 # RS Aggarwal Class 9 Solutions Chapter 14 -Statistics Ex 14G (14.7)

## RS Aggarwal Class 9 Chapter 14 -Statistics Ex 14G (14.7) Solutions Free PDF

Q.1: Find the mode of the following items.

0,6,5,1,6,4,3,0,2,6,5,6

Sol:

Arrange the given data in ascending order, we have:

0, 0, 1, 2, 3, 4, 5, 5, 6, 6, 6, 6

 Observations (x) 0 1 2 3 4 5 6 Frequency 2 1 1 1 1 2 4

As 6 occurs the maximum number of times i.e. 4, mode = 6

Q.2: Determine the mode of the following values of a variable.

23, 15, 25, 40, 27, 25, 22, 25, 20

Sol:

Arranging the given data in ascending order, we have

15, 20, 22, 23, 25, 25, 25, 27, 40

The frequency table of the data is:

 Observations (x) 15 20 22 23 25 27 40 Frequency 1 1 1 1 3 1 1

As 25 occurs the maximum number of times i.e. 3, mode= 25

Q.3: Calculate the mode of the following sizes of shoes sold by a shop on a particular day.

5, 9, 8, 6, 9, 4, 3, 9, 1, 6, 3, 9, 7, 1, 2, 5, 9

Sol:

Arranging the given data in ascending order, we have: 1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 9, 9, 9, 9, 9

The frequency table of the data is:

 Observations (x) 1 2 3 4 5 6 7 8 9 Frequency 2 1 2 1 2 2 1 1 5

As 9, occurs the maximum number of times i.e. 5, mode = 9

Q.4: A cricket player scored the following runs in 12 one-day matches: 50, 30, 9, 32, 60, 50, 28, 50, 19, 50, 27, 35. Find the modal score.

Sol:

Arranging the given data in ascending order, we have:

9, 19, 27, 28, 30, 32, 35, 50, 50, 50, 50, 60

The frequency table of the data is:

 Observations (x) 9 19 27 28 30 32 35 50 60 Frequency 1 1 1 1 1 1 1 4 1

As 50, occurs the maximum number of times i.e. 4, mode = 50

Thus, the modal score of the cricket player is 50.

Q.5: Calculate the mode of each of the following using the empirical formula: 17, 10, 12, 11, 10, 15, 14, 11, 12, 13

Sol:

Arranging the given data in ascending order, we have: 10, 10, 11, 11, 12, 12, 13, 14, 15, 17.

We prepare the table which is as given below:

 Items (x) Frequency (f) Cumulative frequency f×x$f\times x$ 10 2 2 20 11 2 4 22 12 2 6 24 13 1 7 13 14 1 8 14 15 1 9 15 17 1 10 17 N=10 Σf×x$\Sigma f\times x$=125

Here, n = 10, which is even.

Median = 12[n2]thterm+[n2+1]thterm$\frac{1}{2}\left [ \frac{n}{2} \right ]th \; term \; + \left [ \frac{n}{2} +1\right ]th\; term$

= 12[5thterm+6thterm]$\frac{1}{2}[ 5th \; term + 6th \; term]$  (since n= 10)

= 12(12+12)$\frac{1}{2}(12+12)$

= 12(24)=12$\frac{1}{2}(24)= 12$

Therefore, Median= 12

Now, Σf×x$\Sigma f\times x$ =125 and Σf$\Sigma f$= 10

Therefore, Mean = Σf×xΣf=12510=12.5$\frac{\Sigma f\times x}{\Sigma f}=\frac{125}{10}=12.5$

Mode = 3(Median) – 2(Mean) = 3(12) – 2(12.5) = 36 – 25

Thus, Mode = 11

Q.6:

 Marks 10 11 12 13 14 16 19 20 Number of students 3 5 4 5 2 3 2 1

Sol:

We may prepare the table, given below:

 Marks (x) No. of students (f) Cumulative Frequency f×x$f\times x$ 10 3 3 30 11 5 8 55 12 4 12 48 13 5 17 65 14 2 19 28 16 3 22 48 19 2 24 38 20 1 25 20 N=25 Σf×x$\Sigma f\times x$=332

Here n= 25, which is odd.

