NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 13: Statistics and Probability

Exercise 13.2

Step 1: Understand what "median" means. The median is the middle value of a data set when the numbers are arranged in order.

Step 2: Median for ungrouped data. When the data is written as individual numbers (like 3, 7, 9, 12), we can directly pick the middle number. → This gives the exact value of the median.

Step 3: Median for grouped data. When the same numbers are arranged in class intervals (like 0–5, 5–10, 10–15, etc.), we use a formula to calculate the median. → This formula gives only an approximate value, not the exact one.

Step 4: Compare both. The median from ungrouped data (exact) and the median from grouped data (approximate) are usually close to each other, but they are not always exactly the same.

Final Conclusion: The statement is false because the two medians are not guaranteed to be equal.

Step 1: Recall the formula for mean using the assumed mean method:

\[ \bar{x} = a + \dfrac{\sum f_i d_i}{\sum f_i} \]

• \(a\) = assumed mean
• \(f_i\) = frequency of the \(i^{th}\) class
• \(d_i = x_i - a\), where \(x_i\) is the midpoint of the \(i^{th}\) class

Step 2: The formula only needs us to pick an assumed mean (\(a\)) — this can be any number that makes the calculation easy.

Step 3: In practice, we usually choose \(a\) from one of the class midpoints because:

• It is close to the actual mean.
• It makes the subtraction \(x_i - a\) simpler.

Step 4: However, it is not compulsory to take \(a\) as a midpoint. We could assume any convenient number.

Final Step: Since the question says \(a\) must be a midpoint, this statement is false. The correct idea is that \(a\) can be any convenient value, though a midpoint is often chosen for simplicity.

Step 1: First, recall what each term means:

• Mean: It is the average of the data values.
• Median: It is the middle value of the data when arranged in order.
• Mode: It is the value that occurs most frequently.

Step 2: The statement says that mean, median, and mode of grouped data will always be different.

Step 3: But this is not true. For some types of data, especially when the data is symmetric (evenly spread on both sides), all three can be equal.

Example: In a perfectly symmetric distribution like the marks of students forming a bell-shaped curve, we get:

Mean = Median = Mode

Step 4: Therefore, they are not always different. They can sometimes be the same.

Final Conclusion: The statement is False.

Step 1: Recall the meaning of median class.

The median class is the class interval where the middle value (median) lies. It depends on the cumulative frequency.

Step 2: Recall the meaning of modal class.

The modal class is the class interval with the highest frequency. It depends only on the maximum frequency.

Step 3: Compare them.

Since both depend on frequencies, sometimes the same class may turn out to be both the median class and the modal class.

Step 4: Conclusion.

Therefore, it is not necessary that they are always different. They may be the same or different, depending on the data.

Hence, the statement is false.

Step 1: Each child can be either a boy (B) or a girl (G). So there are 2 choices for each child.

Step 2: For 3 children, the total number of possible outcomes is \(2^3 = 8\).

These outcomes are: BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG.

Step 3: Count how many outcomes have:

• 0 girls: only BBB → 1 outcome
• 1 girl: BBG, BGB, GBB → 3 outcomes
• 2 girls: BGG, GBG, GGB → 3 outcomes
• 3 girls: GGG → 1 outcome

Step 4: Probability = (favourable outcomes) ÷ (total outcomes).

• P(0 girls) = \(\tfrac{1}{8}\)
• P(1 girl) = \(\tfrac{3}{8}\)
• P(2 girls) = \(\tfrac{3}{8}\)
• P(3 girls) = \(\tfrac{1}{8}\)

Step 5: Clearly, these probabilities are not equal. They are different fractions, not all \(\tfrac{1}{4}\).

Final Answer: The statement is false because the number of girls follows the binomial distribution and the probabilities are unequal.

Step 1: To decide whether outcomes are equally likely, we must check if each outcome has the same chance of happening.

Step 2: In a spinner, the chance of landing on a region depends on the area of that region. Bigger area → higher chance. Smaller area → lower chance.

Step 3: Look at the figure. The regions marked 1, 2, and 3 do not have the same size. Their areas are different.

Step 4: Since the regions are unequal, the arrow does not have the same probability of stopping at each region.

Step 5: Therefore, the outcomes 1, 2, and 3 are not equally likely.

Step 1: Understand Peehu's case

Peehu throws only one die. A die has six faces: 1, 2, 3, 4, 5, 6.

She squares the number that comes. To get 36, she must have:

\(6^2 = 36\).

So Peehu gets 36 only when she rolls a 6.

Probability = \(\dfrac{1}{6}\), because only 1 favorable outcome (rolling a 6) out of 6 total outcomes.

Step 2: Understand Apoorv's case

Apoorv throws two dice together. Each die has 6 faces, so total outcomes = \(6 \times 6 = 36\).

He multiplies the two numbers. We want the product to be 36.

Let’s check possible pairs:

• (6, 6): product = 36 ✔️
• (3, 12): not possible because dice go only up to 6 ❌
• (4, 9): not possible because dice go only up to 6 ❌

So only one favorable outcome exists: (6, 6).

Probability = \(\dfrac{1}{36}\).

Step 3: Compare probabilities

Peehu: \(\dfrac{1}{6}\)

Apoorv: \(\dfrac{1}{36}\)

Since \(\dfrac{1}{6} > \dfrac{1}{36}\), Peehu has a higher chance.

