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NCERT Exemplar · Class 10 Mathematics Statistics and Probability

96 questions · 96 still being checked

EXERCISE 13.3 11–20 (part 7 of 11)

  1. Exercise 11

    Form the frequency distribution table from the following data :
    Marks (out of 90\displaystyle 90)Number of candidates
    More than or equal to 80\displaystyle 804\displaystyle 4
    More than or equal to 70\displaystyle 706\displaystyle 6
    More than or equal to 60\displaystyle 6011\displaystyle 11
    More than or equal to 50\displaystyle 5017\displaystyle 17
    More than or equal to 40\displaystyle 4023\displaystyle 23
    More than or equal to 30\displaystyle 3027\displaystyle 27
    More than or equal to 20\displaystyle 2030\displaystyle 30
    More than or equal to 10\displaystyle 1032\displaystyle 32
    More than or equal to 0\displaystyle 034\displaystyle 34

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    NCERT’s answer
    MarksNumber of candidates
    $\displaystyle 0$-$\displaystyle 10$$\displaystyle 2$
    $\displaystyle 10$-$\displaystyle 20$$\displaystyle 2$
    $\displaystyle 20$-$\displaystyle 30$$\displaystyle 3$
    $\displaystyle 30$-$\displaystyle 40$$\displaystyle 4$
    $\displaystyle 40$-$\displaystyle 50$$\displaystyle 6$
    $\displaystyle 50$-$\displaystyle 60$$\displaystyle 6$
    $\displaystyle 60$-$\displaystyle 70$$\displaystyle 5$
    $\displaystyle 70$-$\displaystyle 80$$\displaystyle 2$
    $\displaystyle 80$-$\displaystyle 90$$\displaystyle 4$
    Each class frequency is the difference of consecutive more-than-type cumulative frequencies. \[\begin{array}{|c|c|}\hline \text{Marks} & \text{Number of candidates} \\ \hline 0\text{-}10 & 2 \\ 10\text{-}20 & 2 \\ 20\text{-}30 & 3 \\ 30\text{-}40 & 4 \\ 40\text{-}50 & 6 \\ 50\text{-}60 & 6 \\ 60\text{-}70 & 5 \\ 70\text{-}80 & 2 \\ 80\text{-}90 & 4 \\ \hline \text{Total} & 34 \\ \hline \end{array} \]Answer: frequencies $\displaystyle 2$, $\displaystyle 2$, $\displaystyle 3$, $\displaystyle 4$, $\displaystyle 6$, $\displaystyle 6$, $\displaystyle 5$, $\displaystyle 2$, $\displaystyle 4$ (total $\displaystyle 34$).

  2. Exercise 12

    Find the unknown entries a,b,c,d,e,f\displaystyle a, b, c, d, e, f in the following distribution of heights of students in a class :
    Height (in cm)FrequencyCumulative frequency
    150\displaystyle 150-155\displaystyle 15512\displaystyle 12a\displaystyle a
    155\displaystyle 155-160\displaystyle 160b\displaystyle b25\displaystyle 25
    160\displaystyle 160-165\displaystyle 16510\displaystyle 10c\displaystyle c
    165\displaystyle 165-170\displaystyle 170d\displaystyle d43\displaystyle 43
    170\displaystyle 170-175\displaystyle 175e\displaystyle e48\displaystyle 48
    175\displaystyle 175-180\displaystyle 1802\displaystyle 2f\displaystyle f
    Total50\displaystyle 50

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    NCERT’s answer
    \(\displaystyle a=12, b=13, c=35, d=8, e=5, f=50\)
    \[a = 12 \quad \text{(cumulative frequency of the first class = its own frequency)} \] \[a+b=25 \Rightarrow b=25-12=13 \] \[c=a+b+10=25+10=35 \] \[c+d=43 \Rightarrow d=43-35=8 \] \[(c+d)+e=48 \Rightarrow e=48-43=5 \] \[f=48+2=50 \] \[12+13+10+8+5+2=50 \quad \text{(matches the given total)} \]Answer: \(\displaystyle a=12,\ b=13,\ c=35,\ d=8,\ e=5,\ f=50\).

  3. Exercise 13

    The following are the ages of 300\displaystyle 300 patients getting medical treatment in a hospital on a particular day :
    Age (in years)10\displaystyle 10-20\displaystyle 2020\displaystyle 20-30\displaystyle 3030\displaystyle 30-40\displaystyle 4040\displaystyle 40-50\displaystyle 5050\displaystyle 50-60\displaystyle 6060\displaystyle 60-70\displaystyle 70
    Number of patients60\displaystyle 6042\displaystyle 4255\displaystyle 5570\displaystyle 7053\displaystyle 5320\displaystyle 20
    Form:
    (i)
    Less than type cumulative frequency distribution.
    (ii)
    More than type cumulative frequency distribution.

