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NCERT Exemplar · Class 9 Mathematics Statistics and Probability

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EXERCISE 14.3 11–20 (part 6 of 7)

  1. Exercise 11

    A class consists of 50\displaystyle 50 students out of which 30\displaystyle 30 are girls. The mean of marks scored by girls in a test is 73\displaystyle 73 (out of 100\displaystyle 100 ) and that of boys is 71\displaystyle 71 . Determine the mean score of the whole class.

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    NCERT’s answer
    72.$\displaystyle 2$
    \[\text{Boys} = 50-30 = 20 \] \[\sum(\text{marks}) = 30(73)+20(71) = 2190+1420 = 3610 \] \[\text{Mean of class} = \frac{3610}{50} = 72.2 \]Answer: \(\displaystyle 72.2\).

  2. Exercise 12

    Mean of 50\displaystyle 50 observations was found to be 80.4. But later on, it was discovered that 96\displaystyle 96 was misread as 69\displaystyle 69 at one place. Find the correct mean.

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    80.$\displaystyle 94$
    \[\sum x_i(\text{wrong}) = 50 \times 80.4 = 4020 \] \[\sum x_i(\text{correct}) = 4020-69+96 = 4047 \] \[\text{Correct mean} = \frac{4047}{50} = 80.94 \]Answer: Correct mean \(\displaystyle = 80.94\).

  3. Exercise 13

    Ten observations 6\displaystyle 6, 14\displaystyle 14, 15\displaystyle 15, 17\displaystyle 17, x+1,2x−13,30,32,34,43\displaystyle x+1,2 x-13,30,32,34,43 are written in an ascending order. The median of the data is 24. Find the value of x\displaystyle x.

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    NCERT’s answer
    $\displaystyle 20$
    With \(\displaystyle n=10\), the median is the mean of the 5th and 6th terms. \[24 = \frac{(x+1)+(2x-13)}{2} \] \[48 = 3x-12 \] \[x = 20 \] \[x+1=21,\ 2x-13=27 \quad (17<21<27<30,\ \text{order holds}) \]Answer: \(\displaystyle x = 20\).

  4. Exercise 14

    The points scored by a basket ball team in a series of matches are as follows: 17,2,7,27,25,5,14,18,10,24,48,10,8,7,10,28\displaystyle 17,2,7,27,25,5,14,18,10,24,48,10,8,7,10,28 Find the median and mode for the data.

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    NCERT’s answer
    Median $\displaystyle =12$, mode $\displaystyle =10$
    Arranged in ascending order: \[2,\,5,\,7,\,7,\,8,\,10,\,10,\,10,\,14,\,17,\,18,\,24,\,25,\,27,\,28,\,48 \] With \(\displaystyle n=16\), the median is the mean of the 8th and 9th terms: \[\text{Median} = \frac{10+14}{2} = 12 \] \[\text{Mode} = 10 \quad (\text{highest frequency, 3 times}) \]Answer: Median \(\displaystyle = 12\), mode \(\displaystyle = 10\).

  5. Exercise 15

    In Fig. 14.2\displaystyle 14.2, there is a histogram depicting daily wages of workers in a factory. Construct the frequency distribution table. NCERT_Question_Class9_Maths_Exemplar_Ch14_Ex14-3_Q15

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    Class intervalsFrequency
    $\displaystyle 150$-$\displaystyle 200$$\displaystyle 50$
    $\displaystyle 200$-$\displaystyle 250$$\displaystyle 30$
    $\displaystyle 250$-$\displaystyle 300$$\displaystyle 35$
    $\displaystyle 300$-$\displaystyle 350$$\displaystyle 20$
    $\displaystyle 350$-$\displaystyle 400$$\displaystyle 10$
    Total$\displaystyle 145$
    NCERT_Solution_Class9_Maths_Exemplar_Ch14_Ex14-3_Q15 Bars have equal width \(\displaystyle 50\), so each height read off the figure is that class's frequency. \[\begin{array}{|c|c|}\hline \text{Class interval} & \text{Frequency} \\ \hline 150\text{-}200 & 50 \\ 200\text{-}250 & 30 \\ 250\text{-}300 & 35 \\ 300\text{-}350 & 20 \\ 350\text{-}400 & 10 \\ \hline \text{Total} & 145 \\ \hline \end{array} \]Answer: frequency distribution as tabulated above; total \(\displaystyle 145\) workers.

