13 – 18 = 5  5 
15  15 – 18 = 2  2 
18  18 – 18 = 0  0 
20  20 – 18 = 2  2 
21  21 – 18 = 3  3 
26  26 – 18 = 8  8 
N = 7  ⅀D = 28 
Add all D and divide it by the total number of scores. It gives you mean deviation.
i.e.,
^{⅀}^{D}/_{N} = ^{28}/_{7} = 4
Therefore, mean deviation = 4
Calculation f mean deviation for ungrouped data about mean:
Example 10: Calculate the mean deviation from the mean for the scores given below:
15, 11, 13, 20, 26, 18, 21
Solution:
Arrange the scores in an order:
11, 13, 15, 18, 20, 21, 26
We known mean = ^{sum of all the scores}/_{number of scores} = ^{⅀}^{x}/_{N}
= ^{11+13+15+18+20+21+26}/_{7} = ^{124}/_{7} ≅17.7
Now calculate the deviation D of each score from mean and find out ⅀D.
Scores(X)  Deviations from mean
D = X – mean 
D 
11  11 17.7 = 6.7  6.7 
13  13 17.7 = 4.7  4.7 
15  15 17.7 = 2.7  2.7 
18  18 17.7 = 0.3  0.3 
20  20 17.7 = 2.3  2.3 
21  21 17.7 = 3.3  3.3 
26  26 17.7 = 8.3  8.3 
N = 7  ⅀D = 28.3 
Now mean deviation from mean is = ^{⅀}^{D}/_{N} = ^{28.3}/_{7} = 4.04
Calculation of mean deviation for grouped data about median:
Example : Calculate the mean deviation for the following data about median.
Class interval  0 – 4  5 – 9  10 – 14  15 – 19  20 – 24  25 – 29 
Frequency  11  12  17  12  20  28 
Solution:
First let us find the median
CI  f  fc  midpoint of CI(X)  Deviation
D = X – median 
D  fxD 
0 – 4  11  11  2  2 – 18.7 = 16.7  16.7  183.7 
5 – 9  12  23  7  7 – 18.7 = 11.7  11.7  140.4 
10 – 14  17  40  12  12 – 18.7 = 6.7  6.7  113.9 
15 – 19  12  5  17  17 – 18.7 = 1.7  1.7  20.4 
20 – 24  20  72  22  22 – 18.7 = 3.3  3.3  66.0 
25 – 29  28  100  27  27 – 18.7 = 8.3  8.3  232.4 
N = 100  ⅀fD= 756.8 
= 14.5 + (^{50 – 40}/_{12}) x 5
≅ 14.5 + (0.83 x 5)
= 14.5 + 4.15
= 18.65
≅ 18.7
After finding the median , the deviation of median from midpoint of the class intervals are calculated. This gives D. Frequency (f) is multiplied with D to get fD. By adding all FD, we get⅀fD. these are shown above.
Mean deviation = ^{⅀f}^{D}/_{N} = ^{756.8}/_{100} = 7.57
Calculation of mean deviation for grouped data about mean:
Example 12: Calclate the mean deviation for the data given below:
Class interval  0 – 10  10 – 20  20 – 30  30 – 40  40 – 50 
Frequency  5  3  9  12  6 
Solution:
Class interval  Frequency  midpoint x  fx  Deviation
D = x – X 
D  fD 
0 – 10  5  5  25  5 – 28.1 = – 23.1  23.1  115.5 
10 – 20  3  15  45  15 – 28.1 = – 13.1  13.1  39.3 
20 – 30  9  25  225  25 – 28.1 = – 3.1  3.1  27.9 
30 – 40  12  35  420  35 – 28.1 = 6.9  6.9  82.8 
40 – 50  6  45  270  45 – 28.1 = 16.9  16.9  101.4 
⅀fx  ⅀fD = 366.9 
Mean x = ^{⅀fx}/_{N} = ^{985}/_{35} = 28.1
Mean deviation = ^{⅀f}^{D}/_{N} = ^{366.9}/_{35} = 10.4
Statistics[Class 9] Exercise 1.5.3
 Find the mean deviation about mean for the following date:
a) 14, 21, 28, 21, 18
Solution:
Mean = ^{14+21+28+21+18}/_{5} = ^{102}/_{5} = 20.4
Score  Deviation from mean  D 
14  14 – 10.4 = 6.4  6.4 
18  18 – 20.4 = 2.4  2.4 
21  21 – 20.4 = 0.6  0.6 
21  21 – 20.4 = 0.6  0.6 
28  28 – 20.4 = 7.6  7.6 
(b)
