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

Exercise 13.1

Step 1: In grouped data, we do not take every value separately. Instead, we take the mid-point of each class. These mid-points are also called class marks.

Step 2: In the assumed-mean method, we choose any one mid-point (class mark) as the assumed mean and call it \(a\).

Step 3: We calculate the deviation for each class mark using the formula: \(d_i = x_i - a\), where \(x_i\) is the mid-point of the \(i\)-th class.

Step 4: These deviations \(d_i\) are then multiplied with the corresponding frequencies \(f_i\), and we apply the formula: \(\bar{x} = a + \dfrac{\sum f_i d_i}{\sum f_i}\).

Conclusion: Since \(d_i\) is calculated from the class mark \(x_i\), the correct answer is mid-points (class marks) of the classes.

Step 1: In grouped data, values are arranged in class intervals (like 0–10, 10–20, etc.).

Step 2: Inside each interval, we do not know the exact data values of all items. For example, in the class 0–10, we only know how many values (frequency) are there, but not the exact numbers.

Step 3: To calculate the mean, we need to assume one representative value for each class interval.

Step 4: The most fair representative value of a class is its class mark, which is the midpoint of the interval. For example, for 0–10, the class mark is (0+10)/2 = 5.

Step 5: So, we assume that all the data values (frequency) in that class are concentrated at this midpoint.

Step 6: Therefore, when we compute mean of grouped data, we take frequencies as if they are centred at the class marks.

Answer: Option (B) – centred at the class marks of the classes.

Step 1: Recall the formula for the mean of grouped data.

The mean is given by:

\(\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}\)

Step 2: Rearrange this formula.

Multiply both sides by \(\sum f_i\):

\(\bar{x} \cdot \sum f_i = \sum f_i x_i\)

Step 3: Now look at the given expression:

\(\sum f_i (x_i - \bar{x})\)

Step 4: Expand this expression.

\(\sum f_i (x_i - \bar{x}) = \sum f_i x_i - \sum f_i \bar{x}\)

Step 5: Notice that \(\bar{x}\) is the same for every term, so:

\(\sum f_i \bar{x} = \bar{x} \cdot \sum f_i\)

Step 6: From Step 2, we know that:

\(\sum f_i x_i = \bar{x} \cdot \sum f_i\)

Step 7: Substitute this back.

\(\sum f_i (x_i - \bar{x}) = (\bar{x} \cdot \sum f_i) - (\bar{x} \cdot \sum f_i)\)

Step 8: Simplify.

This becomes: \(0\).

Final Answer: \(\sum f_i (x_i - \bar{x}) = 0\). So the correct option is (A).

Step 1: The formula given is for the step-deviation method of finding the mean.

Step 2: In this method, \(u_i\) is a new variable that makes calculations easier.

Step 3: We define \(u_i\) as the ratio of the deviation of \(x_i\) from \(a\), divided by the class size \(h\).

Step 4: Mathematically, this means:

\[ u_i = \frac{x_i - a}{h} \]

Step 5: Now compare this with the given options. The correct match is option (C): \(\dfrac{x_i - a}{h}\).

Final Answer: Option (C)

Let us carefully understand step by step:

1. Step 1: In statistics, to find the median graphically, we draw two types of cumulative frequency curves:
   • Less-than type cumulative frequency curve.
   • More-than type cumulative frequency curve.
2. Step 2: These two curves usually cut (intersect) each other at one point.
3. Step 3: The x-coordinate (abscissa) of this intersection point represents the median of the data.
4. Step 4: The median is the middle value of the data when all values are arranged in order. It divides the data into two equal parts.

Therefore: The abscissa of the intersection point gives the median. Correct answer = (B).

Step 1: Find the total number of observations (N).

Add all frequencies: \(10 + 15 + 12 + 20 + 9 = 66\).

So, \(N = 66\).

Step 2: Find the median class.

We need \(N/2 = 66/2 = 33\).

Now make the cumulative frequencies:

• For 0–5: 10
• For 5–10: 10 + 15 = 25
• For 10–15: 25 + 12 = 37
• For 15–20: 37 + 20 = 57
• For 20–25: 57 + 9 = 66

We check where the 33rd observation lies. It lies in the class 10–15 because up to 25 it is not reached, but at 37 it crosses 33.

So, the median class is 10–15 and its lower limit = 10.

Step 3: Find the modal class.

The class with the highest frequency is the modal class.

Here the highest frequency is 20, in the class 15–20.