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    (i)
    Running totals up to each upper limit give the less-than-type distribution.
    \[\begin{array}{|c|c|}\hline \text{Age (in years)} & \text{Number of patients} \\ \hline \text{Less than }20 & 60 \\ \text{Less than }30 & 102 \\ \text{Less than }40 & 157 \\ \text{Less than }50 & 227 \\ \text{Less than }60 & 280 \\ \text{Less than }70 & 300 \\ \hline \end{array} \]
    (ii)
    Each entry is $\displaystyle 300$ minus the less-than total at that limit.
    \[\begin{array}{|c|c|}\hline \text{Age (in years)} & \text{Number of patients} \\ \hline \text{More than or equal to }10 & 300 \\ \text{More than or equal to }20 & 240 \\ \text{More than or equal to }30 & 198 \\ \text{More than or equal to }40 & 143 \\ \text{More than or equal to }50 & 73 \\ \text{More than or equal to }60 & 20 \\ \hline \end{array} \]
    Answer: (i) less than $\displaystyle 20$, $\displaystyle 30$, $\displaystyle 40$, $\displaystyle 50$, $\displaystyle 60$, $\displaystyle 70$: $\displaystyle 60$, $\displaystyle 102$, $\displaystyle 157$, $\displaystyle 227$, $\displaystyle 280$, $\displaystyle 300$; (ii) at least $\displaystyle 10$, $\displaystyle 20$, $\displaystyle 30$, $\displaystyle 40$, $\displaystyle 50$, $\displaystyle 60$: $\displaystyle 300$, $\displaystyle 240$, $\displaystyle 198$, $\displaystyle 143$, $\displaystyle 73$, 20.

  4. Exercise 14

    Given below is a cumulative frequency distribution showing the marks secured by 50\displaystyle 50 students of a class :
    MarksBelow 20\displaystyle 20Below 40\displaystyle 40Below 60\displaystyle 60Below 80\displaystyle 80Below 100\displaystyle 100
    Number of students17\displaystyle 1722\displaystyle 2229\displaystyle 2937\displaystyle 3750\displaystyle 50
    Form the frequency distribution table for the data.

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    NCERT’s answer
    MarksNumber of students
    $\displaystyle 0$-$\displaystyle 20$$\displaystyle 17$
    $\displaystyle 20$-$\displaystyle 40$$\displaystyle 5$
    $\displaystyle 40$-$\displaystyle 60$$\displaystyle 7$
    $\displaystyle 60$-$\displaystyle 80$$\displaystyle 8$
    $\displaystyle 80$-$\displaystyle 100$$\displaystyle 13$
    Each class frequency is the difference of consecutive below-type cumulative frequencies. \[\begin{array}{|c|c|}\hline \text{Marks} & \text{Number of students} \\ \hline 0\text{-}20 & 17 \\ 20\text{-}40 & 5 \\ 40\text{-}60 & 7 \\ 60\text{-}80 & 8 \\ 80\text{-}100 & 13 \\ \hline \text{Total} & 50 \\ \hline \end{array} \]Answer: frequencies $\displaystyle 17$, $\displaystyle 5$, $\displaystyle 7$, $\displaystyle 8$, $\displaystyle 13$ (total $\displaystyle 50$).

  5. Exercise 15

    Weekly income of 600\displaystyle 600 families is tabulated below :
    Weekly income (in Rs)Number of families
    0\displaystyle 0-1000\displaystyle 1000250\displaystyle 250
    1000\displaystyle 1000-2000\displaystyle 2000190\displaystyle 190
    2000\displaystyle 2000-3000\displaystyle 3000100\displaystyle 100
    3000\displaystyle 3000-4000\displaystyle 400040\displaystyle 40
    4000\displaystyle 4000-5000\displaystyle 500015\displaystyle 15
    5000\displaystyle 5000-6000\displaystyle 60005\displaystyle 5
    Total600\displaystyle 600
    Compute the median income.

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    NCERT’s answer
    Rs $\displaystyle 1263.15$
    Cumulative frequencies (less-than type): \[0\text{--}1000:250,\ 1000\text{--}2000:440,\ 2000\text{--}3000:540,\ 3000\text{--}4000:580,\ 4000\text{--}5000:595,\ 5000\text{--}6000:600 \] \[\frac{n}{2}=\frac{600}{2}=300 \] The cf first exceeding \(\displaystyle 300\) is \(\displaystyle 440\), so the median class is \(\displaystyle 1000\text{--}2000\). \[\text{Median}=l+\left(\frac{\dfrac{n}{2}-cf}{f}\right)h \] \[=1000+\left(\frac{300-250}{190}\right)\times1000 \] \[=1000+\frac{50000}{190} \] \[\approx1000+263.16 \] \[\approx1263.16 \]Answer: Median income \(\displaystyle \approx\) Rs $\displaystyle 1263.16$

  6. Exercise 16

    The maximum bowling speeds, in km per hour, of 33\displaystyle 33 players at a cricket coaching centre are given as follows :
    Speed (km/h)85\displaystyle 85-100\displaystyle 100100\displaystyle 100-115\displaystyle 115115\displaystyle 115-130\displaystyle 130130\displaystyle 130-145\displaystyle 145
    Number of players11\displaystyle 119\displaystyle 98\displaystyle 85\displaystyle 5
    Calculate the median bowling speed.