Median = =12(n+1)$=\frac{1}{2}(n+1)$th term

= =12(25+1)$=\frac{1}{2}(25+1)$th term

Value of the 13th term = 13

Now, Σf×x$\Sigma f\times x$ =332 and Σf$\Sigma f$=25

So, mean = Σf×xΣf=33225=13.28$\frac{\Sigma f\times x}{\Sigma f}=\frac{332}{25}=13.28$

Mode = 3(Median) – 2(Mean)

= 3(13) – 2(13.28) = 39 – 26.56 = 12.44

Thus, mode= 12.4

Q.7:

 Items 5 7 9 12 14 17 19 21 Frequency 6 5 3 6 5 3 2 4

Sol:

We may prepare the table, given below:

 Items (x) Frequency (f) Cumulative Frequency f×x$f\times x$ 5 6 6 30 7 5 11 35 9 3 14 27 12 6 20 72 14 5 25 70 17 3 28 51 19 2 30 38 21 4 34 84 N=Σf$\Sigma f$ =34 Σf×x$\Sigma f\times x$ =407

Here, n = 34, which is even.

Median = 12[n2]thterm+[n2+1]thterm$\frac{1}{2}\left [ \frac{n}{2} \right ]th \; term \; + \left [ \frac{n}{2} +1\right ]th\; term$

= 12[17thterm+18thterm]$\frac{1}{2}[ 17th \; term + 18th \; term]$  (since n= 34)

= 12(12+12)$\frac{1}{2}(12+12)$

= 12(24)=12$\frac{1}{2}(24)= 12$

Therefore, Median = 12

Now, Σf×x$\Sigma f\times x$ = 407 and Σf$\Sigma f$ = 34

Therefore, Mean  =Σf×xΣf=40734=11.97$\frac{\Sigma f\times x}{\Sigma f}=\frac{407}{34}=11.97$

Mode = 3(Median) – 2(Mean)

= 3(12) – 2(11.97)

= 36-23.94

Thus, Mode =12.06

Q.8:

 x 18 20 25 30 34 38 40 f 6 7 3 7 7 5 5

Sol:

We may prepare the table, given below:

 (x) Frequency (f) Cumulative Frequency f×x$f\times x$ 18 6 6 108 20 7 13 140 25 3 16 75 30 7 23 210 34 7 30 238 38 5 35 190 40 5 40 200 N=Σf$\Sigma f$ =40 Σf×x$\Sigma f\times x$ =1161

Here, n = 40, which is even.

Median = 12[n2]thterm+[n2+1]thterm$\frac{1}{2}\left [ \frac{n}{2} \right ]th \; term \; + \left [ \frac{n}{2} +1\right ]th\; term$

= 12[20thterm+21stterm]$\frac{1}{2}[ 20th \; term + 21st \; term]$  (since n= 40)

= 12(30+30)$\frac{1}{2}(30+30)$

= 12(60)=30$\frac{1}{2}(60)= 30$

Therefore, Median = 30

Now, Σf×x$\Sigma f\times x$ = 1161 and Σf$\Sigma f$ = 40

Therefore, Mean = Σf×xΣf=116140=29.025$\frac{\Sigma f\times x}{\Sigma f}=\frac{1161}{40} =29.025$

Mode = 3(Median) – 2(Mean)

= 3(30) – 2(29.025) = 31.95

Thus, Mode = 32

Q.9: The table given below shows the weight (in kg) of 50 persons:

 Weight (in kg) 42 47 52 57 62 67 72 No. of persons (f) 3 8 6 8 11 5 9

Find the mean, median and mode.

Sol:

We may prepare the table, given below:

 Weight (in kg) No.of persons (f) Cumulative frequency f×x$f\times x$ 42 3 3 126 47 8 11 376 52 6 17 312 57 8 25 456 62 11 36 682 67 5 41 335 72 9 50 648 N=Σf$\Sigma f$ =50 Σf×x$\Sigma f\times x$ =2935

Here, Σf×x$\Sigma f\times x$ =2935, and Σf$\Sigma f$ =50

Mean = Σf×xΣf=293550=58.7$\frac{\Sigma f\times x}{\Sigma f} = \frac{2935}{50}=58.7$

Therefore, mean weight = 58.7 kg

Here, N = 50 which is even.

Therefore, Median = 12[n2]thterm+[n2+1]thterm$\frac{1}{2}\left [ \frac{n}{2} \right ]th \; term \; + \left [ \frac{n}{2} +1\right ]th\; term$

= 12[25thterm+26thterm]$\frac{1}{2}[ 25th \; term + 26th \; term]$  (Since, n= 50)

= 12(57+62)$\frac{1}{2}(57+62)$

= 12(119)=59.5kg$\frac{1}{2}(119)= 59.5 kg$

Therefore, Median weight = 59.5 kg

Mode = 3(Median) – 2(Mean)

= 3(59.5) – 2(58.7) = 178.5 – 117.4 = 61.1

Thus, Mode weight = 61.1 kg

Thus we have: Mean = 58.7 kg, Median = 59.5 kg and Mode = 61.1 kg

Q.10: The marks obtained by 80 students in a test are given below:

 Marks 4 12 20 28 36 44 No. of students 8 10 16 24 15 7

Find the modal marks.