So, the modal class is 15–20 and its lower limit = 15.

Step 4: Add the two lower limits.

Median class lower limit = 10

Modal class lower limit = 15

Sum = 10 + 15 = 25

Final Answer: Option B (25)

Step 1: First, add up all the frequencies to get the total number of observations (N).

\(N = 13 + 10 + 15 + 8 + 11 = 57\)

Step 2: Find \(N/2\).

\(N/2 = 57/2 = 28.5\)

Step 3: Make the cumulative frequencies (keep adding one by one).

• For 0–5: \(13\)
• For 6–11: \(13 + 10 = 23\)
• For 12–17: \(23 + 15 = 38\)
• For 18–23: \(38 + 8 = 46\)
• For 24–29: \(46 + 11 = 57\)

Step 4: Now check where \(28.5\) lies in the cumulative frequencies.

It is greater than 23 but less than or equal to 38. So, the median class is 12–17.

Step 5: The upper limit of this class (12–17) is 17.

Final Answer: The upper limit of the median class is 17. So, option (A) is correct.

Step 1: Understand the table

The given data is in the form of cumulative frequencies (students below 10, below 20, etc.). To find the modal class, we first need to convert it into class frequencies.

Step 2: Find class frequencies

We subtract each cumulative value from the previous one:

• Below 10 → 3 (no subtraction needed)
• 10–20 → 12 – 3 = 9
• 20–30 → 27 – 12 = 15
• 30–40 → 57 – 27 = 30
• 40–50 → 75 – 57 = 18
• 50–60 → 80 – 75 = 5

So the class frequencies are: 3, 9, 15, 30, 18, 5.

Step 3: Identify the highest frequency

The frequencies are 3, 9, 15, 30, 18, and 5. The largest number is 30.

Step 4: Find the class interval with this highest frequency

The frequency 30 belongs to the class 30–40.

Step 5: Write the conclusion

The class with the highest frequency is called the modal class. Therefore, the modal class is 30–40.

Final Answer: Option C (30–40)

Step 1: Find total frequency (N).

Add all frequencies: \(4 + 5 + 13 + 20 + 14 + 7 + 4 = 67\). So, \(N = 67\).

Step 2: Find median class.

Median is at position \(N/2 = 67/2 = 33.5\).

Now write cumulative frequencies: 65–85 → 4, 85–105 → 9, 105–125 → 22, 125–145 → 42, 145–165 → 56, 165–185 → 63, 185–205 → 67.

Since 33.5 lies between 22 and 42, the median class = 125–145.

Upper limit of median class = 145.

Step 3: Find modal class.

The class with the highest frequency is 125–145 (frequency = 20).

So, modal class = 125–145.

Lower limit of modal class = 125.

Step 4: Find the required difference.

Difference = (Upper limit of median class) − (Lower limit of modal class)

= \(145 − 125 = 20\).

Final Answer: Option C (20)

Step 1: Look carefully at the classes (time intervals). We want the number of athletes who finished the race in less than 14.6 seconds.

Step 2: The class intervals below 14.6 are:

• 13.8–14   → Frequency = 2
• 14–14.2   → Frequency = 4
• 14.2–14.4   → Frequency = 5
• 14.4–14.6   → Frequency = 71

Step 3: Add all these frequencies:

\(2 + 4 + 5 + 71 = 82\)

Step 4: Therefore, the total number of athletes who finished in less than 14.6 seconds is 82.

So, the correct answer is Option C (82).

Step 1: The given table is a cumulative frequency table. It tells us how many students scored marks greater than or equal to a certain value.

Step 2: To find the number of students in the interval 30–40, we subtract:

• Students who scored ≥30 (which is 51)
• Students who scored ≥40 (which is 48)

Step 3: Subtract the two values:

\( 51 - 48 = 3 \)

Step 4: Therefore, the frequency of the class 30–40 is 3.

Final Answer: Option (A)

Step 1: Probability means the chance of an event happening. It is always a number between 0 and 1.

Step 2: If the probability is 1, it means the event is certain to happen.

Step 3: If the probability is 0, it means the event is impossible — it can never happen.

Step 4: The question says “an event cannot occur.” That is exactly the meaning of an impossible event.

Final Answer: The probability of an impossible event is 0. Hence, the correct option is (D).

Step 1: Recall the rule of probability. The probability of any event always lies between 0 and 1 (both included). In symbols: \(0 \leq P(E) \leq 1\).