Score(x)  6  20  8  18  16  12  14  10 
Frequency(f)  2  7  11  27  18  13  17  5 
Solution:
x  f  fxD= x – x  D  fD 
6  2  12 – 8.58  8.58  17.16 
20  7  140 + 5.42  5.42  37.94 
8  11  88 – 6.58  6.58  72.38 
18  27  486 – 3.42  3.42  92.34 
16  18  288 – 1.42  1.42  25.56 
12  13  156 – 2.58  2.58  33.56 
14  17  238 – 0.58  0.58  9.86 
10  5  50 – 4.58  4.58  22.9 
N = 100  1458  311.68 
 Find the mean deviation about mean for the following data:
a) 15, 18, 13, 16, 12, 24, 10, 20
Solution:
15+ 18+ 13+ 16+ 12+ 24+ 10+ 20 =128
N = 8
⅀x = 1258
Mean= ^{⅀}^{x}/_{N} =^{128}/_{8} = 16
x  D = x – x  D 
10  1016=6  6 
12  1216 = 4  4 
13  13 – 16 = 3  3 
15  15 – 16 = 1  1 
16  16 – 16 = 0  0 
18  18 – 16 = 2  2 
20  20 – 16 = 4  4 
24  24 – 16 = 8  8 
⅀28 
MD =^{⅀}^{D}/_{N} = ^{28}/_{8} = 3.5
(b)
CI  f 
1019  6 
2029  4 
3039  10 
4049  9 
5059  11 
6069  8 
7079  2 
Solution:
CI  f  x  fx  D = x – x  D 
1019  6  14.5  87  29.5  177 
2029  4  24.5  98  19.5  78 
3039  10  34.5  345  9.5  95 
4049  9  44.5  400.5  0.5  4.5 
5059  11  54.5  599.5  10.5  115.5 
6069  8  64.5  516  20.5  164.0 
7079  2  74.5  149  30.5  61 
N = 50  ⅀fx=2195  ⅀D=695 
(c)
Class interval  Frequency 
05  9 
510  13 
1015  6 
1520  12 
2025  9 
2530  6 
3035  10 
3540  15 
4045  6 
4550  4 
Solution:
Class interval  Frequency  x  fx  D= x – x  D 
05  9  2.5  22.5  20.5  184.5 
510  13  7.5  97.5  15.5  201.5 
1015  6  12.5  75.0  10.5  63.0 
1520  12  17.5  210.0  5.5  66.0 
2025  9  22.5  202.5  0.5  4.5 
2530  6  27.5  165.0  4.5  27.0 
3035  10  32.5  325.0  9.5  95.0 
3540  15  37.5  562.5  14.5  217.5 
4045  6  42.5  255.0  19.5  117.0 
4550  4  47.5  190.0  24.5  98.0 
N = 90  ⅀fx=2105  ⅀D=1074.0 
 Find the mean deviation about median for the following data:
a) 18, 23, 9, 11, 26, 4, 14, 21
Solution:
4, 9, 1, 14, 18, 21, 23, 26
Median = ^{N+1}/_{2} = ^{8+1}/_{2} = ^{9}/_{2} = 4.5^{th}
^{14+18}/_{2} = ^{32}/_{2}= 16
median = 16
x  D = x – median  D 
4  4 – 16 = 12  12 
9  9 – 16 = 7  7 
11  11 – 16 = 2  5 
14  14 – 16 = 2  2 
18  18 – 16 = 2  2 
21  21 – 16 = 05  5 
23  23 – 16 = 07  7 
26  26 – 16 = 10  10 
50 
MD = ^{50}/_{8} = 6.25
(b)
Class interval  Frequency 
8 – 12  14 
13 – 17  8 
18 – 22  20 
23 – 27  7 
28 – 32  11 
33 – 37  10 
38 – 42  24 
43 – 47  6 
Solution:
Class interval  Frequency  x  fx  D = x – median  fD 
8 – 12  14  10  14  10 – 23 = 13  182 
13 – 17  8  15  22  15 – 23 = 8  64 
18 – 22  20  20  42  20 – 23 = 3  60 
23 – 27  7  25  29  25 – 23 = 2  14 
28 – 32  11  30  60  30 – 23 = 7  77 
33 – 37  10  35  70  35 – 23 = 12  120 
38 – 42  24  40  94  40 – 23 = 17  408 
43 – 47  6  45  100  45 – 23 = 22  132 
N = 100  ⅀fD=1057 
(c)
Class interval  frequency 
20 – 30  9 
30 – 40  18 
40 – 50  7 
50 – 60  21 
60 – 70  11 
70 – 80  4 
Solution:
Class interval  frequency  x  fc  D = x – median  fD 
20 – 30  9  25  19  2542=17  153 
30 – 40  18  35  27  3542=7  126 
40 – 50  7  45  34  4542=3  21 
50 – 60  21  55  55  5542=13  273 
60 – 70  11  65  66  6542 = 23  253 
70 – 80  4  75  70  7542 =33  132 
N = 70  ⅀fD=958 
