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NCERT Solutions · Class 11 Mathematics Statistics

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EXERCISE 13.1 1–12 (part 1 of 3)

  1. Find the mean deviation about the mean for the data in Exercises $\displaystyle 1$ and 2.

    Exercise 1

    $\displaystyle 4$, $\displaystyle 7$, $\displaystyle 8$, $\displaystyle 9$, $\displaystyle 10$, $\displaystyle 12$, $\displaystyle 13$, $\displaystyle 17$

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    NCERT’s answer
    $\displaystyle 3$
    Mean deviation about the mean. For raw data, \(\displaystyle \text{M.D.}(\bar{x}) = \dfrac{1}{n}\sum |x_i - \bar{x}| \) — the mean of the deviations from the mean, signs ignored.Here \(\displaystyle n = 8\), so \[\bar{x} = \frac{4+7+8+9+10+12+13+17}{8} = \frac{80}{8} = 10 \]Absolute deviations \(\displaystyle |x_i - 10|\): \[6,\; 3,\; 2,\; 1,\; 0,\; 2,\; 3,\; 7 \qquad \sum |x_i - \bar{x}| = 24 \]\[\text{M.D.}(\bar{x}) = \frac{24}{8} = 3 \]Mean deviation about the mean \(\displaystyle = 3\).

  2. Exercise 2

    $\displaystyle 38$, $\displaystyle 70$, $\displaystyle 48$, $\displaystyle 40$, $\displaystyle 42$, $\displaystyle 55$, $\displaystyle 63$, $\displaystyle 46$, $\displaystyle 54$, $\displaystyle 44$

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    NCERT’s answer
    8.$\displaystyle 4$
    Mean deviation about the mean. \(\displaystyle \text{M.D.}(\bar{x}) = \dfrac{1}{n}\sum |x_i - \bar{x}| \).Here \(\displaystyle n = 10\) and the total is \(\displaystyle 38+70+48+40+42+55+63+46+54+44 = 500\), so \[\bar{x} = \frac{500}{10} = 50 \]Absolute deviations \(\displaystyle |x_i - 50|\): \[12,\; 20,\; 2,\; 10,\; 8,\; 5,\; 13,\; 4,\; 4,\; 6 \qquad \sum |x_i - \bar{x}| = 84 \]\[\text{M.D.}(\bar{x}) = \frac{84}{10} = 8.4 \]Mean deviation about the mean \(\displaystyle = 8.4\).
  3. Find the mean deviation about the median for the data in Exercises $\displaystyle 3$ and 4.

    Exercise 3

    $\displaystyle 13$, $\displaystyle 17$, $\displaystyle 16$, $\displaystyle 14$, $\displaystyle 11$, $\displaystyle 13$, $\displaystyle 10$, $\displaystyle 16$, $\displaystyle 11$, $\displaystyle 18$, $\displaystyle 12$, $\displaystyle 17$

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    NCERT’s answer
    2.$\displaystyle 33$
    Mean deviation about the median. Arrange the data in ascending order, read off the median \(\displaystyle M\), then \(\displaystyle \text{M.D.}(M) = \dfrac{1}{n}\sum |x_i - M| \).Arranged (\(\displaystyle n = 12\)): \[10,\;11,\;11,\;12,\;13,\;13,\;14,\;16,\;16,\;17,\;17,\;18 \]Since \(\displaystyle n\) is even, the median is the mean of the \(\displaystyle \left(\tfrac{n}{2}\right)^{\text{th}}\) and \(\displaystyle \left(\tfrac{n}{2}+1\right)^{\text{th}}\) observations — the 6th and the 7th: \[M = \frac{13+14}{2} = 13.5 \]Absolute deviations \(\displaystyle |x_i - 13.5|\): \[3.5,\;2.5,\;2.5,\;1.5,\;0.5,\;0.5,\;0.5,\;2.5,\;2.5,\;3.5,\;3.5,\;4.5 \qquad \sum = 28 \]\[\text{M.D.}(M) = \frac{28}{12} = \frac{7}{3} \approx 2.33 \]Mean deviation about the median \(\displaystyle = \dfrac{7}{3} \approx 2.33\).