Step 2: Check each option one by one:

• (A) \(\tfrac{1}{3}\): This is equal to 0.333… which lies between 0 and 1. ✔ Valid probability.
• (B) 0.1: This is 0.1, which lies between 0 and 1. ✔ Valid probability.
• (C) 3%: Percentage means “out of 100.” So, 3% = \(\tfrac{3}{100} = 0.03\). This also lies between 0 and 1. ✔ Valid probability.
• (D) \(\tfrac{17}{16}\): Divide 17 by 16 = 1.0625. This is greater than 1. ✘ Not possible for probability.

Step 3: Since probability cannot be more than 1, option (D) \(\tfrac{17}{16}\) is not possible.

Final Answer: Option (D)

Step 1: Probability of an event is always between 0 and 1.

Step 2: If the probability is close to 0, the event is very unlikely (almost impossible).

Step 3: Compare the given numbers:

• 0.0001 = one in ten thousand chance (extremely small).
• 0.001 = one in a thousand chance (still small, but bigger than 0.0001).
• 0.01 = one in a hundred chance (more likely than above).
• 0.1 = one in ten chance (not “very unlikely,” actually fairly possible).

Step 4: Out of the options, 0.0001 is the smallest and therefore shows the “very unlikely” situation.

Final Answer: Option (A) 0.0001

Step 1: The probability of an event (say event A) is given as \(p\).

Step 2: The complementary event means the event does not happen.

Step 3: In probability, the sum of an event and its complement is always 1. Mathematically: \( P(A) + P(A') = 1 \).

Step 4: Here, \(P(A) = p\). So, \( p + P(A') = 1 \).

Step 5: Rearranging, \( P(A') = 1 - p \).

Final Answer: The probability of the complementary event is \(1-p\). Hence, option (C).

Step 1: Probability is a number that tells us how likely something is to happen.

Step 2: The lowest probability value is 0 (meaning the event is impossible).

Step 3: The highest probability value is 1 (meaning the event is certain).

Step 4: When we express probability as a percentage, we multiply it by 100.

Example:
Probability = 0 → Percentage = 0%
Probability = 1 → Percentage = 100%

Step 5: So probability (in percentage) can only range from 0% to 100%.

Step 6: It can never be less than 0% because probability cannot be negative.

Answer: (B) less than 0.

Step 1: Probability is a number that tells us how likely an event is to happen.

Step 2: The smallest value of probability is 0. This means the event is impossible (cannot happen at all).

Step 3: The largest value of probability is 1. This means the event is certain (will definitely happen).

Step 4: For all real situations, the probability of any event lies between 0 and 1. So, it cannot be negative (less than 0), and it cannot be greater than 1.

Step 5: Therefore, the correct condition is: \(0 \leq P(A) \leq 1\).

Final Answer: Option (C).

Step 1: A standard deck has 52 cards in total.

Step 2: The cards are divided into two colors:

• 26 red cards (Hearts ♥ and Diamonds ♦)
• 26 black cards (Clubs ♣ and Spades ♠)

Step 3: Face cards are Jack (J), Queen (Q), and King (K).

Step 4: Each suit (♥, ♦, ♣, ♠) has 3 face cards (J, Q, K).

Step 5: We only want red face cards. There are 2 red suits (♥ and ♦). So:

Number of red face cards = \(3 + 3 = 6\).

Step 6: Probability formula is:

\( P(E) = \dfrac{\text{Favourable outcomes}}{\text{Total outcomes}} \)

Here, favourable outcomes = 6, total outcomes = 52.

Step 7: So, \( P = \dfrac{6}{52} = \dfrac{3}{26} \).

Final Answer: Option (A).

Step 1: A non-leap year has 365 days.

Step 2: Divide 365 days by 7 (since 1 week = 7 days).

\(365 = 52 \times 7 + 1\)

This means: 52 full weeks and 1 extra day.

Step 3: In 52 full weeks, there are exactly 52 Sundays (one in each week).

Step 4: Whether we get a 53rd Sunday depends on what that extra day is.

Step 5: The extra day can be any one of the 7 days: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, or Saturday.

Step 6: Out of these 7 possibilities, only 1 case (when the extra day is Sunday) gives us 53 Sundays.

Step 7: Therefore, the probability is:

\(\dfrac{\text{Favourable outcomes}}{\text{Total outcomes}} = \dfrac{1}{7}\)

Final Answer: Option (A) \(\dfrac{1}{7}\)

Step 1: A die has 6 faces. The possible outcomes are: 1, 2, 3, 4, 5, 6.