Final Answer: Peehu has the better chance of getting 36.

Step 1: Recall what probability means.

Probability of an event = \( \dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \).

Step 2: Write the sample space for tossing a coin.

The sample space is the set of all possible outcomes when we toss the coin.
\(S = \{ H, T \}\), where H = Head and T = Tail.

Step 3: Count the total number of outcomes.

There are 2 outcomes in the sample space (H or T).

Step 4: Find the probability of getting a Head.

Number of favourable outcomes for Head = 1 (only H).
Total outcomes = 2.
\(P(H) = \dfrac{1}{2}\).

Step 5: Find the probability of getting a Tail.

Number of favourable outcomes for Tail = 1 (only T).
Total outcomes = 2.
\(P(T) = \dfrac{1}{2}\).

Step 6: Conclusion.

Both outcomes are equally likely and each has probability \(\dfrac{1}{2}\).
So, the statement is True.

Step 1: When we throw a die, the sample space (all possible outcomes) is:

\(S = \{1, 2, 3, 4, 5, 6\}\)

So there are a total of 6 equally likely outcomes.

Step 2: Probability of getting a 1:

Only one outcome (1) is favorable.

So, \(P(1) = \dfrac{1}{6}\).

Step 3: Probability of getting ‘not 1’:

Here the favorable outcomes are \(\{2, 3, 4, 5, 6\}\).

That means there are 5 favorable outcomes.

So, \(P(\text{not 1}) = \dfrac{5}{6}\).

Step 4: Compare the two probabilities:

\(P(1) = \dfrac{1}{6}\), but \(P(\text{not 1}) = \dfrac{5}{6}\).

These are not equal.

Final Step: The student’s claim that both probabilities are \(\dfrac{1}{2}\) is false. The correct probabilities are \(\dfrac{1}{6}\) and \(\dfrac{5}{6}\).

Step 1: When 3 coins are tossed together, each coin can show either Head (H) or Tail (T).

Step 2: Total number of possible outcomes = \(2^3 = 8\).

Step 3: Write the sample space (all outcomes): {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.

Step 4: "No heads" means all coins must be Tails. That happens only in one case: TTT.

Step 5: Number of favourable outcomes = 1 (only TTT).

Step 6: Probability formula is:

\( P(E) = \dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \)

Step 7: Substituting values:

\( P(\text{no heads}) = \dfrac{1}{8} \).

Step 8: The wrong reasoning was treating the 4 cases (no head, 1 head, 2 heads, 3 heads) as equally likely. But they are not. Each event has a different number of outcomes.

Final Answer: Probability of no heads is \(\dfrac{1}{8}\), not \(\dfrac{1}{4}\).

Step 1: Recall the basic rule of probability. For a fair coin, there are two equally likely outcomes: Head (H) or Tail (T).

Step 2: The probability of getting a head in one toss is:

\( P(H) = \dfrac{1}{2} = 0.5 \)

Step 3: Every coin toss is independent. This means that what happened in the past (for example, 6 heads in a row) does not change the probability of the next toss.

Step 4: Even if you got 6 heads in a row, the probability of getting a head in the 7th toss is still:

\( P(H) = \dfrac{1}{2} \)

Step 5: If the probability of head were 1, that would mean it is certain to get a head every time, which is not true for a fair coin.

Conclusion: The probability of getting a head is always \(0.5\), not 1, regardless of how many heads appeared before.

Step 1: When we toss a coin, there are only two possible outcomes – Head (H) or Tail (T).

Step 2: Each toss of a coin is called an independent event. This means that the result of one toss does not change or affect the result of the next toss.

Step 3: Even if Sushma got Tail 3 times in a row, the coin has no memory of what happened before.

Step 4: On the 4th toss, the probability of getting Head is still \(\tfrac{1}{2}\) and the probability of getting Tail is also \(\tfrac{1}{2}\).

Step 5: So, we cannot say that Tail will surely come next. Both Head and Tail are equally likely.

Therefore, the statement is False.

Step 1: A coin has two possible outcomes when tossed — Head (H) or Tail (T).

Step 2: For a fair coin, the probability of getting a Head is \(\dfrac{1}{2}\), and the probability of getting a Tail is also \(\dfrac{1}{2}\).

Step 3: Each coin toss is an independent event. This means the result of one toss does not affect the result of another toss.

Step 4: Even if you got Head three times in a row, the probability of getting Tail in the next (4th) toss is still \(\dfrac{1}{2}\).

Step 5: Therefore, the chance of getting Tail does not become higher in the 4th toss.

Final Reason: The probability remains the same (\(\dfrac{1}{2}\)) for both Head and Tail, no matter what happened in the earlier tosses.

Step 1: Total slips in the bag = 100 (from 1 to 100).

Step 2: We need to see how many slips are odd and how many are even.

• Odd numbers between 1 and 100 = 50 (for example: 1, 3, 5, ..., 99).
• Even numbers between 1 and 100 = 50 (for example: 2, 4, 6, ..., 100).

Step 3: Probability formula (in SI):

\( P(E) = \dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \)

Step 4: Probability of getting an odd slip = \(\dfrac{50}{100} = \dfrac{1}{2}\).

Step 5: Probability of getting an even slip = \(\dfrac{50}{100} = \dfrac{1}{2}\).

Final Step: Since both probabilities are equal and each is \(\dfrac{1}{2}\), the statement is justified.