 Find the mean deviation about mean and median for the following data:
a)
Cl  15  610  1115  1620  2125 
f  2  9  5  4  10 
Solution:
Cl  f  fc  x  fx  D = x – x  fD 
15  2  2  3  06  3 – 15 = 12  24 
610  9  11  8  72  8 – 15 = 7  63 
1115  5  16  13  65  13 – 15 = 2  10 
1620  4  20  18  72  1815 = 3  12 
2125  10  30  23  230  23 – 15 = 8  80 
N=30  ⅀fx=445  189 
Mean = ^{445}/_{30} = 14.83 = 15
MD from Mean = ^{189}/_{30} = 6.3
CI  f  fc  x  D  fx 
15  2  2  3  11.5  23 
610  9  11  8  6.5  58.5 
1115  5  16  13  1.5  7.5 
1620  4  20  18  ±3.5  14.0 
2125  10  30  23  ±8.5  85.0 
N=30  ⅀fx = 188 
(b)
CI  5 – 10  1015  1520  2025  2530 
f  5  12  3  11  9 
Solution:
CI  f  fc  x  fx  D=xx  fD  xmedian  fD 
510  5  5  7.5  37.5  7.518=10.5  52.5  7.5  62.5 
1015  12  17  12.5  150  5.5  66  2.5  90 
1520  3  20  17.5  52.5  0.5  1.5  2.5  7.5 
2025  11  31  22.5  247.5  4.5  49.5  7.5  29.5 
2530  9  40  27.5  247.5  9.5  85.5  12.5  0.5 
N= 40  ⅀fx=735  235  255 
Mean = ^{735}/_{40} = 18.375 ≈18
1.5.4 Graphical Representation – Statistics[Class 9]:
1. Construction and Interpretation of Histogram Statistics[Class 9]:
 Histogram is the most properly and widely used methods of graphical representations.
 Histogram is a two dimensional graphical representation of a continuous frequency distribution.
 In a histogram the area of rectangular are proportional to the frequencies.
Class intervals are marked on the x – axis and frequencies on the Y – axis. Class intervals must be exclusive. IF the class intervals are in inclusive form, they are to be converted into exclusive form. Rectangles of width equal to class interval and length equal to frequencies re drawn.
(a) Histograms of varying width:
The width of each class interval is calculated by the corresponding frequencies. This is done by using a concept Frequency Density.
Frequency density is the ratio of frequency and its class width. when we consider the frequency density the length of the rectangle is to be modified accordingly. Length of the rectangle is the product of frequency density and the minimum class width of given data.
Length of the rectangle = ^{frequency}/_{class width} X C
Here C is the minimum class width of the given data.

Commutative Frequency Curve – Statistics[Class 9]:
Commutative frequency curve is a graph drawn with commutative frequency against the upper limit of class interval. The points are joined by a smooth curve and the curve is joined to the lower limit of the first class interval.

Frequency Polygon – Statistics[Class 9]:
Frequency polygon is also a graphical representation of data, where the frequency is plotted against midpoint of the class interval. The frequencies corresponding to the mid points of class intervals are joined by line segments to get the frequency polygon.
A frequency polygon is drawn by drawing a histogram for the given data and joining the midpoints of the top of the rectangles. It can be drawn by marking the midpoints of the class intervals corresponding to their respective frequencies and joining them by line segments.
