NCERT Solutions for Class 11 Economics Chapter 5 - Measures Of Central Tendency

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    1. Which average would be suitable in the following cases?
    (i) Average size of readymade garments.
    (ii) Average intelligence of students in a class.
    (iii) Average production in a factory per shift.
    (iv) Average wage in an industrial concern.
    (v) When the sum of absolute deviations from average is least.
    (vi) When quantities of the variable are in ratios.
    (vii) In case of open-ended frequency distribution.

    Ans. (i) Mode: Average size of readymade garments should be the size for which demand is the maximum. As the modal value represents the value with the highest frequency should be taken as the average size to be produced.
    (ii) Median: It is the middlemost term of the series. Therefore, the Median will be the best measure to calculate the average intelligence of students in a class as it will give the average intelligence such that there is an equal number of students above and below this average. It will not be affected by extreme values.
    (iii) Arithmetic Mean: The average production in a factory per shift can best be calculated by arithmetic mean as it will acquires all types of fluctuations in production during the shifts.
    (iv) Arithmetic Mean: Arithmetic mean will be the most suitable measure to calculate average wages. It is calculated by dividing the sum of the wages of all the workers by the total number of workers in the industrial concern. It gives a fair idea of the average wage bill taking into account all the workers.
    (v) Arithmetic Mean: The algebraic sum of the deviations of values about the arithmetic mean is zero. Hence, when the sum of absolute deviations from the average is the least, then mean could be used to calculate the average.
    (vi) Median: Median will be the most suitable measure when the variables are in ratios as it is least affected by the extreme values.
    (vii) Median: Median is the most suitable measure because it can be easily computed even in an open-ended frequency distribution and will not get affected by extreme values.

    2. Indicate the most appropriate alternative from the multiple choices provided against each question.
    (i) The most suitable average for qualitative measurement is:
    (a) arithmetic mean
    (b) median
    (c) mode
    (d) geometric mean
    (e) none of the above
    Ans. (b) median

    Explanation:

    Median is the most relevant average for qualitative measurement because it divides a series into two equal parts representing the average qualitative measure without being affected by extreme values.

    (ii) Which average is affected most by the presence of extreme items?
    (a) median
    (b) mode
    (c) arithmetic mean
    (d) none of the above
    Ans. (c) arithmetic mean

    Explanation:

    Arithmetic mean is mostly affected by the presence of extreme items. It is easily affected by extreme values. The value of the arithmetic mean may not figure out at all in the series.
    (iii) The algebraic sum of deviation of a set of n values from A.M. is:
    (a) n
    (b) 0
    (c) 1
    (d) none of these
    Ans. (b) 0

    Explanation:

    The algebraic sum of deviation of a set of n values from A.M. is zero. This is one of the mathematical properties of the arithmetic mean.

    3. Comment whether the following statements are true or false.
    (i) The sum of deviation of items from median is zero.
    (ii) An average alone is not enough to compare series.
    (iii) Arithmetic mean is a positional value.
    (iv) Upper quartile is the low est value of top 25% of items.
    (v) Median is unduly affected by extreme observations.

    Ans. (i) False
    The mathematical property stated here applies to the arithmetic mean and not to the median.
    (ii) True
    The average is not enough to compare the series as it does not explain the scope of deviation of different items from the central tendency and the difference in the frequency of values. These are measured by the measures of dispersion.
    (iii) False
    The given statement is false as arithmetic mean is not a positional value, but median and mode are because the calculation of them is based on the position of the items.
    (iv) True
    The upper quartile also called the third quartile has 75% of the items below it and 25% of the items above it.
    (v) False
    Median is a positional average as it is calculated by observation of a series and not by extereme values. Hence, arithmetic mean is unduly affected by extreme obseravtions.