  4. Exercise 4

    $\displaystyle 36$, $\displaystyle 72$, $\displaystyle 46$, $\displaystyle 42$, $\displaystyle 60$, $\displaystyle 45$, $\displaystyle 53$, $\displaystyle 46$, $\displaystyle 51$, $\displaystyle 49$

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    NCERT’s answer
    $\displaystyle 7$
    Mean deviation about the median. \(\displaystyle \text{M.D.}(M) = \dfrac{1}{n}\sum |x_i - M| \).Arranged in ascending order (\(\displaystyle n = 10\)): \[36,\;42,\;45,\;46,\;46,\;49,\;51,\;53,\;60,\;72 \]\(\displaystyle n\) is even, so the median is the mean of the 5th and 6th observations: \[M = \frac{46+49}{2} = 47.5 \]Absolute deviations \(\displaystyle |x_i - 47.5|\): \[11.5,\;5.5,\;2.5,\;1.5,\;1.5,\;1.5,\;3.5,\;5.5,\;12.5,\;24.5 \qquad \sum = 70 \]\[\text{M.D.}(M) = \frac{70}{10} = 7 \]Mean deviation about the median \(\displaystyle = 7\).
  5. Find the mean deviation about the mean for the data in Exercises $\displaystyle 5$ and 6.

    Exercise 5

    \(\displaystyle x_i\)$\displaystyle 5$$\displaystyle 10$$\displaystyle 15$$\displaystyle 20$$\displaystyle 25$
    \(\displaystyle f_i\)$\displaystyle 7$$\displaystyle 4$$\displaystyle 6$$\displaystyle 3$$\displaystyle 5$

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    NCERT’s answer
    6.$\displaystyle 32$
    Mean deviation about the mean (discrete frequency distribution). With \(\displaystyle N = \sum f_i\), \[\bar{x} = \frac{\sum f_i x_i}{N}, \qquad \text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N} \]First get the mean: \(\displaystyle \sum f_i x_i = 35+40+90+60+125 = 350 \) and \(\displaystyle N = 25 \), so \[\bar{x} = \frac{350}{25} = 14 \]Now measure every \(\displaystyle x_i\) from \(\displaystyle 14\):
    \(\displaystyle x_i\)\(\displaystyle f_i\)\(\displaystyle f_i x_i\)\(\displaystyle |x_i-\bar{x}|\)\(\displaystyle f_i|x_i-\bar{x}|\)
    $\displaystyle 5$$\displaystyle 7$$\displaystyle 35$$\displaystyle 9$$\displaystyle 63$
    $\displaystyle 10$$\displaystyle 4$$\displaystyle 40$$\displaystyle 4$$\displaystyle 16$
    $\displaystyle 15$$\displaystyle 6$$\displaystyle 90$$\displaystyle 1$$\displaystyle 6$
    $\displaystyle 20$$\displaystyle 3$$\displaystyle 60$$\displaystyle 6$$\displaystyle 18$
    $\displaystyle 25$$\displaystyle 5$$\displaystyle 125$$\displaystyle 11$$\displaystyle 55$
    Total$\displaystyle 25$$\displaystyle 350$$\displaystyle 158$
    \[\text{M.D.}(\bar{x}) = \frac{158}{25} = 6.32 \]Mean deviation about the mean \(\displaystyle = 6.32\).

  6. Exercise 6

    \(\displaystyle x_i\)$\displaystyle 10$$\displaystyle 30$$\displaystyle 50$$\displaystyle 70$$\displaystyle 90$
    \(\displaystyle f_i\)$\displaystyle 4$$\displaystyle 24$$\displaystyle 28$$\displaystyle 16$$\displaystyle 8$

