CBSE Class 10 Maths Important Questions: Statistics

Statistics can be understood as the study that deals with the collection, organisation, analysis, interpretation, and presentation of any sort of data or information. The word “statistics” originated from the German word ‘Statistik’. Data can be graphically presented in the form of bar graphs, histograms, and frequency polygons and can also be numerically represented as grouped or ungrouped data. Mean, median and mode are three measures of the central tendency of ungrouped data.

The important question for class 10 maths Chapter 14 Statistics can be related to these entities

• Define data
• Graphical presentation of data in form of bar graphs, histograms, and frequency polygons
• Three measures of central tendency of ungrouped data
• The Mean measure of central tendency
• The Median measure of central tendency
• The Mode measure of central tendency

Important Questions Chapter 14 – Statistics

Q1. 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

Q2. 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

Q3. 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

Q4. Construct the cumulative frequency distribution of the following distribution : 

Solution:

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

Q5. 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.

Q6. 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

Q7. Show that the mode of the series obtained by combining the two series S1 and S2 given below is different from that of S1 and S2 taken separately

S1 : 3, 5, 8, 8, 9, 12, 13, 9, 9

S2 : 7, 4, 7, 8, 7, 8, 13

Solution:

In S1 : Number 9 occurs 3 times (maximum)

∴ Mode of S1 Series = 9

In S2 : Number 7 occurs 3 times (maximum)

∴ Mode of S, Series = 7

After combination:

In S1 & S2 : No. 8 occurs 4 times (maximum)

∴ Mode of S1 & S2 taken combined = 8

So, mode of S1 & S2 combined is different from that of S1 & S2 taken separately.

Q8. Find the median of the data using an empirical formula, when it is given that mode = 35.3 and mean = 30.5

Solution:

Mode = 3(Median) – 2(Mean)

35.3 = 3(Median) – 2(30.5)

35.3 = 3(Median) – 61

96.3 = 3 Median

Median = 96.3/3 = 32.1

Q9. In a continuous frequency distribution, the median of the data is 21. If each observation is increased by 5, then find the new median.

Solution:

New median = 21 + 5 = 26

Q10. If the value of mean and mode are respectively 30 and 15, then find median.

• 22.5

• 24.5

• 25

• 26

Solution:

1. c) 25

Mode can be calculated as;

Mode = 3 median — 2 mean

Median = (Mode + 2 Mean)/3

Median = (15 + (2×30))/3

Median = 25