Class 10 Maths Chapter 13 Statistics Most Important Questions By PW

Physics Wallah brings you a well-designed set of Chapter 13 Statistics CBSE Class 10 Maths important questions to help you strengthen your data-handling skills. As you practice, you will work on finding the mean (using direct, assumed mean, and step deviation methods), median, and mode for grouped data.

The CBSE Class 10 Maths syllabus requires careful calculation and a clear understanding of formulas, and this set helps you develop both. With regular practice, you will improve your speed, reduce errors, and gain the confidence to solve different types of data-based questions accurately in your upcoming board exams.

Most Important Questions Class 10 Maths Chapter 13 Statistics

Short Answer Type Questions

Below, we have provided Important Questions For Class 10 Maths Chapter 14 to help students better understand the chapter's concepts.

Q.1. Find the mean of the 32 numbers, such that if the mean of 10 of them is 15 and the mean of 20 of them is 11. The last two numbers are 10.

Solution: The given mean of 10 numbers = 15

So, Mean of 10 numbers = sum of observations/ no. of observations 15 = sum of observations / 10 Sum of observations of 10 numbers = 150 Similarly, Mean of 20 numbers = sum of observations/ no. of observations 11 = sum of observations / 20 Sum of observations of 20 numbers = 220 Hence, Mean of 32 numbers = (sum 10 numbers + sum of 20 numbers + sum of last two numbers)/ no. of observations Mean of 32 numbers = (150 + 220 + 20 ) / 32 = 390 /32 = 12.188

Q.2. Find the mean of the first 10 natural numbers.

Solution: The first 10 natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Mean = (1 + 2 +3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) / 10 = 55/10 = 5.5

Q.3 . Find the value of y from the following observations if these are already arranged in ascending order. The Median is 63.

20, 24, 42, y, y + 2, 73, 75, 80, 99

Solution:

As the number of observations made is odd, so the median will be the middle term, i.e. y + 2. Therefore, y + 2 = 63 y = 63 – 2 = 61

Q.4. While checking the value of 20 observations, it was noted that 125 was wrongly noted as 25 while calculating the mean and then the mean was 60. Find the correct mean.

Solution:

Let y be the sum of observation of 19 (20 – 1) numbers leaving 125. So, y + 25 = 20 × 60 = 1200 {Mean = (sum of observations/ no. of observations)} As we know, x + 25 = 20 × 60 = 1200 Also x + 125 = 20 × y = 20y Next, Subtract 125 − 25 = 20y − 1200 20y = 1300 y = 65

Q.5. Find the mode of the following items.

0, 5, 5, 1, 6, 4, 3, 0, 2, 5, 5, 6

Solution: On arranging the items in ascending order, we get:

0, 0, 1, 2, 3, 4, 5, 5, 5, 5, 6, 6 As we can see 5 occurs the maximum number of times. Therefore, the mode of the given items = 5

Q.6. A student scored the following marks in 6 subjects:

30, 19, 25, 30, 27, 30

Find his modal score.

Solution: If we arrange his marks in ascending order

19, 25, 27, 30, 30, 30 As we can see, 30 occurs a maximum number of times. Therefore, the modal score of the student = 30

Q.7.The daily minimum steps climbed by a man during a week were as under:

Find the mean of the steps.

Solution: Number of steps climbed in a week: 35, 30, 27, 32, 23, 28.

So, we get, Mean = sum of observation (steps) / total no of observations = (35 + 30 + 27 + 32 + 23 + 28) / 6 = 175/6 = 29.17

Q. 8. If the mean of 4 numbers, 2,6,7 and a is 15 and also the mean of other 5 numbers, 6, 18, 1, a, b is 50. What is the value of b?

Solution:

Mean = sum of observations / no. of observations 15 = (2 + 6 + 7 +a)/4 15 = (15 + a) / 4 15 x 4 = 15 + a 60 – 15 = a a = 45 Similarly, Mean = sum of observations / no. of observations 50 = (18 + 6 + 1 +a + b)/5 50 = (18 + 6 + 1 +45 + b)/5 50 = (70 + b)/5 250 = 70 + b b = 250 – 70 = 180 So, The value of b = 180. Q.9: If the mean of first n natural numbers is 15, then find n. Solution: We know that the sum of first n natural numbers = n(n + 1)/2 Mean of the first n natural numbers = Sum of first n natural numbers/n = [n(n + 1)/2]/ n = (n + 1)/2 According to the given, (n + 1)/2 = 15 n + 1 = 30 n = 29 Q.10: Construct the cumulative frequency distribution of the following distribution:

Solution: The cumulative frequency distribution of the given distribution is given below :

Long Answer Type Questions

Below we have provided Important Questions For Class 10 Maths Chapter 14 to help students understand the concepts of the chapter better. Students are advised to practice these Important Questions For Class 10 Maths Chapter 14 statistics to ace their math exam.

Q. 1: Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution:

Find the midpoint of the given interval using the formula. Midpoint (x _i ) = (upper limit + lower limit)/2 In this case, the value of mid-point (x _i ) is very large, so let us assume the mean value, A = 150 and class interval is h = 20. So, u _i = (x _i – A)/h = u _i = (x _i – 150)/20 Substitute and find the values as follows:

So, the formula to find out the mean is: Mean = x̄ = A + h (∑f _i u _i /∑f _i ) =150 + (20 × -12/50) = 150 – 4.8 = 145.20 Therefore, mean daily wage of the workers = Rs. 145.20

Q.2: Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute was recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Solution:

From the given data, let us assume the mean as A = 75.5 x _i = (Upper limit + Lower limit)/2 Class size (h) = 3 Now, find the u _i and f _i u _i as follows:

Mean = x̄ = A + h∑f _i u _i /∑f _i = 75.5 + 3×(4/30) 75.5 + 4/10 = 75.5 + 0.4 = 75.9 Therefore, the mean heartbeats per minute for these women is 75.9

Q. 3: The following data gives the distribution of the total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :

Solution:

From the given data: Modal class = 1500-2000 l = 1500 Frequencies: f _m = 40 f ₁ = 24, f< = 33 and h = 500 Mode formula: Substitute the values in the formula, we get;

Mode = 1500 + ((16 x 500)/23) Mode = 1500+(8000/23) = 1500 + 347.83 Therefore, the modal monthly expenditure of the families= Rupees 1847.83 Calculation for mean: First find the midpoint using the formula, x _i =(upper limit +lower limit)/2 Let us assume a mean, A be 2750

The formula to calculate the mean, Mean = x̄ = a + (∑f _i u _i /∑f _i ) х h Substitute the values in the given formula = 2750 + (-35/200) х 500 = 2750 – 87.50 = 2662.50 So, the mean monthly expenditure of the families = Rupees 2662.50

Q. 4: A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:

Solution:

From the given data: Modal class = 40 – 50, l = 40, class width (h) = 10, f _m = 20, f ₁ = 12 and f ₂ = 11 Substitute the values

Mode = 40 + (80/17) = 40 + 4.7 = 44.7 Thus, the mode of the given data is 44.7 cars

Q. 5: An aircraft has 120 passenger seats. The number of seats occupied during 100 flights are given in the following table :

Determine the mean number of seats occupied over the flights.

Solution:

∴ Assumed mean, a = 110 Class width, h = 4 And total observations, N = 100 = 110 + (-8/100) = 110 – 0.08 = 109.92 But we know that the seats cannot be in decimal. Therefore, the number of seats = 109 (approx).