TOPICS
Exercise - 14.3

Question-1 :-  The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them. Statistics

Solution :-
    
Monthly consuption (in units)Number of consumersCumulative Frequency (cf)
65 - 8544
85 - 10554 + 5 = 9
105 - 125139 + 13 = 22
125 - 1452022 + 20 = 42
145 - 1651442 + 14 = 56
165 - 185856 + 8 = 64
185 - 205464 + 4 = 68
Totaln = 68
Now, n = 68 n/2 = 68/2 = 34 is 42. So, this observation lies in the class 125 - 145. Then, l (the lower limit) = 125, cf (the cumulative frequency of the class preceding 125 - 145) = 22, f (the frequency of the median class 125 - 145) = 20, h (the class size) = 20. Therefore, Median = l + [(n/2 - cf)/f] x h = 125 + [(68/2 - 22)/20] x 20 = 125 + [(34 - 22)/20] x 20 = 125 + (12/20 x 20) = 125 + 12 = 137 So, the median of data is 137.
  From given table, to find the class mark (xi) for each inerval, the following relation is used.
  Class mark (xi) = (upper class limit + lower class limit)/2
  Class size (h) = 20
  Taking assumed mean (a) = 135
    
Monthly consuption (in units)Number of consumers (fi)xidi = xi - 135ui = di/20fiui
65 - 85475-60-3-12
85 - 105595-40-2-10
105 - 12513115-20-1-13
125 - 14520135000
145 - 1651415520114
165 - 185817540216
185 - 205419560312
TotalΣfi = 68Σfiui = 7
Now, x = a + (Σfiui/Σfi) x h = 135 + (7/68) x 20 = 135 + 140/68 = 135 + 2.058 = 137.058 Therefore, the mean of data is 137.058.
  Here the maximum class frequency is 20, and the class corresponding to this frequency is 125 – 145. 
  So, the modal class is 125 – 145.
  Now, Modal class = 125 – 145, 
  lower limit(l) of modal class = 125,
  class size(h) = 20
  frequency(f1) of modal class = 20,
  frequency(f0) of modal class = 13,
  frequency(f2) of modal class = 14,

  Therefore, Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
  = 125 + [(20-13)/{(2x20)-13-14}] x 20
  = 125 + [7/(40-27)] x 20
  = 125 + (7/13 x 20)
  = 125 + 140/13
  = 125 + 10.76
  = 135.76

  Therefore, the mode of this data is 135.76.  

  Hence, median, mean and mode of the given data is 137, 137.05 and 135.76 respectively.
  The three measures are approximately the same in this case. 
    

Question-2 :-  If the median of the distribution given below is 28.5, find the values of x and y. Statistics

Solution :-
    
Class intervalFrequencyCumulative Frequency (cf)
0 - 1055
10 - 20x5 + x
20 - 302025 + x
30 - 401540 + x
40 - 50y40 + x + y
50 - 60545 + x + y
Totaln = 45 + x + y
Given that: n = 60 Now, n = 45 + x + y 60 = 45 + x + y x + y = 60 - 45 x + y = 15 ........(i) The median is 28.5, which lies in the class 20 – 30 So, l = 20, f = 20, cf = 5 + x, h = 10 Therefore, Median = l + [(n/2 - cf)/f] x h 28.5 = 20 + [{(60/2) - (5 + x)}/20] x 10 28.5 = 20 + [(30 - 5 - x)/20] x 10 28.5 - 20 = [(25 - x)/20] x 10 8.5 = (25 - x)/2 17 = 25 - x x = 25 - 17 x = 8 Therefore, from (i), we get 8 + y = 15 y = 15 - 8 y = 7 Hence, the values of x and y are 8 and 7 respectively.

Question-3 :-  A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year. Statistics

Solution :-
  Here, class width is not the same. 
  There is no requirement of adjusting the frequencies according to class intervals. 
  The given frequency table is of less than type represented with upper class limits. 
  The policies were given only to the persons with age 18 years onwards but less than 60 years.
    
Age (in years)Number of policy holdersCumulative Frequency (cf)
18 - 2022
20 - 256 - 2 = 42 + 4 = 6
25 - 3024 - 6 = 186 + 18 = 24
30 - 3545 - 24 = 2124 + 21 = 45
35 - 4078 - 45 = 3345 + 33 = 78
40 - 4589 - 78 = 1178 + 11 = 89
45 - 5092 - 89 = 389 + 3 = 92
50 - 5598 - 92 = 692 + 6 = 98
55 - 60100 - 98 = 298 + 2 = 100
Totaln = 100
Now, n = 100 n/2 = 100/2 = 50 is 78. So, this observation lies in the class 35 - 40. Then, l (the lower limit) = 35, cf (the cumulative frequency of the class preceding 35 - 40) = 45, f (the frequency of the median class 35 - 40) = 33, h (the class size) = 5. Therefore, Median = l + [(n/2 - cf)/f] x h = 35 + [(100/2 - 45)/33] x 5 = 35 + [(50 - 45)/33] x 5 = 35 + (5/33 x 5) = 35 + 25/33 = 35 + 0.76 = 35.76 So, the median of data is 35.76 years.

Question-4 :-  The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table: Statistics Find the median length of the leaves. (Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

Solution :-
  It can be observed that class intervals are not continuous. There is a gap of 1 between two class intervals.
  Therefore, 1/2 has to be added to upper class limit and 1/2 has to be substracted from the lower class limit of each interval.
    