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    NCERT’s answer
    109.$\displaystyle 17$ km/h
    Cumulative frequencies: \[85\text{--}100:11,\ 100\text{--}115:20,\ 115\text{--}130:28,\ 130\text{--}145:33 \] \[\frac{n}{2}=\frac{33}{2}=16.5 \] Median class: \(\displaystyle 100\text{--}115\) (cf \(\displaystyle 20\) is the first to exceed \(\displaystyle 16.5\)). \[\text{Median}=l+\left(\frac{\dfrac{n}{2}-cf}{f}\right)h=100+\left(\frac{16.5-11}{9}\right)\times15 \] \[=100+\frac{82.5}{9} \] \[\approx100+9.17 \] \[\approx109.17 \]Answer: Median bowling speed \(\displaystyle \approx 109.17\) km/h

  7. Exercise 17

    The monthly income of 100\displaystyle 100 families are given as below :
    Income (in Rs)Number of families
    0\displaystyle 0-5000\displaystyle 50008\displaystyle 8
    5000\displaystyle 5000-10000\displaystyle 1000026\displaystyle 26
    10000\displaystyle 10000-15000\displaystyle 1500041\displaystyle 41
    15000\displaystyle 15000-20000\displaystyle 2000016\displaystyle 16
    20000\displaystyle 20000-25000\displaystyle 250003\displaystyle 3
    25000\displaystyle 25000-30000\displaystyle 300003\displaystyle 3
    30000\displaystyle 30000-35000\displaystyle 350002\displaystyle 2
    35000\displaystyle 35000-40000\displaystyle 400001\displaystyle 1
    Calculate the modal income.

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    NCERT’s answer
    Rs $\displaystyle 11875$
    Highest frequency \(\displaystyle 41\) is in class \(\displaystyle 10000\text{--}15000\); this is the modal class, with \(\displaystyle f_0=26,\ f_1=41,\ f_2=16,\ h=5000\). \[\text{Mode}=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)h \] \[=10000+\left(\frac{41-26}{2(41)-26-16}\right)\times5000 \] \[=10000+\left(\frac{15}{40}\right)\times5000 \] \[=10000+1875=11875 \]Answer: Modal income = Rs $\displaystyle 11875$

  8. Exercise 18

    The weight of coffee in 70\displaystyle 70 packets are shown in the following table :
    Weight (in g)Number of packets
    200\displaystyle 200-201\displaystyle 20112\displaystyle 12
    201\displaystyle 201-202\displaystyle 20226\displaystyle 26
    202\displaystyle 202-203\displaystyle 20320\displaystyle 20
    203\displaystyle 203-204\displaystyle 2049\displaystyle 9
    204\displaystyle 204-205\displaystyle 2052\displaystyle 2
    205\displaystyle 205-206\displaystyle 2061\displaystyle 1
    Determine the modal weight.

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    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    Highest frequency \(\displaystyle 26\) is in class \(\displaystyle 201\text{--}202\); \(\displaystyle f_0=12,\ f_1=26,\ f_2=20,\ h=1\). \[\text{Mode}=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)h \] \[=201+\left(\frac{26-12}{52-12-20}\right)\times1 \] \[=201+\frac{14}{20}=201+0.7 \] \[=201.7 \]Answer: Modal weight = $\displaystyle 201.7$ g

  9. Exercise 19

    Two dice are thrown at the same time. Find the probability of getting
    (i)
    same number on both dice.
    (ii)
    different numbers on both dice.

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{1}{6}\)
    (ii)
    \(\displaystyle \frac{5}{6}\)
    \[n(S)=6\times6=36 \]
    (i)
    Same number: \(\displaystyle (1,1),(2,2),\dots,(6,6)\), \(\displaystyle 6\) outcomes.
    \[P(\text{same})=\frac{6}{36}=\frac{1}{6} \]
    (ii)
    Different numbers is the complement of same numbers.
    \[P(\text{different})=1-\frac{1}{6}=\frac{5}{6} \]
    Answer: (i) \(\displaystyle \frac{1}{6}\) (ii) \(\displaystyle \frac{5}{6}\)

  10. Exercise 20

    Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is
    (i)
    7\displaystyle 7? (ii) a prime number?
    (iii)
    1\displaystyle 1?

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{1}{6}\)
    (ii)
    \(\displaystyle \frac{5}{12}\)
    (iii)
    $\displaystyle 0$
    \[n(S)=36 \]
    (i)
    Sum \(\displaystyle 7\): \(\displaystyle (1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\), \(\displaystyle 6\) outcomes.
    \[P(7)=\frac{6}{36}=\frac{1}{6} \]
    (ii)
    Prime sums \(\displaystyle 2,3,5,7,11\) have outcome counts \(\displaystyle 1,2,4,6,2\).
    \[P(\text{prime})=\frac{1+2+4+6+2}{36}=\frac{15}{36}=\frac{5}{12} \]
    (iii)
    The least possible sum is \(\displaystyle 1+1=2\), so sum \(\displaystyle 1\) is impossible.
    \[P(1)=\frac{0}{36}=0 \]
    Answer: (i) \(\displaystyle \frac{1}{6}\) (ii) \(\displaystyle \frac{5}{12}\) (iii) \(\displaystyle 0\)