  6. Exercise 16

    A company selected 4000\displaystyle 4000 households at random and surveyed them to find out a relationship between income level and the number of television sets in a home. The information so obtained is listed in the following table:
    2Number of Televisions/household
    0\displaystyle 01\displaystyle 12\displaystyle 2Above 2\displaystyle 2
    < 10000\displaystyle 1000020\displaystyle 2080\displaystyle 8010\displaystyle 100\displaystyle 0
    10000\displaystyle 10000-14999\displaystyle 1499910\displaystyle 10240\displaystyle 24060\displaystyle 600\displaystyle 0
    15000\displaystyle 15000-19999\displaystyle 199990\displaystyle 0380\displaystyle 380120\displaystyle 12030\displaystyle 30
    20000\displaystyle 20000-24999\displaystyle 249990\displaystyle 0520\displaystyle 520370\displaystyle 37080\displaystyle 80
    25000\displaystyle 25000 and above0\displaystyle 01100\displaystyle 1100760\displaystyle 760220\displaystyle 220
    Find the probability: (i) of a household earning Rs 10000\displaystyle 10000 - Rs 14999\displaystyle 14999 per year and having exactly one television. (ii) of a household earning Rs 25000\displaystyle 25000 and more per year and owning 2\displaystyle 2 televisions. (iii) of a household not having any television.

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    NCERT’s answer
    (i)
    0.$\displaystyle 06$ (ii) $\displaystyle 0.19$ (iii) $\displaystyle \frac{3}{400}$
    (i)
    \[P(10000\text{-}14999,\ 1\text{ TV}) = \frac{240}{4000} = 0.06 \]
    (ii)
    \[P(\ge 25000,\ 2\text{ TV}) = \frac{760}{4000} = 0.19 \]
    (iii)
    \[P(\text{no TV}) = \frac{20+10+0+0+0}{4000} = \frac{30}{4000} = \frac{3}{400} = 0.0075 \]
    Answer: (i) \(\displaystyle 0.06\) (ii) \(\displaystyle 0.19\) (iii) \(\displaystyle \tfrac{3}{400}\).

  7. Exercise 17

    Two dice are thrown simultaneously 500\displaystyle 500 times. Each time the sum of two numbers appearing on their tops is noted and recorded as given in the following table:
    SumFrequency
    2\displaystyle 214\displaystyle 14
    3\displaystyle 330\displaystyle 30
    4\displaystyle 442\displaystyle 42
    5\displaystyle 555\displaystyle 55
    6\displaystyle 672\displaystyle 72
    7\displaystyle 775\displaystyle 75
    8\displaystyle 870\displaystyle 70
    9\displaystyle 953\displaystyle 53
    10\displaystyle 1046\displaystyle 46
    11\displaystyle 1128\displaystyle 28
    12\displaystyle 1215\displaystyle 15
    If the dice are thrown once more, what is the probability of getting a sum (i) 3\displaystyle 3? (ii) more than 10\displaystyle 10? (iii) less than or equal to 5\displaystyle 5? (iv) between 8\displaystyle 8 and 12\displaystyle 12?

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    NCERT’s answer
    (i)
    0.$\displaystyle 06$ (ii) $\displaystyle 0.086$ (iii) $\displaystyle 0.282$ (iv) $\displaystyle 0.254$
    \[n = 500 \]
    (i)
    sum \(\displaystyle =3\):
    \[P(3) = \frac{30}{500} = \frac{3}{50} \]
    (ii)
    sum \(\displaystyle >10\) means \(\displaystyle 11\) or \(\displaystyle 12\):
    \[P(\text{sum}>10) = \frac{28+15}{500} = \frac{43}{500} \]
    (iii)
    sum \(\displaystyle \le 5\) means \(\displaystyle 2,3,4,5\):
    \[P(\text{sum}\le 5) = \frac{14+30+42+55}{500} = \frac{141}{500} \]
    (iv)
    sum between \(\displaystyle 8\) and \(\displaystyle 12\) means \(\displaystyle 9,10,11\):
    \[P(8<\text{sum}<12) = \frac{53+46+28}{500} = \frac{127}{500} \]
    Answer: \(\displaystyle \frac{3}{50}\), \(\displaystyle \frac{43}{500}\), \(\displaystyle \frac{141}{500}\), \(\displaystyle \frac{127}{500}\)

  8. Exercise 18

    Bulbs are packed in cartons each containing 40\displaystyle 40 bulbs. Seven hundred cartons were examined for defective bulbs and the results are given in the following table:
    Number of defective bulbs0\displaystyle 01\displaystyle 12\displaystyle 23\displaystyle 34\displaystyle 45\displaystyle 56\displaystyle 6more than 6\displaystyle 6
    Frequency400\displaystyle 400180\displaystyle 18048\displaystyle 4841\displaystyle 4118\displaystyle 188\displaystyle 83\displaystyle 32\displaystyle 2
    One carton was selected at random. What is the probability that it has (i) no defective bulb? (ii) defective bulbs from 2\displaystyle 2 to 6\displaystyle 6? (iii) defective bulbs less than 4\displaystyle 4?