Length (in mm)Number of leavesCumulative Frequency (cf)
117.5 - 126.533
126.5 - 135.553 + 5 = 8
135.5 - 144.598 + 9 = 17
144.5 - 153.51217 + 12 = 29
153.5 - 162.5529 + 5 = 34
162.5 - 171.5434 + 4 = 38
171.5 - 180.5238 + 2 = 40
Totaln = 40
Now, n = 40 n/2 = 40/2 = 20 is 29. So, this observation lies in the class 144.5 - 153.5. Then, l (the lower limit) = 144.5, cf (the cumulative frequency of the class preceding 144.5 - 153.5) = 17, f (the frequency of the median class 144.5 - 153.5) = 12, h (the class size) = 9. Therefore, Median = l + [(n/2 - cf)/f] x h = 144.5 + [(40/2 - 17)/12] x 9 = 144.5 + [(20 - 17)/12] x 9 = 144.5 + (3/12 x 9) = 144.5 + 9/4 = 144.5 + 2.25 = 146.75 Therefore, the median length of leaves is 146.75 mm.

Question-5 :-  The following table gives the distribution of the life time of 400 neon lamps : Statistics Find the median life time of a lamp.

Solution :-
    
Life time (in hours)Number of lampsCumulative Frequency (cf)
1500 - 20001414
2000 - 25005614 + 56 = 70
2500 - 30006070 + 60 = 130
3000 - 350086130 + 86 = 216
3500 - 400074216 + 74 = 290
4000 - 450062290 + 62 = 352
4500 - 500048352 + 48 = 400
Totaln = 400
Now, n = 400 n/2 = 400/2 = 200 is 216. So, this observation lies in the class 3000 - 3500. Then, l (the lower limit) = 3000, cf (the cumulative frequency of the class preceding 3000 - 3500) = 130, f (the frequency of the median class 3000 - 3500) = 86, h (the class size) = 500. Therefore, Median = l + [(n/2 - cf)/f] x h = 3000 + [(400/2 - 130)/86] x 500 = 3000 + [(200 - 130)/86] x 500 = 3000 + (70/86 x 500) = 3000 + 3500/86 = 3000 + 406.976 = 3406.976 So, the median life time of lamps is 3406.976 hours.

Question-6 :- 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows: Statistics Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

Solution :-
    
Number of lettersNumber of surnamesCumulative Frequency (cf)
1 - 466
4 - 7306 + 30 = 36
7 - 104036 + 40 = 76
10 - 131676 + 16 = 92
13 - 16492 + 4 = 96
16 - 19496 + 4 = 100
Totaln = 100
Now, n = 100 n/2 = 100/2 = 50 is 76. So, this observation lies in the class 7 - 10. Then, l (the lower limit) = 7, cf (the cumulative frequency of the class preceding 7 - 10) = 36, f (the frequency of the median class 7 - 10) = 40, h (the class size) = 3. Therefore, Median = l + [(n/2 - cf)/f] x h = 7 + [(100/2 - 36)/40] x 3 = 7 + [(50 - 36)/40] x 3 = 7 + (14/40 x 3) = 7 + 42/40 = 7 + 1.05 = 8.05 So, the median of data is 8.05.
  From given table, to find the class mark (xi) for each inerval, the following relation is used.
  Class mark (xi) = (upper class limit + lower class limit)/2
  Class size (h) = 3
  Taking assumed mean (a) = 11.5
    
Number of lettersNumber of surnames (fi)xidi = xi - 11.5ui = di/3fiui
1 - 462.5-9-3-18
4 - 7305.5-6-2-60
7 - 10408.5-3-1-40
10 - 131611.5000
13 - 16414.5314
16 - 19417.5628
TotalΣfi = 100Σfiui = -106
Now, x = a + (Σfiui/Σfi) x h = 11.5 + (-106/100) x 3 = 11.5 - 318/100 = 11.5 - 3.18 = 8.32 Therefore, the mean of data is 8.32.
  Here the maximum class frequency is 40, and the class corresponding to this frequency is 7 – 10. 
  So, the modal class is 7 – 10.
  Now, Modal class = 7 – 10, 
  lower limit(l) of modal class = 7,
  class size(h) = 3
  frequency(f1) of modal class = 40,
  frequency(f0) of modal class = 30,
  frequency(f2) of modal class = 16,

  Therefore, Mode = l + [(f1 - f0)/(2f1 - f0 - f2)] x h
  = 7 + [(40-30)/{(2x40)-30-16}] x 3
  = 7 + [10/(80-46)] x 3
  = 7 + (10/34 x 3)
  = 7 + 30/34
  = 7 + 0.88
  = 7.88

  Therefore, the mode of this data is 7.88.  

  Hence, median and mean of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.88. 
    

Question-7 :- The distribution below gives the weights of 30 students of a class. Find the median weight of the students. Statistics

Solution :-
    
Weights (in kg)Number of studentsCumulative Frequency (cf)
40 - 4522
45 - 5032 + 3 = 5
50 - 5585 + 8 = 13
55 - 60613 + 6 = 19
60 - 65619 + 6 = 25
65 - 70325 + 3 = 28
70 - 75228 + 2 = 30
Totaln = 30
Now, n = 30 n/2 = 30/2 = 15 is 19. So, this observation lies in the class 55 - 60. Then, l (the lower limit) = 55, cf (the cumulative frequency of the class preceding 55 - 60) = 13, f (the frequency of the median class 55 - 60) = 6, h (the class size) = 5. Therefore, Median = l + [(n/2 - cf)/f] x h = 55 + [(30/2 - 13)/6] x 5 = 55 + [(15 - 13)/6] x 5 = 55 + (2/6 x 5) = 55 + 10/6 = 55 + 1.67 = 56.67 Therefore, the median weight is 56.67 kg.
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