Statistics[Class 9] – Full Chapter with Exercise

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,

Statistics[Class 9]

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 multi-mode.

1.5.2 Measures of dispersion – Statistics[Class 9]:

There are four measures of dispersion viz.

  1. range(R)
  2. Quartile Deviation(QD)
  3. Mean Deviation(MD)
  4. 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:

Marks263854657288
No. of students51015202530

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 wards12345678
No. of houses32572896138906658

Solution:

Here, H = 8 ; L = 1

Hence, coefficient of range = H L/H + L = 8-1/8+1 = 7/8


Merits and demerits of range[Class 9]:

Merits:

  1. It is the simplest measure of dispersion and easy to calculate.
  2. It does not require special knowledge to understand.
  3. Its calculation takes less time.

Demerits:

  1. It does not take into account  all the scores/items of distribution.
  2. It is affected by extreme scores.
  3. 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

Q3 – Q1. Half of this is the Semi-Inter Quartile Range or quartile deviation which is (Q3 – Q1)/2 . Quartile deviation is also called the semi-inter 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 Q1, Q2, Q3 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

  1. First quartile(Q1) = n+1/4 th score = 7+1/4 = 2nd score = 28
  2. Median(Q2) = n+1/2 th score = 7+1/2 = 4th score = 60
  3. Third quartile(Q3) = 3(n+1)/4 = 3(7+1)/4 th score = 6th score = 85
  4. 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:

x34567
f1235524118

Solution:

Let us find the commutative frequency for the data given:

xfCommutative frequencyRemarks
311212, the first 12 scores correspond to x = 3
43547(12 + 35) from 13th to 47th scores correspond to x = 5
55299(47+52) from 48th to 99th scores correspond to x = 5
641140(99+41) from 100th to 140th scores correspond to x = 6
718158(140+18) from 141th to 158th scores correspond to x = 7

Observe that, n = 158, the last commutative frequency.

Median = n+1/2 th score = 79.5th score ≅ 80th score

From the column fc, from 48th to 99th score corresponds to x = 5. Hence 80th position = 5. Therefore, median = 5

To find the Quartile Deviation:

Q1 = n/4 th score = 158/4 th score = 39.5 i.e., 40th score

From column fc, 13th to 47th scores correspond to x = 4. Hence the 40th position is 4. Therefore,

Q1 = 4

Similarly, 3n/4 = 3×158/4 = 118.5

Hence, Q3 is 119th score. The column for fc shows that 100th to 140th scores correspond to x = 6. Therefore, Q3 = 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 9th standard are given below:

Height(cm)100 – 110110 – 120120 – 130130 – 140140 – 150150 – 160
No. of Students(f)101216301220

Find quartile deviation.

Solution:

Step 1: First let us find the commutative frequency corresponding to the frequencies given.

Height(cm)f1fc
100 – 1101010
110 – 1201222
120 – 1301638(Q1 Class)
130 – 1403068
140 – 1501280
150 – 16020100(Q3 Class)
N = 100

Step 2: To find Q1 : Recall the formula to find the median

Statistics[Class 9]

where LRL = lower classs limit, fc = commutative frequency just above the median class fm = frequency corresponding the median class and I = size of the class interval. Now, to find Q1 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 Q1­ class.

From this class, LRL = 120, fc = 22 , fm = 16 and I = 10. Substituting these values in Q1 , we get,

Statistics[Class 9]

= 120 + (0.1875  x 10)

= 120 + 1.875

= 121.88

Thus, Q1 ≅ 122

Step 3: To find Q3:

In the median formula, replace, N/2 by 3N/4 and follow the same steps as in step 2.

Statistics[Class 9]

Here, 3N/4 = 3×100/4 = 75. In the commutative frequency column 75 corresponds to class interval 140 – 150. Therefore,

LRL = 140, fc = 68 , fm = 12 and I = 10

Statistics[Class 9]

= 140 + (0.58 x 10)

= 120 + 5.8

= 145.8

Thus, Q3 ≅ 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

  1. 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


  1. 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)

Age3691215
No. of children4811712

Solution:

xfcommutative frequency
344
6812
91123
12730
151242

\ 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 scored102030405060
No. of students12716081822

Solution:

xffc
101212
200719
301635
400843
501861
602283

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


  1. Compute quartile deviation for each of the following tables.

a)

Class intervalFrequencyfc
5 – 151111
15 – 25516
25 – 351531
35  – 45940
45 – 552262
55 – 65870
65 – 751787

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 intervalfrequencyfc
1 – 944
10 – 1937
20 – 292027
30 – 391239
40 – 49544
50 – 59852
60 – 691466
70 – 792793
80 – 89295
90 – 995100