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    NCERT’s answer
    $\displaystyle 16$
    Mean deviation about the mean (discrete frequency distribution). With \(\displaystyle N = \sum f_i\), \[\bar{x} = \frac{\sum f_i x_i}{N}, \qquad \text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N} \]Here \(\displaystyle \sum f_i x_i = 40+720+1400+1120+720 = 4000 \) and \(\displaystyle N = 4+24+28+16+8 = 80 \), so \[\bar{x} = \frac{4000}{80} = 50 \]
    \(\displaystyle x_i\)\(\displaystyle f_i\)\(\displaystyle f_i x_i\)\(\displaystyle |x_i-\bar{x}|\)\(\displaystyle f_i|x_i-\bar{x}|\)
    $\displaystyle 10$$\displaystyle 4$$\displaystyle 40$$\displaystyle 40$$\displaystyle 160$
    $\displaystyle 30$$\displaystyle 24$$\displaystyle 720$$\displaystyle 20$$\displaystyle 480$
    $\displaystyle 50$$\displaystyle 28$$\displaystyle 1400$$\displaystyle 0$$\displaystyle 0$
    $\displaystyle 70$$\displaystyle 16$$\displaystyle 1120$$\displaystyle 20$$\displaystyle 320$
    $\displaystyle 90$$\displaystyle 8$$\displaystyle 720$$\displaystyle 40$$\displaystyle 320$
    Total$\displaystyle 80$$\displaystyle 4000$$\displaystyle 1280$
    \[\text{M.D.}(\bar{x}) = \frac{1280}{80} = 16 \]Mean deviation about the mean \(\displaystyle = 16\).
  7. Find the mean deviation about the median for the data in Exercises $\displaystyle 7$ and 8.

    Exercise 7

    \(\displaystyle x_i\)$\displaystyle 5$$\displaystyle 7$$\displaystyle 9$$\displaystyle 10$$\displaystyle 12$$\displaystyle 15$
    \(\displaystyle f_i\)$\displaystyle 8$$\displaystyle 6$$\displaystyle 2$$\displaystyle 2$$\displaystyle 2$$\displaystyle 6$

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    NCERT’s answer
    3.$\displaystyle 23$
    Mean deviation about the median (discrete frequency distribution). Build the cumulative frequency to locate the median \(\displaystyle M\), then \[\text{M.D.}(M) = \frac{\sum f_i |x_i - M|}{N} \]The cumulative frequencies against \(\displaystyle x_i = 5,\,7,\,9,\,10,\,12,\,15\) are \[8,\; 14,\; 16,\; 18,\; 20,\; 26 \qquad (N = 26) \]\(\displaystyle N\) is even, so the median is the mean of the \(\displaystyle 13^{\text{th}}\) and \(\displaystyle 14^{\text{th}}\) observations. The cumulative frequency first reaches \(\displaystyle 13\) — and \(\displaystyle 14\) as well — at the entry with c.f. \(\displaystyle 14\), whose value is \(\displaystyle 7\). Both observations are therefore \(\displaystyle 7\), so \[M = \frac{7+7}{2} = 7 \]
    \(\displaystyle x_i\)\(\displaystyle f_i\)c.f.\(\displaystyle |x_i-M|\)\(\displaystyle f_i|x_i-M|\)
    $\displaystyle 5$$\displaystyle 8$$\displaystyle 8$$\displaystyle 2$$\displaystyle 16$
    $\displaystyle 7$$\displaystyle 6$$\displaystyle 14$$\displaystyle 0$$\displaystyle 0$
    $\displaystyle 9$$\displaystyle 2$$\displaystyle 16$$\displaystyle 2$$\displaystyle 4$
    $\displaystyle 10$$\displaystyle 2$$\displaystyle 18$$\displaystyle 3$$\displaystyle 6$
    $\displaystyle 12$$\displaystyle 2$$\displaystyle 20$$\displaystyle 5$$\displaystyle 10$
    $\displaystyle 15$$\displaystyle 6$$\displaystyle 26$$\displaystyle 8$$\displaystyle 48$
    Total$\displaystyle 26$$\displaystyle 84$
    \[\text{M.D.}(M) = \frac{84}{26} = \frac{42}{13} \approx 3.23 \]Mean deviation about the median \(\displaystyle = \dfrac{42}{13} \approx 3.23\).