Sol:

We may prepare the table, given below:

 Marks (x) No. of students (f) Cumulative frequency f×x$f\times x$ 4 8 8 32 12 10 18 120 20 16 34 320 28 24 58 672 36 15 73 540 44 7 80 308 N=Σf$\Sigma f$ =80 Σf×x$\Sigma f\times x$ = 1992

Here, n = 80, which is even

Median = 12[n2]thterm+[n2+1]thterm$\frac{1}{2}\left [ \frac{n}{2} \right ]th \; term \; + \left [ \frac{n}{2} +1\right ]th\; term$

= 12[40thterm+41stterm]$\frac{1}{2}[ 40th \; term + 41st \; term]$  (since n= 80)

= 12(28+28)$\frac{1}{2}(28+28)$

= 12(56)=28$\frac{1}{2}(56)= 28$

Therefore, Median= 28

Now, Σf×x$\Sigma f\times x$ = 1992 and Σf$\Sigma f$ = 80

Therefore, Mean =Σf×xΣf=199280=24.9$\frac{\Sigma f\times x}{\Sigma f}=\frac{1992}{80} =24.9$

Mode = 3(Median) – 2(Mean)

= 3(28) – 2(24.9) = 84 – 49.8 = 34.2

Thus, Mode = 34.2

Q.11: The ages of the employees of a company are given below:

 Age (in years) 19 21 23 25 27 29 31 No. of persons 13 15 16 18 16 15 13

Find the mean, median and mode of the above data.

Sol:

We may prepare the table, given below:

 Age (in years) (x) No. of persons (f) Cumulative frequency f×x$f\times x$ 19 13 13 247 21 15 28 315 23 16 44 368 25 18 62 450 27 16 78 432 29 15 93 435 31 13 106 403 N=Σf$\Sigma f$ =106 Σf×x$\Sigma f\times x$ = 2650

Here, Σf×x$\Sigma f\times x$ =2650, and Σf$\Sigma f$ = 106

Mean = Σf×xΣf=2650106=25$\frac{\Sigma f\times x}{\Sigma f} = \frac{2650}{106}=25$

Therefore, mean =25

Here, N = 106 which is even.

Therefore, Median = 12[n2]thterm+[n2+1]thterm$\frac{1}{2}\left [ \frac{n}{2} \right ]th \; term \; + \left [ \frac{n}{2} +1\right ]th\; term$

= 12[53thterm+54thterm]$\frac{1}{2}[ 53th \; term + 54th \; term]$  (since n= 50)

= 12(25+25)$\frac{1}{2}(25+25)$

= 12(50)=25$\frac{1}{2}(50)= 25$

Therefore, Median = 25

Mode = 3(Median) – 2(Mean)

= 3(25) – 2(25)

= 75 – 50 =25

Mode = 25

Thus we have, Mean = 25, Median = 25 and Mode = 25

Q.12: The following table shows the weight of 12 students:

 Weight in kg 47 50 53 56 60 No. of students 4 3 2 2 4

Find the mean, median and mode for the above data.

Sol:

We may prepare the table, given as:

 Weight in kg (x) No. of students (f) Cumulative frequency f×x$f\times x$ 47 4 4 188 50 3 7 150 53 2 9 106 56 2 11 112 60 4 15 240 N=Σf$\Sigma f$ =15 Σf×x$\Sigma f\times x$ = 796

Here, Σf×x$\Sigma f\times x$ = 796, and Σf$\Sigma f$ = 15

Mean = Σf×xΣf=79615=53.06$\frac{\Sigma f\times x}{\Sigma f} = \frac{796}{15}= 53.06$

Therefore, mean = 53.06

Here, N= 15 which is odd.

Therefore, Median = 12[n+12]thterm$\frac{1}{2}\left [ \frac{n+1}{2} \right ]th \; term$

= 12[15+12]thterm$\frac{1}{2}\left [ \frac{15+1}{2} \right ]th \; term$

= value of 8th term = 53

Therefore, Median = 53

Mode = 3 (Median) -2 (Mean)

= 3(53) – 2(53.06)

= 159 – 106.12 = 52.88

Mode = 52.88

Thus we have: Mean = 53.06, Median = 53 and Mode = 52.88