    4. If the arithmetic mean of the data given below is 28, find
    (a) the missing frequency, and
    (b) the median of the series:

    Profit per retail shop (in Rs) Number of retail shops
    0 – 10 12
    10 – 20 18
    20 – 30 27
    30 – 40
    40 – 50 17
    50 – 60 6

    Ans. (a) The missing frequency

    Profit per Retail Shop (in ₹) No. of Shops (f) Mid-value (m) fm
    0 – 10 12 5 60
    10 – 20 18 15 270
    20 – 30 27 25 675
    30 – 40 f1 35 35f1
    40 – 50 17 45 765
    50 – 60 6 55 330
    Σf = 80 + f1 2100 + 35f1

    $$ \text{Arithmetic mea} \text{\={(X)}}= 28 \text{(given)}\\\text{\={(X)}}=\frac{\Sigma f m}{\Sigma f}\\28=\frac{2100+35f_1}{80+f_1}$$

    28(80 + f1) = 2100 + 35f1
    2240 + 28f1 = 2100 + 35f1
    2240 – 2100 = 35f1 – 28f1
    140 = 7f1
    20 = f
    Hence, the missing frequency is 20.

    (b) The median of the series 

    CI Frequency (f) Cumulative Frequency (cf)
    0 – 10 12 12
    10 – 20 18 30
    20 – 30 27 57
    30 – 40 20 77
    40 – 50 17 94
    50 – 60 6 100
    Σf = 100

    N = Σf = 100
    Median class (M) = Size of (N/2)th term
    M = 50th term
    Hence, median class is 20 – 30 as 50th item lies in 57th cumulative frequnecy.
    $$M=L+\frac{\frac{N}{2}-cf}{f}×i\\=20+\frac{\frac{100}{2}-30}{27}×10\\=20+\frac{50-30}{27}×10\\M=20+\frac{20}{27}×10\\M=20+7·40\\M=27·41.$$

    5. The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.

    Workers Daily Income (in ₹)
    A 120
    B 150
    C 180
    D 200
    E 250
    F 300
    G 220
    H 350
    I 370
    J 260

    Ans. 

    Workers Daily Income (in ₹)
    A 120
    B 150
    C 180
    D 200
    E 250
    F 300
    G 220
    H 350
    I 370
    J 260
    Total ΣX = 2400

    N = 10
    $$\text{\={X}}=\frac{\Sigma X}{N}\\\text{\={X}}=\frac{2400}{10}=₹240\\\text{Arithmetic mean} = ₹240.$$

    6. Following information pertains to the daily income of 150 families. Calculate the arithmetic mean.

    Income (in ₹) Number of families
    More than 75 150
    More than 85 140
    More than 95 115
    More than 105 95
    More than 115 70
    More than 125 60
    More than 135 40
    More than 145 25

    Ans.

    C.I. cf f m fm
    75 – 85 150 10 80 800
    85 – 95 140 25 90 2250
    95 – 105 115 20 100 2000
    105 – 115 95 25 110 2750
    115 – 125 70 10 120 1200
    125 – 135 60 20 130 2600
    135 – 145 40 15 140 2100
    145 – 155 25 25 150 3750
    150 17450

    $$\text{\={X}}=\frac{\Sigma fm}{\Sigma f}\\\text{\={X}}=\frac{17450}{150}\\\text{\={X}}= 116·33\\\text{Arithmetic mean} = ₹116·33.$$

    7. The size of land holdings of 380 families in a village is given below. Find the median size of land holdings.

    Size of Land Holdings (in acres) Number of families
    Less than 100 40
    100 – 200 89
    200 – 300 148
    300 – 400 64
    400 and above 39

    Ans.

    Size of Land Holdings C.I. Number of families (f) Cumulative Frequency (c.f.)
    0 – 100 40 40
    100 – 200 89 129
    200 – 300 148 277
    300 – 400 64 341
    400 – 500 39 380
    Σf = 380

    Σ f = N = 380
    M class = Size of(N/2) th item
    Median class = 190th item
    Hence, median class interval is 200 – 300. 
    $$M=L_1+\frac{\frac{N}{2}+cf}{f}\times i\\=200+\frac{\frac{380}{2}-129}{148}\times 100\\M=200+(\frac{190-129}{148})\times 100\\M = 200 + 41·22\\M = 241·22.$$

    8. The following series relates to the daily income of workers employed in a firm. Compute
    (a) highest income of lowest 50% workers
    (b) minimum income earned by the top 25% workers and
    (c) maximum income earned by lowest 25% workers.
    [Hint: Compute median, low er quartile and upper quartile.]