  8. Exercise 8

    \(\displaystyle x_i\)$\displaystyle 15$$\displaystyle 21$$\displaystyle 27$$\displaystyle 30$$\displaystyle 35$
    \(\displaystyle f_i\)$\displaystyle 3$$\displaystyle 5$$\displaystyle 6$$\displaystyle 7$$\displaystyle 8$

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    NCERT’s answer
    5.$\displaystyle 1$
    Mean deviation about the median (discrete frequency distribution). Build the cumulative frequency to locate the median \(\displaystyle M\), then \[\text{M.D.}(M) = \frac{\sum f_i |x_i - M|}{N} \]The cumulative frequencies against \(\displaystyle x_i = 15,\,21,\,27,\,30,\,35\) are \[3,\; 8,\; 14,\; 21,\; 29 \qquad (N = 29) \]\(\displaystyle N\) is odd, so the median is the \(\displaystyle \left(\dfrac{N+1}{2}\right)^{\text{th}} = 15^{\text{th}}\) observation. The cumulative frequency first reaches \(\displaystyle 15\) at the entry with c.f. \(\displaystyle 21\), whose value is \(\displaystyle 30\), so \(\displaystyle M = 30\).
    \(\displaystyle x_i\)\(\displaystyle f_i\)c.f.\(\displaystyle |x_i-M|\)\(\displaystyle f_i|x_i-M|\)
    $\displaystyle 15$$\displaystyle 3$$\displaystyle 3$$\displaystyle 15$$\displaystyle 45$
    $\displaystyle 21$$\displaystyle 5$$\displaystyle 8$$\displaystyle 9$$\displaystyle 45$
    $\displaystyle 27$$\displaystyle 6$$\displaystyle 14$$\displaystyle 3$$\displaystyle 18$
    $\displaystyle 30$$\displaystyle 7$$\displaystyle 21$$\displaystyle 0$$\displaystyle 0$
    $\displaystyle 35$$\displaystyle 8$$\displaystyle 29$$\displaystyle 5$$\displaystyle 40$
    Total$\displaystyle 29$$\displaystyle 148$
    \[\text{M.D.}(M) = \frac{148}{29} \approx 5.1 \]Mean deviation about the median \(\displaystyle = \dfrac{148}{29} \approx 5.1\).
  9. Find the mean deviation about the mean for the data in Exercises $\displaystyle 9$ and 10.

    Exercise 9

    Income per day in ₹$\displaystyle 0$-$\displaystyle 100$$\displaystyle 100$-$\displaystyle 200$$\displaystyle 200$-$\displaystyle 300$$\displaystyle 300$-$\displaystyle 400$$\displaystyle 400$-$\displaystyle 500$$\displaystyle 500$-$\displaystyle 600$$\displaystyle 600$-$\displaystyle 700$$\displaystyle 700$-$\displaystyle 800$
    Number of persons$\displaystyle 4$$\displaystyle 8$$\displaystyle 9$$\displaystyle 10$$\displaystyle 7$$\displaystyle 5$$\displaystyle 4$$\displaystyle 3$

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    NCERT’s answer
    157.$\displaystyle 92$
    Mean deviation about the mean (continuous frequency distribution). Replace each class by its mid-point \(\displaystyle x_i\) and treat the data as a discrete distribution: \[\bar{x} = \frac{\sum f_i x_i}{N}, \qquad \text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N} \]Here \(\displaystyle \sum f_i x_i = 17900 \) and \(\displaystyle N = 50 \), so \[\bar{x} = \frac{17900}{50} = 358 \]
    Income (₹)\(\displaystyle f_i\)mid-point \(\displaystyle x_i\)\(\displaystyle f_i x_i\)\(\displaystyle |x_i-\bar{x}|\)\(\displaystyle f_i|x_i-\bar{x}|\)
    $\displaystyle 0$–$\displaystyle 100$$\displaystyle 4$$\displaystyle 50$$\displaystyle 200$$\displaystyle 308$$\displaystyle 1232$
    $\displaystyle 100$–$\displaystyle 200$$\displaystyle 8$$\displaystyle 150$$\displaystyle 1200$$\displaystyle 208$$\displaystyle 1664$
    $\displaystyle 200$–$\displaystyle 300$$\displaystyle 9$$\displaystyle 250$$\displaystyle 2250$$\displaystyle 108$$\displaystyle 972$
    $\displaystyle 300$–$\displaystyle 400$$\displaystyle 10$$\displaystyle 350$$\displaystyle 3500$$\displaystyle 8$$\displaystyle 80$
    $\displaystyle 400$–$\displaystyle 500$$\displaystyle 7$$\displaystyle 450$$\displaystyle 3150$$\displaystyle 92$$\displaystyle 644$
    $\displaystyle 500$–$\displaystyle 600$$\displaystyle 5$$\displaystyle 550$$\displaystyle 2750$$\displaystyle 192$$\displaystyle 960$
    $\displaystyle 600$–$\displaystyle 700$$\displaystyle 4$$\displaystyle 650$$\displaystyle 2600$$\displaystyle 292$$\displaystyle 1168$
    $\displaystyle 700$–$\displaystyle 800$$\displaystyle 3$$\displaystyle 750$$\displaystyle 2250$$\displaystyle 392$$\displaystyle 1176$
    Total$\displaystyle 50$$\displaystyle 17900$$\displaystyle 7896$
    \[\text{M.D.}(\bar{x}) = \frac{7896}{50} = 157.92 \]Mean income \(\displaystyle = ₹358\) and mean deviation about the mean \(\displaystyle = ₹157.92\).

