Class 10 Maths: Statistics – Extra Questions with Answers
Q. 1 If 35 students have the following blood groups: 10 A, 9 B, 11 O, and the rest are AB, what is the percentage of students with AB blood group?
Check Solution
Ans: D
Solution:
1. Find the number of students with AB blood type: 35 (total) – 10 (A) – 9 (B) – 11 (O) = 5
2. Calculate the percentage: (5 / 35) * 100% = 14.2857… %
Answer: 14.29%
Q. 2 A frequency distribution table shows the scores of 20 students on a quiz. The intervals are 0-10, 11-20, 21-30, 31-40, and 41-50. The corresponding frequencies are 2, 5, 7, 4, and 2, respectively. Using the midpoint of each interval to represent all values within that interval, calculate the estimated mean score of the quiz.
Check Solution
Ans: B
Solution:
Midpoints: 5, 15, 25, 35, 45
Weighted sum: (2*5) + (5*15) + (7*25) + (4*35) + (2*45) = 10 + 75 + 175 + 140 + 90 = 490
Estimated mean: 490 / 20 = 24.5
Answer: 24.5
Q. 3 Let $\bar{x}$ be the mean of $x_1, x_2, x_3$ where $x_1=2, x_2=4, x_3=6$ and $\bar{y}$ be the mean of $y_1, y_2$ where $y_1=1, y_2=3$. If $\bar{z}$ is the mean of $x_1, x_2, x_3, y_1, y_2$, then $\bar{z}$ is equal to ______.
Check Solution
Ans: C
Solution:
$\bar{x} = \frac{2+4+6}{3} = \frac{12}{3} = 4$
$\bar{y} = \frac{1+3}{2} = \frac{4}{2} = 2$
$\bar{z} = \frac{2+4+6+1+3}{5} = \frac{16}{5} = 3.2$
Answer: 3.2
Q. 4 The standard deviation of a dataset is 5, and the mean is 25. Calculate the coefficient of variation, expressed as a percentage.
Check Solution
Ans: D
Solution:
Coefficient of Variation (CV) = (Standard Deviation / Mean) * 100
CV = (5 / 25) * 100
CV = 0.2 * 100
CV = 20%
Answer: 20%
Q. 5 The average weight of 15 apples is 120 grams, and the average weight of 25 oranges is 180 grams. What is the combined average weight of all the fruits?
Check Solution
Ans: B
Solution:
Total weight of apples: 15 * 120 = 1800 grams
Total weight of oranges: 25 * 180 = 4500 grams
Total weight of all fruits: 1800 + 4500 = 6300 grams
Total number of fruits: 15 + 25 = 40
Combined average weight: 6300 / 40 = 157.5 grams
Answer: 157.5
Q. 6 A shipment of 800 boxes of screws was inspected for defective screws. The results are as follows: | Number of defective screws | 0 | 1 | 2 | 3 | 4 | 5 | 6 | more than 6 | | ————————- | — | — | — | — | — | — | — | ———– | | Frequency | 450 | 200 | 50 | 40 | 30 | 15 | 10 | 5 | If one box is selected at random, what is the probability that it contains exactly 3 defective screws?
Check Solution
Ans: C
Solution:
The frequency of boxes with exactly 3 defective screws is 40. The total number of boxes is 800. The probability is the frequency divided by the total number of boxes.
Answer: 40/800
Q. 7 The following table shows the distribution of marks obtained by 40 students in a mathematics test: | Marks (Class Interval) | Number of Students (Frequency) | |—|—| | 0-10 | 3 | | 10-20 | 5 | | 20-30 | 12 | | 30-40 | 10 | | 40-50 | 8 | | 50-60 | 2 | Find the mean marks obtained by the students.
Check Solution
Ans: B
Solution:
1. Find the midpoint (x) of each class interval: 5, 15, 25, 35, 45, 55
2. Multiply the midpoint (x) by the frequency (f) for each interval: 15, 75, 300, 350, 360, 110
3. Sum the products (fx): 15+75+300+350+360+110 = 1210
4. Sum the frequencies (f): 3 + 5 + 12 + 10 + 8 + 2 = 40
5. Calculate the mean: Σfx / Σf = 1210 / 40 = 30.25
Answer: 30.25
Q. 8 If the arithmetic mean of the first $n$ natural numbers is 21, then $n$ is __________.
Check Solution
Ans: B
Solution:
The arithmetic mean of the first $n$ natural numbers is $\frac{1+2+3+…+n}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2}$.
We are given that $\frac{n+1}{2} = 21$. Multiplying both sides by 2, we have $n+1 = 42$. Subtracting 1 from both sides, we get $n=41$.
Answer: 41
Q. 9 Two fair six-sided dice are rolled. What is the probability that the sum of the numbers rolled is a prime number?
Check Solution
Ans: D
Solution:
The possible sums range from 2 to 12. The prime sums are 2, 3, 5, 7, and 11.
The number of ways to achieve each sum are:
2: (1,1) – 1 way
3: (1,2), (2,1) – 2 ways
5: (1,4), (4,1), (2,3), (3,2) – 4 ways
7: (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) – 6 ways
11: (5,6), (6,5) – 2 ways
The total number of ways to get a prime sum is 1 + 2 + 4 + 6 + 2 = 15.
There are 36 total possible outcomes when rolling two dice (6*6).
The probability is 15/36 = 5/12.
Answer: 5/12
Q. 10 If the mean of group A with 10 observations is 15, the mean of group B with 20 observations is 20, and the mean of group C with 30 observations is 25, what is the mean of all three groups combined?
Check Solution
Ans: A
Solution:
Total sum of A = 10 * 15 = 150
Total sum of B = 20 * 20 = 400
Total sum of C = 30 * 25 = 750
Total sum of all groups = 150 + 400 + 750 = 1300
Total observations = 10 + 20 + 30 = 60
Combined mean = 1300 / 60 = 65/3 = 21.666…
Answer: 21.67
Next Topic: Probability
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