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Class 10 RD Sharma Solution – Chapter 7 Statistics – Exercise 7.4 | Set 1

  • Last Updated : 03 Mar, 2021

Question 1. Following are the lives in hours of 15 pieces of the components of aircraft engine. Find the median:

715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719.

Solution:

On arranging the observations in ascending order, we have

694, 696, 699, 705, 710, 712, 715, 716, 719, 721, 725, 728, 729, 734, 745

Number of terms in the observation sequence is odd, i.e., N = 15



Now,

Median = (N + 1)/2th term

= (15 + 1)/2th term

= 8th term

Therefore, 716, which is the 8th term is the median of the data.

Question 2. The following is the distribution of height of students of a certain class in a certain city:

Height (in cm): 160 – 162163 – 165 166 – 168 169 – 171  172 – 174
No of students:1511814212718

Find the median height.

Solution:

Class interval (exclusive)  Class interval  (inclusive) Class interval frequencyCumulative frequency
160 – 162 159.5 – 162.51515
163 – 165162.5 – 165.5118 133(F)
166 – 168 165.5 – 168.5142(f)275
169 – 171168.5 – 171.5127402
172 – 174 171.5 – 174.518420
  N =420 

We have N = 420,

So, N/2 = 420/ 2 = 210



Now, The cumulative frequency just greater than N/2 is 275 

Therefore, 165.5 – 168.5 is the median class s.t

L = 165.5, f = 142, F = 133 and h = (168.5 – 165.5) = 3

Median = L + \frac{\frac{N}{2} - F}{f} * h \\ = 165.5 +  \frac{210 - 133}{142} * 3 \\ = 165.5 + \frac{231}{142}

= 165.5 + 1.63

= 167.13

Question 3. Following is the distribution of I.Q of 100 students. Find the median I.Q.

I.Q:        55 – 6465 – 7475 – 8485 – 94 95 – 104105 – 114115 – 124125 – 134135 – 144
No of students: 129223322821

Solution:

Class interval (exclusive)  Class interval  (inclusive) Class interval frequencyCumulative frequency
55 – 6454.5 – 64.511
65 – 7464.5 – 74.523
75 – 8474.5 – 84.5912
85 – 9484.5 – 94.52234(F)
95 – 10494.5 – 104.533(f)67
105 – 114104.5 – 114.52289
115 – 124114.5 – 124.5897
125 – 134 124.5 – 134.5298
135 – 144134.5 – 144.51100
  N = 100 

N = 100,

Therefore, N/2 = 100/ 2 = 50

The cumulative frequency just greater than N/ 2 is 67 then the median class is (94.5 – 104.5) s.t,



L = 94.5, F = 33, h = (104.5 – 94.5) = 10

Median = L + \frac{\frac{N}{2} - F}{f} * h \\ = 94.5 +  \frac{50 - 34}{33} * 10

= 94.5 + 4.85

= 99.35

Question 4. Calculate the median from the following data:

Rent (in Rs): 15 – 2525-3535-4545-5555-6565-7575-8585-95
No of houses: 81015254020157

Solution:

Class interval  FrequencyCumulative frequency
15 – 25 88
25 – 351018
35 – 451533
45 – 552558 (F)
55 – 6540(f)98
65 – 7520118
75 – 8515133
85 – 957140
 N = 140  

N = 140,

And, N/2 = 140/ 2 = 70

The cumulative frequency just greater than N/ 2 is 98 then median class is 55 – 65 s.t,

L = 55, f = 40, F = 58, h = 65 – 55 = 10

Median = L + \frac{\frac{N}{2} - F}{f} * h \\ = 55 +  \frac{70 - 58}{40} * 10 \\ = 58



Question 5. Calculate the median from the following data:

Marks below: 0-1010 – 2020 – 3030-4040-5050-6060-7070-80
No of students:1535608496127198250

Solution:

Marks below    No. of studentsClass intervalFrequencyCumulative frequency
10150-101515
203510-202035
306020-302560
408430-402484
509640-501296(F)
6012750-6031(f)127
7019860-7071198
8025070-8052250
   N = 250 

N = 250,

And, N/2 = 250/ 2 = 125

The cumulative frequency just greater than N/ 2 is 127 then median class is 50 – 60 s.t,

 L = 50, f = 31, F = 96, h = 60 -50 = 10

Median = L + \frac{\frac{N}{2} - F}{f} * h \\ = 50 +  \frac{125 - 96}{31} * 10 \\ = 59.35

Question 6. Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.

Age in years:  0 – 10 10 – 20 20 – 30 30 – 4040 – 50
No of persons:25?187

Solution:

Let us assume the unknown frequency to be x.