  10. Exercise 10

    Height in cms$\displaystyle 95$-$\displaystyle 105$$\displaystyle 105$-$\displaystyle 115$$\displaystyle 115$-$\displaystyle 125$$\displaystyle 125$-$\displaystyle 135$$\displaystyle 135$-$\displaystyle 145$$\displaystyle 145$-$\displaystyle 155$
    Number of boys$\displaystyle 9$$\displaystyle 13$$\displaystyle 26$$\displaystyle 30$$\displaystyle 12$$\displaystyle 10$

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    NCERT’s answer
    11.$\displaystyle 28$
    Mean deviation about the mean (continuous frequency distribution). Use the class mid-points \(\displaystyle x_i\): \[\bar{x} = \frac{\sum f_i x_i}{N}, \qquad \text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{N} \]Here \(\displaystyle \sum f_i x_i = 12530 \) and \(\displaystyle N = 100 \), so \[\bar{x} = \frac{12530}{100} = 125.3 \]
    Height (cm)\(\displaystyle f_i\)mid-point \(\displaystyle x_i\)\(\displaystyle f_i x_i\)\(\displaystyle |x_i-\bar{x}|\)\(\displaystyle f_i|x_i-\bar{x}|\)
    $\displaystyle 95$–$\displaystyle 105$$\displaystyle 9$$\displaystyle 100$$\displaystyle 900$$\displaystyle 25.3$$\displaystyle 227.7$
    $\displaystyle 105$–$\displaystyle 115$$\displaystyle 13$$\displaystyle 110$$\displaystyle 1430$$\displaystyle 15.3$$\displaystyle 198.9$
    $\displaystyle 115$–$\displaystyle 125$$\displaystyle 26$$\displaystyle 120$$\displaystyle 3120$$\displaystyle 5.3$$\displaystyle 137.8$
    $\displaystyle 125$–$\displaystyle 135$$\displaystyle 30$$\displaystyle 130$$\displaystyle 3900$$\displaystyle 4.7$$\displaystyle 141.0$
    $\displaystyle 135$–$\displaystyle 145$$\displaystyle 12$$\displaystyle 140$$\displaystyle 1680$$\displaystyle 14.7$$\displaystyle 176.4$
    $\displaystyle 145$–$\displaystyle 155$$\displaystyle 10$$\displaystyle 150$$\displaystyle 1500$$\displaystyle 24.7$$\displaystyle 247.0$
    Total$\displaystyle 100$$\displaystyle 12530$$\displaystyle 1128.8$
    \[\text{M.D.}(\bar{x}) = \frac{1128.8}{100} = 11.288 \]Mean height \(\displaystyle = 125.3\) cm and mean deviation about the mean \(\displaystyle = 11.288\) cm.

