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NCERT Exemplar · Class 10 Mathematics Statistics and Probability

96 questions · 96 still being checked

EXERCISE 13.2 1–10 (part 4 of 11)

  1. Exercise 1

    The median of an ungrouped data and the median calculated when the same data is grouped are always the same. Do you think that this is a correct statement? Give reason.

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    NCERT’s answer
    Not always, because for calculating median of a grouped data, the formula used is based on the assumption that the observations in the classes are uniformly distributed (or equally spaced).
    False. Grouping assumes values are spread uniformly within each class interval, so the formula below only estimates the median; it need not match the exact value found from the raw data. \[\text{Median} = l + \left( \dfrac{\frac{n}{2}-cf}{f} \right) h \]

  2. Exercise 2

    In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula xˉ=a+∑fidi∑fi\bar{x}=a+\frac{\sum f_i d_i}{\sum f_i} where a\displaystyle a is the assumed mean. a\displaystyle a must be one of the mid-points of the classes. Is the last statement correct? Justify your answer.

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    NCERT’s answer
    Not necessary, the mean of the data does not depend on the choice of \(\displaystyle a\) (assumed mean).
    False. \(\displaystyle a\) is merely an assumed value used to shift the origin; the algebra below holds for every real \(\displaystyle a\), so it need not be a class midpoint, though a central value keeps the deviations small. \[d_i = x_i - a \implies \sum f_i d_i = \sum f_i x_i - a\sum f_i \] \[a + \frac{\sum f_i d_i}{\sum f_i} = \frac{\sum f_i x_i}{\sum f_i} = \bar{x} \]

  3. Exercise 3

    Is it true to say that the mean, mode and median of grouped data will always be different? Justify your answer.

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    NCERT’s answer
    No, it is not always the case. The values of these three measures can be the same. It depends on the type of data.
    False. Mean, mode and median need not differ: for a symmetric distribution all three coincide. \[\text{Mode} = 3\,\text{Median} - 2\,\text{Mean} \] This empirical relation permits \(\displaystyle \text{Mode}=\text{Median}=\text{Mean}\) as one valid case.

  4. Exercise 4

    Will the median class and modal class of grouped data always be different? Justify your answer.

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    NCERT’s answer
    Not always. It depends on the data.
    False. The median class is fixed by cumulative frequency, the modal class by highest frequency — two independent criteria that can single out the same class. \[cf_{\text{median class}} \ge \frac{n}{2} \quad \text{(first class reaching this)} \] \[f_{\text{modal class}} = \max_i f_i \] Nothing stops one class from satisfying both conditions at once.

  5. Exercise 5

    In a family having three children, there may be no girl, one girl, two girls or three girls. So, the probability of each is 14\displaystyle \frac{1}{4}. Is this correct? Justify your answer.

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    NCERT’s answer
    No, the outcomes are not equally likely. For example, outcome ‘one girl’ means gbb, bgb, bbg ‘three girls’ means ggg and so on.
    False. The four cases aren't equally likely — treating birth order as equally likely outcomes gives a sample space of size \(\displaystyle 2^3\), and only one arrangement gives no girls or three girls. \[n(S) = 2^3 = 8 \] \[P(0\text{ girls}) = \frac18, \ P(1\text{ girl}) = \frac38, \ P(2\text{ girls}) = \frac38, \ P(3\text{ girls}) = \frac18 \]

  6. Exercise 6

    A game consists of spinning an arrow which comes to rest pointing at one of the regions (1\displaystyle 1, 2\displaystyle 2 or 3\displaystyle 3) (Fig. 13.1\displaystyle 13.1). Are the outcomes 1\displaystyle 1, 2\displaystyle 2 and 3\displaystyle 3 equally likely to occur? Give reasons. NCERT_Question_Class10_Maths_Exemplar_Ch13_Ex13-2_Q6

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    NCERT’s answer
    No, the outcomes are not equally likely. The outcome '$\displaystyle 3$' is more likely than the others.
    False. Reading the angles at the centre from the figure, regions $\displaystyle 1$ and $\displaystyle 2$ each span a right angle while region $\displaystyle 3$ spans a straight angle, so the three sectors are not equal in size. \[\theta_1=\theta_2=90^\circ,\quad \theta_3=180^\circ,\quad \theta_1+\theta_2+\theta_3=360^\circ \] \[P(1)=\frac{90^\circ}{360^\circ}=\frac14,\quad P(2)=\frac14,\quad P(3)=\frac{180^\circ}{360^\circ}=\frac12 \]

  7. Exercise 7

    Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has the better chance of getting the number 36\displaystyle 36? Why?