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 = 4th 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|
1111 – 18 = -77
1313 – 18 = -55
1515 – 18 = -22
1818 – 18 = 00
2020 – 18 = 22
2121 – 18 = 33
2626 – 18 = 88
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|
1111 -17.7 = -6.76.7
1313 -17.7 = -4.74.7
1515 -17.7 = -2.72.7
1818 -17.7 = 0.30.3
2020 -17.7 = 2.32.3
2121 -17.7 = 3.33.3
2626 -17.7 = 8.38.3
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 interval0 – 45 – 910 – 1415 – 1920 – 2425 – 29
Frequency111217122028

Solution:

First let us find the median

CIffcmidpoint of CI(X)Deviation

D = X – median

|D|fx|D|
0 – 4111122 – 18.7 = -16.716.7183.7
5 – 9122377 – 18.7 = -11.711.7140.4
10 – 1417401212 – 18.7 = -6.76.7113.9
15 – 191251717 – 18.7 = -1.71.720.4
20 – 2420722222 – 18.7 = 3.33.366.0
25 – 29281002727 – 18.7 = 8.38.3232.4
N = 100⅀f|D|= 756.8

Statistics[Class 9]

= 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 f|D|. By adding all F|D|, we get⅀f|D|. 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 interval0 – 1010 – 2020 – 3030 – 4040 – 50
Frequency539126

Solution:

Class intervalFrequencymidpoint xfxDeviation

D = x – X

|D|f|D|
0 – 1055255 – 28.1 = – 23.123.1115.5
10 – 203154515 – 28.1 = – 13.113.139.3
20 – 3092522525 – 28.1 = – 3.13.127.9
30 – 40123542035 – 28.1 = 6.96.982.8
40 – 5064527045 – 28.1 = 16.916.9101.4
⅀fx⅀f|D| = 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

  1. 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

ScoreDeviation from mean|D|
1414 – 10.4 = -6.46.4
1818 – 20.4 = -2.42.4
2121 – 20.4 = 0.60.6
2121 – 20.4 = 0.60.6
2828 – 20.4 = 7.67.6

Statistics - Exercise 1.5.3 – Class IX


(b)

Score(x)62081816121410
Frequency(f)2711271813175

Solution:

xffxD= x – x|D|f|D|
6212 – 8.588.5817.16
207140 + 5.425.4237.94
81188 – 6.586.5872.38
1827486 – 3.423.4292.34
1618288 – 1.421.4225.56
1213156 – 2.582.5833.56
1417238 – 0.580.589.86
10550 – 4.584.5822.9
N = 1001458311.68

Statistics - Exercise 1.5.3 – Class IX


  1. 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

xD = x – x|D|
1010-16=-66
1212-16 = -44
1313 – 16 = -33
1515 – 16 = -11
1616 – 16 = 00
1818 – 16 = 22
2020 – 16 = 44
2424 – 16 = 88
⅀28

MD =|D|/N = 28/8 = 3.5


(b)

CIf
10-196
20-294
30-3910
40-499
50-5911
60-698
70-792

Solution:

CIfxfxD = x – x|D|
10-19614.587-29.5177
20-29424.598-19.578
30-391034.5345-9.595
40-49944.5400.50.54.5
50-591154.5599.510.5115.5
60-69864.551620.5164.0
70-79274.514930.561
N  = 50⅀fx=2195⅀|D|=695

Statistics[Class 9]


(c)

Class intervalFrequency
0-59
5-1013
10-156
15-2012
20-259
25-306
30-3510
35-4015
40-456
45-504

Solution:

Class intervalFrequencyxfxD= x – x|D|
0-592.522.5-20.5184.5
5-10137.597.5-15.5201.5
10-15612.575.0-10.563.0
15-201217.5210.0-5.566.0
20-25922.5202.5-0.54.5
25-30627.5165.0-4.527.0
30-351032.5325.09.595.0
35-401537.5562.514.5217.5
40-45642.5255.019.5117.0
45-50447.5190.024.598.0
N = 90⅀fx=2105⅀|D|=1074.0

Statistics[Class 9]


  1. 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 =  4.5th

14+18/2 = 32/2= 16

median = 16

xD =  x – median|D|
44 – 16 = -1212
99 – 16 = -77
1111 – 16 = -25
1414  – 16 = -22
1818 – 16 =  22
2121 – 16 = 055
2323 – 16 = 077
2626 – 16 = 1010
50

MD = 50/8 = 6.25


(b)