    Daily Income (in ₹) 10-14 15-19 20-24 25-29 30-34 35-39
    Number of workers 5 10 15 20 10 5

    Ans. 

    Daily Income (in `) Exclusive Group Number of workers (f) Cumulative frequency (cf)
    10 – 14 9.5 – 14.5 5 5
    15 – 19 14.5 – 19.5 10 15
    20 – 24 19.5 – 24.5 15 30
    25 – 29 24.5 – 29.5 20 50
    30 – 34 29.5 – 34.5 10 60
    35 – 39 34.5 – 39.5 5 65
    n = Σf = 65

    (a) Highest income of lowest 50% workers will be given by the median.
    Σf = N = 65
    $$\text{Median class} =\frac{n}{2}\text{th item}=\frac{65}{2}=32.5\\M=L_1+(\frac{\frac{N}{2}-cf}{f})\times i\\=24·5 +\frac{65}{2}\\M=24·5 +\frac{32.5-30}{20}\times100\\M=24·5 + 0·625 = 25·125.$$

    (b) Minimum income earned by the lowest 25% workers will be given by the Quartile (Q1).
    $$\text{Class interval of}\space Q_1=\frac{N}{2}\text{th item}\\Q_1=\frac{65}{4}\text{th item}\\Q_1 = 16·25^{th} item\\\text{Quartile class}=19·5–24·5\\Q_1=L_1+(\frac{\frac{N}{4}-cf}{f})\times i\\Q_1=19.5+(\frac{16.25-15}{15})\times 5\\Q_1=19.5+(\frac{1.2\times5}{15})\\Q_1 = 19·5 + 0·42\\Q_1 = 19·92.$$

    (c) Maximum income earned by the top 25% workers will be given by the upper quartiler Q3.
    $$\text{Class interval of}\space Q_3=3(\frac{N}{2})\text{th item}\\Q_3=3(\frac{65}{4})\text{th item}\\Q_3 = 3\times16·25^{th} item\\Q_3 = 48·75^{th} item\\\text{It lies in}\space 24·5 – 29·5\\\text{Quartile class}=19·5–24·5\\Q_3=L_1+(\frac{\frac{N}{4}-cf}{f})\times i\\Q_3=24.5+(\frac{48.75-30}{20})\times 5\\Q_3=24.5+(\frac{18.75\times5}{20})\\Q_3 = 24·5 + 4.69\\Q_3 = 29·19.$$

    9. The following table gives production yield in kg. per hectare of wheat of 150 farms in a village. Calculate the mean, median and mode values.

    Production yield (kg. per hectare) Number of farms
    50 – 53 3
    53 – 56 8
    56 – 59 14
    59 – 62 30
    62 – 65 36
    65 – 68 28
    68 – 71 16
    71 – 74 10
    74 – 77 5

    Ans. (i) Mean

    (Production Yield) C.I. Number of Farms (f) Mid-value (m) fm
    50 – 53 3 51.5 154.5
    53 – 56 8 54.5 436.0
    56 – 59 14 57.5 805.0
    59 – 62 30 60.5 1815.0
    62 – 65 36 63.5 2286.0
    65 – 68 28 66.5 1862.0
    68 – 71 16 69.5 1112.0
    71 – 74 10 72.5 725.0
    74 – 77 5 75.5 377.5
    Σf = 150 Σfm = 9573.0

    (ii) Median 

    CI f cf
    50 – 53 3 3
    53 – 56 8 11
    56 – 59 14 25
    59 – 62 30 55
    62 – 65 36 91
    65 – 68 28 119
    68 – 71 16 135
    71 – 74 10 145
    74 – 77 5 150

    (iii) Mode
    Mode
    = 3 Median – 2 Mean
    = 3 × 63·67 – 2 × 63·82
    = 191·01 – 127·64
    = 63·37.

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