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    NCERT’s answer
    Peehu; probability of Apoorv’s getting \(\displaystyle 36=\frac{1}{36}\) while probability of Peehu’s getting \(\displaystyle 36=\frac{1}{6}=\frac{6}{36}\).
    Apoorv's two dice give \(\displaystyle 6\times6=36\) equally likely outcomes, and only \(\displaystyle (6,6)\) has product \(\displaystyle 36\). \[P(\text{Apoorv}) = \frac{1}{36} \] Peehu's one die gives \(\displaystyle 6\) equally likely outcomes, and only the face \(\displaystyle 6\) squares to \(\displaystyle 36\). \[P(\text{Peehu}) = \frac{1}{6} \] Since \(\displaystyle \frac16 > \frac{1}{36}\), Peehu is more likely to get \(\displaystyle 36\). Answer: Peehu has the better chance.

  8. Exercise 8

    When we toss a coin, there are two possible outcomes - Head or Tail. Therefore, the probability of each outcome is 12\displaystyle \frac{1}{2}. Justify your answer.

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    NCERT’s answer
    Yes, the probability of each outcome is \(\displaystyle \frac{1}{2}\), since the two outcomes are equally likely.
    For a fair coin, Head and Tail are equally likely. \[P(\text{Head}) = \frac{\text{favourable outcomes}}{\text{total outcomes}} = \frac{1}{2} \] \[P(\text{Tail}) = \frac{1}{2} \] Answer: Correct: the two outcomes are equally likely, so each has probability \(\displaystyle \tfrac12\).

  9. Exercise 9

    A student says that if you throw a die, it will show up 1\displaystyle 1 or not 1. Therefore, the probability of getting 1\displaystyle 1 and the probability of getting 'not 1\displaystyle 1' each is equal to 12\displaystyle \frac{1}{2}. Is this correct? Give reasons.

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    NCERT’s answer
    No, outcomes '$\displaystyle 1$' and 'not $\displaystyle 1$' are not equally likely, \(\displaystyle \mathrm{P}(1)=\frac{1}{6}, \mathrm{P}(\) not $\displaystyle 1$\(\displaystyle )=\frac{5}{6}\),
    \(\displaystyle \text{Not }1\) covers five faces of the die, while \(\displaystyle 1\) covers only one face — the two events do not have equal counts of favourable outcomes. \[P(1) = \frac{1}{6} \] \[P(\text{not }1) = \frac{5}{6} \] \[P(1)+P(\text{not }1) = \tfrac16+\tfrac56 = 1 \] Two possible outcomes does not by itself mean equal probability; equal likelihood is what is required, and it fails here. Answer: Incorrect — \(\displaystyle P(1)=\tfrac16\) and \(\displaystyle P(\text{not }1)=\tfrac56\), not \(\displaystyle \tfrac12\) each.

  10. Exercise 10

    I toss three coins together. The possible outcomes are no heads, 1\displaystyle 1 head, 2\displaystyle 2 heads and 3\displaystyle 3 heads. So, I say that probability of no heads is 14\displaystyle \frac{1}{4}. What is wrong with this conclusion?

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    NCERT’s answer
    No, the outcomes are not equally likely. Outcome 'no head' means 'TTT'; outcome 'one head' means THT, HTT, TTH and so on. \(\displaystyle \mathrm{P}(\mathrm{TTT})=\frac{1}{8}, \mathrm{P}(\) one head \(\displaystyle )=\frac{3}{8}\) and so on.
    The equally likely elementary outcomes of tossing $\displaystyle 3$ coins are \[\{HHH,\ HHT,\ HTH,\ THH,\ HTT,\ THT,\ TTH,\ TTT\} \] which number $\displaystyle 8$, occurring in the ratio $\displaystyle 1$ : $\displaystyle 3$ : $\displaystyle 3$ : $\displaystyle 1$ across no-heads, $\displaystyle 1$-head, $\displaystyle 2$-heads, $\displaystyle 3$-heads. \[P(\text{no heads}) = P(TTT) = \frac{1}{8} \] Treating the four head-count groups as four equally likely outcomes is wrong, since they are not equinumerous. Answer: Wrong — the sample space has $\displaystyle 8$ equally likely outcomes, so \(\displaystyle P(\text{no heads})=\tfrac18\), not \(\displaystyle \tfrac14\).