Class intervalFrequency
8 – 1214
13   – 178
18 –  2220
23 – 277
28 – 3211
33 – 3710
38 – 4224
43  – 476

Solution:

Class intervalFrequencyxfxD = x –  medianf|D|
8 – 1214101410 – 23 = -13182
13   – 178152215 – 23 =  -864
18 –  2220204220 – 23 = -360
23 – 277252925 – 23 = 214
28 – 3211306030 – 23 = 777
33 – 3710357035  –  23 = 12120
38 – 4224409440  – 23 = 17408
43  – 4764510045 – 23 = 22132
N = 100⅀f|D|=1057

Statistics[Class 9]


(c)                                                       

Class intervalfrequency
20 – 309
30 – 4018
40 – 507
50 – 6021
60 – 7011
70 – 804

Solution:

Class intervalfrequencyxfcD = x – medianf|D|
20 – 309251925-42=-17153
30 – 4018352735-42=-7126
40 – 507453445-42=321
50 – 6021555555-42=13273
60 – 7011656665-42 = 23253
70 – 804757075-42 =33132
N = 70⅀f|D|=958

Statistics[Class 9]


  1. Find the mean deviation about mean and median for the following data:

a)

Cl1-56-1011-1516-2021-25
f295410

Solution:

ClffcxfxD = x – xf|D|
1-5223063 – 15 = -1224
6-109118728 – 15 = -763
11-15516136513 – 15 = -210
16-20420187218-15 = 312
21-2510302323023 – 15 = 880
N=30⅀fx=445189

Mean = 445/30 = 14.83 = 15

MD from Mean = 189/30 = 6.3

CIffcxDfx
1-5223-11.523
6-109118-6.558.5
11-1551613-1.57.5
16-2042018±3.514.0
21-25103023±8.585.0
N=30⅀fx = 188

Statistics[Class 9]


(b)

CI5 – 1010-1515-2020-2525-30
f5123119

Solution:

CIffcxfxD=x-xf|D|x-medianf|D|
5-10557.537.57.5-18=10.552.5-7.562.5
10-15121712.5150-5.566-2.590
15-2032017.552.5-0.51.52.57.5
20-25113122.5247.54.549.57.529.5
25-3094027.5247.59.585.512.50.5
N= 40⅀fx=735235255

Mean = 735/40 = 18.375 ≈18

Statistics[Class 9]


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.

  1. 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.

  1. 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

  1. Construct histogram of variable width for the following data:

a)

CI      25-2930-3536-4041-5051-5657-60
f102415201216

Solution:

CI24.5-2929.5-35.535.5 – 40.540.5-50.550.5-56.556.5-60.5
f102415201216
class width5651064
length of  the rectangles102015101020

Statistics[Class 9]


(b)

CI0 – 1010- 1515 – 2020 – 3030 – 4040 – 6060-70
f2015102553050

Solution:

CI0 – 1010- 1515 – 2020 – 3030 – 4040 – 6060-70
f2015102553050
class width105510102010
Length of the rectangles10151012.52.57.525

Statistics[Class 9]


  1. Draw given (cumulative frequency curve) for the data given below:
Class intervalfrequency
1000 – 110052
1100 – 120035
1200 – 130025
1300 – 140014
1400 – 150041
1500 – 160033

Solution:

Class intervalfrequencyCumulative frequency
1000 – 11005252
1100 – 12003587
1200 – 130025112
1300 – 140014126
1400 – 150041167
1500 – 160033200

Statistics[Class 9]


(b)

Class intervalFrequency
5 – 144
15 – 248
25 – 3412
35 – 4414
45 – 546
55 – 644
65 – 7418
75 – 8424

Solution:

Class intervalcorrective factorFrequencycumulative frequency
5 – 144.5 – 14.544
15 – 2414.5 – 24.5812
25 – 3424.5 – 34.51224
35 – 4434.5 – 44.51438
45 – 5444.5 – 54.5644
55 – 6454.5 – 64.5448
65 – 7464.5 – 74.51866
75 – 8474.5 – 84.52490

Statistics[Class 9]


  1. Construct frequency polygon for the following data :

a)

CI5 – 1010 – 1515 – 2020 – 2525 – 3030 – 3535 – 4040 – 45
f257619148

Solution:

Statistics[Class 9]

(b)

CI30-3535-4040-4545-5050-55
f122016814

Solution:

Statistics[Class 9]


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:

  1. rolling an unbiased die.
  2. Picking up a red card from a deck of playing cards.
  3. 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

  1. Two unbiased 6 – faced die are thrown. What is the total number out comes?

Solution:

total number of outcomes = 6 × 6 = 36


  1. 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


  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.


  1. 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