Step 2: We want an odd number that is also less than 3.

Odd numbers on a die are: 1, 3, 5.

Among these, the numbers less than 3 are only: 1.

Step 3: So, the favorable outcome is only {1}. That means total favorable outcomes = 1.

Step 4: Total possible outcomes when throwing a die = 6.

Step 5: Probability = (Number of favorable outcomes) ÷ (Total outcomes).

= \(\dfrac{1}{6}\).

Final Answer: Option (A).

Step 1: A deck of playing cards has 52 cards in total.

Step 2: Out of these 52 cards, there is exactly 1 card which is the "Ace of Hearts".

Step 3: Event \(E\) is that the card drawn is not the Ace of Hearts. This means we must count all the cards except that single Ace of Hearts.

Step 4: Number of total cards = 52. Number of cards removed (Ace of Hearts) = 1. So, number of favourable outcomes = \(52 - 1 = 51\).

Step 5: Therefore, there are 51 favourable outcomes for event \(E\).

Final Answer: Option D (51)

Step 1: We are told that the probability of getting a bad egg is 0.035. This means, out of 1 egg, on average, 0.035 eggs are expected to be bad.

Step 2: The total number of eggs in the lot is 400.

Step 3: To find the expected number of bad eggs, we multiply the probability by the total number of eggs:

\[ ext{Expected bad eggs} = ext{Probability} imes ext{Total eggs} \]

\[ = 0.035 imes 400 \]

Step 4: Do the multiplication: \(0.035 imes 400 = 14\).

Final Answer: The expected number of bad eggs is 14. So, the correct option is (B).

Step 1: Recall the formula for probability:

\( \text{Probability} = \dfrac{\text{Favourable outcomes}}{\text{Total outcomes}} \)

Step 2: Here,

• Probability = \(0.08\)
• Total tickets = \(6000\)
• Favourable outcomes = Number of tickets she bought (this is what we need to find).

Step 3: Put the values into the formula:

\( 0.08 = \dfrac{\text{Number of tickets bought}}{6000} \)

Step 4: Multiply both sides by 6000 to get the number of tickets:

\( \text{Number of tickets bought} = 0.08 \times 6000 \)

Step 5: Do the multiplication:

\( 0.08 \times 6000 = 480 \)

Final Answer: She has bought 480 tickets. Hence, the correct option is (C).

Step 1: Total number of tickets

The tickets are numbered from 1 to 40. So, the total number of possible outcomes = 40.

Step 2: Favourable outcomes

We need tickets that are multiples of 5. Let’s list them:

5, 10, 15, 20, 25, 30, 35, 40

So, there are 8 favourable tickets.

Step 3: Formula for probability

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

Step 4: Substitute the values

\( P = \dfrac{8}{40} \)

Step 5: Simplify the fraction

\( \dfrac{8}{40} = \dfrac{1}{5} \)

Final Answer: Option (A) \(\dfrac{1}{5}\)

Step 1: Probability means

\[ P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \]

Step 2: Here, the experiment is: choosing one number from 1 to 100.

Total outcomes = 100 (because there are 100 numbers from 1 to 100).

Step 3: Favourable outcomes = numbers that are prime.

Prime numbers are those which have exactly 2 factors: 1 and itself.

Step 4: Count prime numbers between 1 and 100. There are 25 prime numbers in this range.

Step 5: So, probability = \( \tfrac{25}{100} \).

Step 6: Simplify the fraction: \( \tfrac{25}{100} = \tfrac{1}{4} \).

Final Answer: Option C (\( \tfrac{1}{4} \)).

Step 1: Total students in the class

Total = 23 students.

Step 2: Students from each house

• House A = 4
• House B = 8
• House C = 5
• House D = 2
• House E = Remaining students

Step 3: Find number of students in House E

Add students from A, B, C, D: \(4 + 8 + 5 + 2 = 19\).

So, House E = Total – (A + B + C + D) = \(23 - 19 = 4\).

Step 4: Which houses are not allowed?

The student should NOT be from A, B, C. That means the student must be from D or E only.

Step 5: Favourable outcomes

Students in D = 2

Students in E = 4

Total favourable = \(2 + 4 = 6\).

Step 6: Probability formula

Probability = (Favourable outcomes) ÷ (Total outcomes)

Probability = \(\dfrac{6}{23}\).

Final Answer: Option B (\(\dfrac{6}{23}\)).