  11. Exercise 11

    Find the mean deviation about median for the following data :
    Marks$\displaystyle 0$-$\displaystyle 10$$\displaystyle 10$-$\displaystyle 20$$\displaystyle 20$-$\displaystyle 30$$\displaystyle 30$-$\displaystyle 40$$\displaystyle 40$-$\displaystyle 50$$\displaystyle 50$-$\displaystyle 60$
    Number of Girls$\displaystyle 6$$\displaystyle 8$$\displaystyle 14$$\displaystyle 16$$\displaystyle 4$$\displaystyle 2$

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    NCERT’s answer
    10.$\displaystyle 34$
    Mean deviation about the median (continuous frequency distribution). Find the median class from the cumulative frequency, then \[M = l + \frac{\dfrac{N}{2} - C}{f} \times h \] where \(\displaystyle l\) is the lower limit of the median class, \(\displaystyle C\) the cumulative frequency before it, \(\displaystyle f\) its frequency and \(\displaystyle h\) its width. Then \(\displaystyle \text{M.D.}(M) = \dfrac{\sum f_i |x_i - M|}{N} \), taking \(\displaystyle x_i\) to be the class mid-points.Cumulative frequencies for the classes \(\displaystyle 0\)–\(\displaystyle 10, \dots, 50\)–\(\displaystyle 60\): \[6,\; 14,\; 28,\; 44,\; 48,\; 50 \qquad (N = 50) \]Since \(\displaystyle \dfrac{N}{2} = 25\) and the cumulative frequency first crosses \(\displaystyle 25\) in the class \(\displaystyle 20\)–\(\displaystyle 30\), that is the median class: \(\displaystyle l = 20\), \(\displaystyle C = 14\), \(\displaystyle f = 14\), \(\displaystyle h = 10\). \[M = 20 + \frac{25 - 14}{14} \times 10 = 20 + \frac{110}{14} = \frac{195}{7} \approx 27.86 \]Keeping the sevenths exact avoids rounding error in the column total:
    Marks\(\displaystyle f_i\)c.f.mid-point \(\displaystyle x_i\)\(\displaystyle |x_i-M|\)\(\displaystyle f_i|x_i-M|\)
    $\displaystyle 0$–$\displaystyle 10$$\displaystyle 6$$\displaystyle 6$$\displaystyle 5$\(\displaystyle \frac{160}{7}\)\(\displaystyle \frac{960}{7}\)
    $\displaystyle 10$–$\displaystyle 20$$\displaystyle 8$$\displaystyle 14$$\displaystyle 15$\(\displaystyle \frac{90}{7}\)\(\displaystyle \frac{720}{7}\)
    $\displaystyle 20$–$\displaystyle 30$$\displaystyle 14$$\displaystyle 28$$\displaystyle 25$\(\displaystyle \frac{20}{7}\)\(\displaystyle \frac{280}{7}\)
    $\displaystyle 30$–$\displaystyle 40$$\displaystyle 16$$\displaystyle 44$$\displaystyle 35$\(\displaystyle \frac{50}{7}\)\(\displaystyle \frac{800}{7}\)
    $\displaystyle 40$–$\displaystyle 50$$\displaystyle 4$$\displaystyle 48$$\displaystyle 45$\(\displaystyle \frac{120}{7}\)\(\displaystyle \frac{480}{7}\)
    $\displaystyle 50$–$\displaystyle 60$$\displaystyle 2$$\displaystyle 50$$\displaystyle 55$\(\displaystyle \frac{190}{7}\)\(\displaystyle \frac{380}{7}\)
    Total$\displaystyle 50$\(\displaystyle \mathbf{\frac{3620}{7}}\)
    \[\text{M.D.}(M) = \frac{1}{50}\cdot\frac{3620}{7} = \frac{362}{35} \approx 10.34 \]Median \(\displaystyle = \dfrac{195}{7} \approx 27.86\) marks and mean deviation about the median \(\displaystyle = \dfrac{362}{35} \approx 10.34\).