Class interval  FrequencyCumulative frequency
0 – 1055
10-202530 (F)
20-30x(f)30 + x 
30-401848 + x 
40-50755 + x
 N=170 

Given: Median = 24

Therefore,



Median class = 20 – 30; L = 20, h = 30 -20 = 10, f = x, F = 30

Median = L + \frac{\frac{N}{2} - F}{f} * h \\24  = 20 +  \frac{\frac{55+x}{2} - 30}{x} * 10 \\4 =  \frac{\frac{55+x}{2} - 30}{x} * 10 \\ 4x = (\frac{55+x}{2} - 30) * 10

4x = 275 + 5x – 300

4x – 5x = – 25

– x = – 25

x = 25

Therefore, x = 25

Question 7. The following table gives the frequency distribution of married women by age at marriage.

Age (in years)FrequencyAge (in years)Frequency
15 – 195340 – 449
20 – 2414045 – 495
25 – 29 9845 – 493
30 – 343255 – 593
35 – 391260 and above2

Calculate the median and interpret the results.

Solution:

Class interval (exclusive) Class interval (inclusive)FrequencyCumulative frequency
15 – 1914.5 – 19.55353(F)
20 – 24 19.5 – 24.5 140(f)193
25 – 29 24.5 – 29.598291
30 – 34 29.5 – 34.532323
35 – 3934.5 – 39.512335
40 – 44 39.5 – 44.59344
45 – 49 44.5 – 49.55349
50 – 5449.5 – 54.53352
55 – 5454.5 – 59.53355
60 and above 59.5 and above2357
  N =357 

N = 357,

And, N/2 = 357/ 2 = 178.5



The cumulative frequency just greater than N/2 is 193, 

Therefore, median class is 19.5 – 24.5 s.t

l = 19.5, f = 140, F = 53, h = 25.5 – 19.5 = 5

Median = L + \frac{\frac{N}{2} - F}{f} * h \\ = 19.5 +  \frac{178.5 - 53}{140} * 5

Median = 23.98, that implies that nearly half of the women are married between the ages of 15 and 25.

Question 8. The following table gives the distribution of the life time of 400 neon lamps:

Life time: (in hours) Number of lamps
1500 – 200014
2000 – 250056
2500 – 300060
3000 – 350086
3500 – 400074
4000 – 450062
4500 – 500048

Find the median life.

Solution:

Life time Number of lamps fiCumulative frequency (cf)
1500 – 20001414
2000 – 25005670
2500 – 3000 60130(F)
3000 – 350086216
3500 – 4000 74290
4000 – 450062352
4500 – 500048400
 N = 400 

Now

N = 400 

And the cumulative frequency just greater than n/2 (= 200) is 216, which belongs to the class interval 3000 – 3500 

Median class = 3000 – 3500. Therefore,



(l)  = 3000 and,(f) of median class = 86, (cf) of class preceding median class = 130 and (h) = 500

We have,

Median = l+ \frac{\frac{n}{2} - cf}{f} * h \\ = 3000 +  \frac{200 - 130}{86} * 500

= 3000 + (35000/86)

= 3406.98 hrs, which is the median time of lamps. 

Question 9. The distribution below gives the weight of 30 students in a class. Find the median weight of students:

Weight (in kg): 40 – 45 45 – 50 50 – 55 55 – 6060 – 65  65 – 70 70 – 75
No of students:2386632

Solution:

Weight (in kg) Number of students fi Cumulative frequency (cf)
40 – 4522
45-5035
50-55813
55-60619
60-65625
65-70328
70-75230

The cf value just greater than n/ 2 (i.e. 30/ 2 = 15) is 19, belongs to class interval 55 – 60.

Therefore,

Median class = 55 – 60

where,

(l) of median class = 55, (f) of median class = 6, (cf) = 13 and (h) = 5

We have,

Median = l+ \frac{\frac{n}{2} - cf}{f} * h \\ = 55 +  \frac{15 - 13}{6} * 5

= 55 + 10/6 = 56.666 which is approximately 56.67 kg.

Question 10. Find the missing frequencies and the median for the following distribution if the mean is 1.46

No. of accidents: 012345Total
Frequencies (no. of days):46??25105200

Solution:

No. of accidents (x) No. of days (f) fx
0460
1xx
2y2y
32575
41040
5525
 N = 200 Sum = x + 2y + 140

Since, we know,

 N = 200

Substituting values, we get, 

⇒ 46 + x + y + 25 + 10 + 5 = 200

⇒ x + y = 200 – 46 – 25 – 10 – 5



⇒ x + y = 114…… (i)

Also, Mean = 1.46

⇒ Sum/ N = 1.46

Substituting values, 

⇒ (x + 2y + 140)/ 200 = 1.46

⇒ x + 2y = 292 – 140

⇒ x + 2y = 152 …….(ii)

Solving from (i) and (ii), we get

x + 2y – x – y = 152 – 114

⇒ y = 38

And, x = 114 – 38 = 76 (from equation (i))

Now, putting the values, we get,

N = 200 N/2 = 200/2 = 100

So, the cumulative frequency just greater than N/2 is 122

And, therefore, the median is 1.

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