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NCERT Exemplar · Class 9 Mathematics Statistics and Probability

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EXERCISE 14.2 1–10 (part 4 of 7)

  1. Exercise 1

    The frequency distribution :
    Marks0\displaystyle 0-20\displaystyle 2020\displaystyle 20-40\displaystyle 4040\displaystyle 40-60\displaystyle 6060\displaystyle 60-100\displaystyle 100
    Number of Students10\displaystyle 1015\displaystyle 1520\displaystyle 2025\displaystyle 25
    has been represented graphically as follows : Do you think this representation is correct? Why?

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    NCERT’s answer
    Not correct. The classes are of varying widths, not of uniform widths.
    No. The class widths are unequal, so each rectangle's height must be fixed by area (frequency), not read off directly — using the smallest width, $\displaystyle 20$, as the standard.\[h_{60-100} = \frac{25}{40}\times 20 = 12.5 \]NCERT_Solution_Class9_Maths_Exemplar_Ch14_Ex14-2_Q1Answer: No; the \(\displaystyle 60\)–\(\displaystyle 100\) bar's height must be \(\displaystyle 12.5\), not \(\displaystyle 25\).

  2. Exercise 2

    In a diagnostic test in mathematics given to students, the following marks (out of 100\displaystyle 100) are recorded : 46\displaystyle 46, 52\displaystyle 52, 48\displaystyle 48, 11\displaystyle 11, 41\displaystyle 41, 62\displaystyle 62, 54\displaystyle 54, 53\displaystyle 53, 96\displaystyle 96, 40\displaystyle 40, 98\displaystyle 98, 44\displaystyle 44 Which 'average' will be a good representative of the above data and why?

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    NCERT’s answer
    Median will be a good representative of the data, because (i) each value occcurs once, (ii) The data is influenced by extreme values.
    Sorted: $\displaystyle 11$, $\displaystyle 40$, $\displaystyle 41$, $\displaystyle 44$, $\displaystyle 46$, $\displaystyle 48$, $\displaystyle 52$, $\displaystyle 53$, $\displaystyle 54$, $\displaystyle 62$, $\displaystyle 96$, 98.\[\text{Mean}=\frac{645}{12}=53.75,\qquad \text{Median}=\frac{48+52}{2}=50 \]$\displaystyle 11$, $\displaystyle 96$, $\displaystyle 98$ are outliers that pull the mean up; the median ignores them.Answer: the median ($\displaystyle 50$) — not distorted by the outliers $\displaystyle 11$, $\displaystyle 96$, 98.

  3. Exercise 3

    A child says that the median of 3,14,18,20,5\displaystyle 3,14,18,20,5 is 18\displaystyle 18 . What doesn't the child understand about finding the median?

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    NCERT’s answer
    Data has to be arranged in ascending (or descending ) order before finding the median.
    The child took the middle term of the list as printed — $\displaystyle 3$, $\displaystyle 14$, $\displaystyle 18$, $\displaystyle 20$, $\displaystyle 5$ — without first arranging the values in order.\[3,5,14,18,20 \quad\text{(ascending order)} \] \[\text{Median}=14 \]Answer: the median is $\displaystyle 14$; finding it requires sorting the data first.

  4. Exercise 4

    A football player scored the following number of goals in the 10\displaystyle 10 matches : 1\displaystyle 1, 3\displaystyle 3, 2\displaystyle 2, 5\displaystyle 5, 8\displaystyle 8, 6\displaystyle 6, 1\displaystyle 1, 4\displaystyle 4, 7\displaystyle 7, 9\displaystyle 9 Since the number of matches is 10\displaystyle 10 (an even number), therefore, the median =5th  observation +6th  observation 2\displaystyle =\frac{5^{\text {th }} \text { observation }+6^{\text {th }} \text { observation }}{2} =8+62=7\displaystyle =\frac{8+6}{2}=7 Is it the correct answer and why?

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    NCERT’s answer
    No, the data have first to be arranged in ascending (or descending) order before finding the median.
    No. The 5th and 6th terms must come from the data sorted in ascending order, not the order the goals were scored.\[1,1,2,3,4,5,6,7,8,9 \] \[\text{Median}=\frac{5^{\text{th}}+6^{\text{th}}}{2}=\frac{4+5}{2}=4.5 \]Answer: No; the median is \(\displaystyle 4.5\), not \(\displaystyle 7\).

  5. Exercise 5

    Is it correct to say that in a histogram, the area of each rectangle is proportional to the class size of the corresponding class interval? If not, correct the statement.