  12. Exercise 12

    Calculate the mean deviation about median age for the age distribution of $\displaystyle 100$ persons given below:
    Age (in years)$\displaystyle 16$-$\displaystyle 20$$\displaystyle 21$-$\displaystyle 25$$\displaystyle 26$-$\displaystyle 30$$\displaystyle 31$-$\displaystyle 35$$\displaystyle 36$-$\displaystyle 40$$\displaystyle 41$-$\displaystyle 45$$\displaystyle 46$-$\displaystyle 50$$\displaystyle 51$-$\displaystyle 55$
    Number$\displaystyle 5$$\displaystyle 6$$\displaystyle 12$$\displaystyle 14$$\displaystyle 26$$\displaystyle 12$$\displaystyle 16$$\displaystyle 9$
    [Hint Convert the given data into continuous frequency distribution by subtracting $\displaystyle 0.5$ from the lower limit and adding $\displaystyle 0.5$ to the upper limit of each class interval]

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    NCERT’s answer
    7.$\displaystyle 35$
    Mean deviation about the median (continuous frequency distribution). The classes \(\displaystyle 16\)–\(\displaystyle 20\), \(\displaystyle 21\)–\(\displaystyle 25\), \(\displaystyle \dots\) are inclusive, so they leave gaps between them. Following the hint, subtract \(\displaystyle 0.5\) from every lower limit and add \(\displaystyle 0.5\) to every upper limit to make them continuous: \(\displaystyle 15.5\)–\(\displaystyle 20.5\), \(\displaystyle 20.5\)–\(\displaystyle 25.5\), and so on. This correction does not move the mid-points.Then use \[M = l + \frac{\dfrac{N}{2} - C}{f} \times h, \qquad \text{M.D.}(M) = \frac{\sum f_i |x_i - M|}{N} \]Cumulative frequencies: \[5,\; 11,\; 23,\; 37,\; 63,\; 75,\; 91,\; 100 \qquad (N = 100) \]\(\displaystyle \dfrac{N}{2} = 50\), and the cumulative frequency first crosses \(\displaystyle 50\) in the class \(\displaystyle 35.5\)–\(\displaystyle 40.5\), so \(\displaystyle l = 35.5\), \(\displaystyle C = 37\), \(\displaystyle f = 26\), \(\displaystyle h = 5\): \[M = 35.5 + \frac{50 - 37}{26} \times 5 = 35.5 + \frac{65}{26} = 35.5 + 2.5 = 38 \]
    Age (years)\(\displaystyle f_i\)c.f.mid-point \(\displaystyle x_i\)\(\displaystyle |x_i-M|\)\(\displaystyle f_i|x_i-M|\)
    $\displaystyle 15.5$–$\displaystyle 20.5$$\displaystyle 5$$\displaystyle 5$$\displaystyle 18$$\displaystyle 20$$\displaystyle 100$
    $\displaystyle 20.5$–$\displaystyle 25.5$$\displaystyle 6$$\displaystyle 11$$\displaystyle 23$$\displaystyle 15$$\displaystyle 90$
    $\displaystyle 25.5$–$\displaystyle 30.5$$\displaystyle 12$$\displaystyle 23$$\displaystyle 28$$\displaystyle 10$$\displaystyle 120$
    $\displaystyle 30.5$–$\displaystyle 35.5$$\displaystyle 14$$\displaystyle 37$$\displaystyle 33$$\displaystyle 5$$\displaystyle 70$
    $\displaystyle 35.5$–$\displaystyle 40.5$$\displaystyle 26$$\displaystyle 63$$\displaystyle 38$$\displaystyle 0$$\displaystyle 0$
    $\displaystyle 40.5$–$\displaystyle 45.5$$\displaystyle 12$$\displaystyle 75$$\displaystyle 43$$\displaystyle 5$$\displaystyle 60$
    $\displaystyle 45.5$–$\displaystyle 50.5$$\displaystyle 16$$\displaystyle 91$$\displaystyle 48$$\displaystyle 10$$\displaystyle 160$
    $\displaystyle 50.5$–$\displaystyle 55.5$$\displaystyle 9$$\displaystyle 100$$\displaystyle 53$$\displaystyle 15$$\displaystyle 135$
    Total$\displaystyle 100$$\displaystyle 735$
    \[\text{M.D.}(M) = \frac{735}{100} = 7.35 \]Median age \(\displaystyle = 38\) years and mean deviation about the median age \(\displaystyle = 7.35\) years.