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    NCERT’s answer
    (i)
    $\displaystyle \frac{4}{7}$ (ii) $\displaystyle \frac{59}{350}$ (iii) $\displaystyle \frac{669}{700}$
    \[n = 700 \]
    (i)
    \[P(\text{no defective}) = \frac{400}{700} = \frac{4}{7} \]
    (ii)
    \(\displaystyle 2\) to \(\displaystyle 6\) defective:
    \[P(2\text{ to }6) = \frac{48+41+18+8+3}{700} = \frac{118}{700} = \frac{59}{350} \]
    (iii)
    fewer than \(\displaystyle 4\) defective means \(\displaystyle 0,1,2,3\):
    \[P(<4) = \frac{400+180+48+41}{700} = \frac{669}{700} \]
    Answer: \(\displaystyle \frac{4}{7}\), \(\displaystyle \frac{59}{350}\), \(\displaystyle \frac{669}{700}\)

  9. Exercise 19

    Over the past 200\displaystyle 200 working days, the number of defective parts produced by a machine is given in the following table:
    Number of defective parts0\displaystyle 01\displaystyle 12\displaystyle 23\displaystyle 34\displaystyle 45\displaystyle 56\displaystyle 67\displaystyle 78\displaystyle 89\displaystyle 910\displaystyle 1011\displaystyle 1112\displaystyle 1213\displaystyle 13
    Days50\displaystyle 5032\displaystyle 3222\displaystyle 2218\displaystyle 1812\displaystyle 1212\displaystyle 1210\displaystyle 1010\displaystyle 1010\displaystyle 108\displaystyle 86\displaystyle 66\displaystyle 62\displaystyle 22\displaystyle 2
    Determine the probability that tomorrow's output will have (i) no defective part (ii) atleast one defective part (iii) not more than 5\displaystyle 5 defective parts (iv) more than 13\displaystyle 13 defective parts

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    NCERT’s answer
    (i)
    0.$\displaystyle 25$ (ii) $\displaystyle 0.75$ (iii) $\displaystyle 0.73$ (iv) $\displaystyle 0$
    \[n = 200 \]
    (i)
    \[P(0) = \frac{50}{200} = \frac{1}{4} \]
    (ii)
    at least one is the complement of none:
    \[P(\ge 1) = 1 - P(0) = 1 - \frac{1}{4} = \frac{3}{4} \]
    (iii)
    not more than \(\displaystyle 5\) means \(\displaystyle 0\) to \(\displaystyle 5\):
    \[P(\le 5) = \frac{50+32+22+18+12+12}{200} = \frac{146}{200} = \frac{73}{100} \]
    (iv)
    the table records no day past \(\displaystyle 13\) defective parts:
    \[P(>13) = \frac{0}{200} = 0 \]
    Answer: \(\displaystyle \frac{1}{4}\), \(\displaystyle \frac{3}{4}\), \(\displaystyle \frac{73}{100}\), \(\displaystyle 0\)

  10. Exercise 20

    A recent survey found that the ages of workers in a factory is distributed as follows:
    Age (in years)20\displaystyle 20-29\displaystyle 2930\displaystyle 30-39\displaystyle 3940\displaystyle 40-49\displaystyle 4950\displaystyle 50-59\displaystyle 5960\displaystyle 60 and above
    Number of workers38\displaystyle 3827\displaystyle 2786\displaystyle 8646\displaystyle 463\displaystyle 3
    If a person is selected at random, find the probability that the person is: (i) 40\displaystyle 40 years or more (ii) under 40\displaystyle 40 years (iii) having age from 30\displaystyle 30 to 39\displaystyle 39 years (iv) under 60\displaystyle 60 but over 39\displaystyle 39 years (E) Long Answer Questions Sample Question 1\displaystyle 1: Following is the frequency distribution of total marks obtained by the students of different sections of Class VIII.
    Marks100\displaystyle 100-150\displaystyle 150150\displaystyle 150-200\displaystyle 200200\displaystyle 200-300\displaystyle 300300\displaystyle 300-500\displaystyle 500500\displaystyle 500-800\displaystyle 800
    Number of students60\displaystyle 60100\displaystyle 100100\displaystyle 10080\displaystyle 80180\displaystyle 180
    Draw a histogram for the distribution above. Solution: In the given frequency distribution, the class intervals are not of equal width.

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    (i)
    0.$\displaystyle 675$ (ii) $\displaystyle 0.325$ (iii) $\displaystyle 0.135$ (iv) $\displaystyle 0.66$
    \[n = 38+27+86+46+3 = 200 \]
    (i)
    \(\displaystyle 40\) or more spans the last three groups:
    \[P(\ge 40) = \frac{86+46+3}{200} = \frac{135}{200} = \frac{27}{40} \]
    (ii)
    under \(\displaystyle 40\) is the complement:
    \[P(<40) = 1 - \frac{27}{40} = \frac{13}{40} \]
    (iii)
    \[P(30\text{-}39) = \frac{27}{200} \]
    (iv)
    under \(\displaystyle 60\) but over \(\displaystyle 39\) is the middle two groups:
    \[P(40\text{-}59) = \frac{86+46}{200} = \frac{132}{200} = \frac{33}{50} \]
    Answer: \(\displaystyle \frac{27}{40}\), \(\displaystyle \frac{13}{40}\), \(\displaystyle \frac{27}{200}\), \(\displaystyle \frac{33}{50}\)