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    NCERT’s answer
    It is not correct. In a histogram, the area of each rectangle is propotional to the frequency of its class.
    No. Area, not height, is proportional to a class's frequency; height alone equals frequency only for equal-width classes.Answer: No — area is proportional to frequency, not class size.

  6. Exercise 6

    The class marks of a continuous distribution are : 1.04\displaystyle 1.04, 1.14\displaystyle 1.14, 1.24\displaystyle 1.24, 1.34\displaystyle 1.34, 1.44\displaystyle 1.44, 1.54\displaystyle 1.54 and 1.64\displaystyle 1.64 Is it correct to say that the last interval will be 1.55\displaystyle 1.55-1.73\displaystyle 1.73? Justify your answer.

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    NCERT’s answer
    It is not correct. Reason is that differnce between two consecutive marks should be equal to the class size.
    No. Consecutive class marks differ by a constant amount, fixing one common class width for the whole table.\[1.14-1.04=1.24-1.14=\cdots=0.10 \] \[1.64\pm\frac{0.10}{2}=1.59 \text{ to } 1.69 \]The stated interval $\displaystyle 1.55$–$\displaystyle 1.73$ is $\displaystyle 0.18$ wide — wider than every other class.Answer: No — the last interval is \(\displaystyle 1.59\)–\(\displaystyle 1.69\), not \(\displaystyle 1.55\)–\(\displaystyle 1.73\).

  7. Exercise 7

    30\displaystyle 30 children were asked about the number of hours they watched TV programmes last week. The results are recorded as under :
    Number of hours0\displaystyle 0-5\displaystyle 55\displaystyle 5-10\displaystyle 1010\displaystyle 10-15\displaystyle 1515\displaystyle 15-20\displaystyle 20
    Frequency8\displaystyle 816\displaystyle 164\displaystyle 42\displaystyle 2
    Can we say that the number of children who watched TV for 10\displaystyle 10 or more hours a week is 22\displaystyle 22? Justify your answer.

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    NCERT’s answer
    No. Infact the number of children who watch TV for $\displaystyle 10$ or more hours a week is $\displaystyle 4$ + $\displaystyle 2$, i.e., 6.
    No. Ten or more hours means only the last two classes; $\displaystyle 22$ wrongly includes the $\displaystyle 5$-$\displaystyle 10$ hours class too.\[4+2=6 \] \[16+4+2=22 \]Answer: No — $\displaystyle 6$ children watched TV for $\displaystyle 10$ or more hours a week, not 22.

  8. Exercise 8

    Can the experimental probability of an event be a negative number? If not, why?

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    NCERT’s answer
    No, since the number of trials in which the event can happen cannot be negative, and the total number of trials is always positive.
    No. Experimental probability is favourable trials over total trials; both counts are non-negative, so their ratio can never be negative.\[P(E)=\frac{\text{trials with }E}{\text{total trials}}\ge 0 \]Answer: No — experimental probability is never negative.

  9. Exercise 9

    Can the experimental probability of an event be greater than 1\displaystyle 1? Justify your anwer.

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    NCERT’s answer
    No, since the number of trials in which the event can happen cannot be greater than the total number of trials.
    \[P(E) = \frac{\text{number of trials with outcome } E}{\text{total number of trials}} \] The favourable count can never exceed the total count, so \[0 \le \text{number of trials with outcome } E \le \text{total number of trials} \implies P(E) \le 1. \]Answer: No — the experimental probability of an event can never exceed 1.

  10. Exercise 10

    As the number of tosses of a coin increases, the ratio of the number of heads to the total number of tosses will be 12\displaystyle \frac{1}{2}. Is it correct? If not, write the correct one.

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    NCERT’s answer
    No. As the number of tosses of a coin increases, the ratio of the number of heads to the total number of tosses will be nearer to $\displaystyle \frac{1}{2}$, not exactly $\displaystyle \frac{1}{2}$. \begin{table} \captionsetup{labelformat=empty}
    \[P(\text{heads}) = \frac{\text{number of heads}}{\text{number of tosses}} \] \[P(\text{heads}) \to \frac{1}{2} \quad \text{as number of tosses} \to \infty \] The ratio only tends toward \(\displaystyle \tfrac12\) as tosses grow; it need not equal \(\displaystyle \tfrac12\) exactly at any finite number of tosses.Answer: No — the ratio approaches \(\displaystyle \tfrac12\) as tosses increase; it need not be exactly \(\displaystyle \tfrac12\).