Statistics[Class 9] Exercise 1.5.4
 Construct histogram of variable width for the following data:
a)
CI  2529  3035  3640  4150  5156  5760 
f  10  24  15  20  12  16 
Solution:
CI  24.529  29.535.5  35.5 – 40.5  40.550.5  50.556.5  56.560.5 
f  10  24  15  20  12  16 
class width  5  6  5  10  6  4 
length of the rectangles  10  20  15  10  10  20 
(b)
CI  0 – 10  10 15  15 – 20  20 – 30  30 – 40  40 – 60  6070 
f  20  15  10  25  5  30  50 
Solution:
CI  0 – 10  10 15  15 – 20  20 – 30  30 – 40  40 – 60  6070 
f  20  15  10  25  5  30  50 
class width  10  5  5  10  10  20  10 
Length of the rectangles  10  15  10  12.5  2.5  7.5  25 
 Draw given (cumulative frequency curve) for the data given below:
Class interval  frequency 
1000 – 1100  52 
1100 – 1200  35 
1200 – 1300  25 
1300 – 1400  14 
1400 – 1500  41 
1500 – 1600  33 
Solution:
Class interval  frequency  Cumulative frequency 
1000 – 1100  52  52 
1100 – 1200  35  87 
1200 – 1300  25  112 
1300 – 1400  14  126 
1400 – 1500  41  167 
1500 – 1600  33  200 
(b)
Class interval  Frequency 
5 – 14  4 
15 – 24  8 
25 – 34  12 
35 – 44  14 
45 – 54  6 
55 – 64  4 
65 – 74  18 
75 – 84  24 
Solution:
Class interval  corrective factor  Frequency  cumulative frequency 
5 – 14  4.5 – 14.5  4  4 
15 – 24  14.5 – 24.5  8  12 
25 – 34  24.5 – 34.5  12  24 
35 – 44  34.5 – 44.5  14  38 
45 – 54  44.5 – 54.5  6  44 
55 – 64  54.5 – 64.5  4  48 
65 – 74  64.5 – 74.5  18  66 
75 – 84  74.5 – 84.5  24  90 
 Construct frequency polygon for the following data :
a)
CI  5 – 10  10 – 15  15 – 20  20 – 25  25 – 30  30 – 35  35 – 40  40 – 45 
f  2  5  7  6  1  9  14  8 
Solution:
(b)
CI  3035  3540  4045  4550  5055 
f  12  20  16  8  14 
Solution:
1.5.5 Random Experiment and the concept of Probability – Statistics[Class 9]:
Based on some assumptions , uncertainity can be measured mathematically by what is called probability. Probability has wide applications in the feild of physical science, commerce, biological sciences, weather forecasting, insurance, economics, sociology, investments and in various such other areas.
(a) Trial – Statistics[Class 9]:
A trial is an action which results in one or more outcomes.
Consider the following examples:
 rolling an unbiased die.
 Picking up a red card from a deck of playing cards.
 Drawing a marble from a bag of different colored marbles.
(b) Random Experiment – Statistics[Class 9]:
A random experiment is one which exact outcomes are not possible to predict. For example, in tossing a coin we cannot predict outcomes head or tail.
(c) Sample Space – Statistics[Class 9]:
Consider the event of throwing an unbiased die. The possible outcomes are 1, 2, 3, 4,5 6. The set of all these possible outcomes is called Sample space.
S = {1, 2, 3,4, 5, 6}
(d) Empirical probability – Statistics[Class 9]:
It is the probability based on actual experiment leading to the possibility of outcomes.
Consider an experiment of tossing a coin 10 times. Let the frequency of head appearing would be 6 and that tail would be 4. Then the empirical probability of appearing head would be ^{6}/_{10} = 0/6 and that of tall would be ^{4}/_{10} = 0.4 .
The empirical probability of a certain event of an experiment is based on the outcomes of actual experiment. For this reason, empirical probability is also known as experimental probability.
If the number of tosses increases, the empirical probability of a head (or also tail)seems to approach the number ^{1}/_{2} or 0.5. This is actually known as the theoretical probability of getting a head (or a tail)
IF n is the number of trails of an event E, then the empirical probability p€ is given by,
P(E) = ^{Number of outcomes favourable to E}/_{Number of possible outcomes to the experiment.}
_{ }
Statistics[Class 9] – Exercise 1.5.5
 Two unbiased 6 – faced die are thrown. What is the total number out comes?
Solution:
total number of outcomes = 6 × 6 = 36
 A die has the faces numbered 2, 4, 6, 8, 10 and 12. It is thrown once. What is the probability that an even numbered face shows up?
Solution:
Probability that even number farm shows up = ^{6}/_{6} = 1
 In a pack of 52 playing cards, a card was selected at random. What is the probability that the card selected was both red and black?
Solution:
Zero: A card cannot be both red and black.
 Weather forecast made for 30 days in a month was recorded and found that it was correct for 21 days. What is the probability that on a randomly selected day, the forecast is (i) Correct and (ii) Not correct?
Solution:
Probability of correct forecast = ^{21}/_{30} = ^{7}/_{10}
Probability of correct forecast= ^{9}/_{30} = ^{3}/_{10}