Important Questions For Class 10 Maths Chapter 14- Statistics

Important questions for Class 10 Maths Chapter 14 Statistics are available at BYJU’S, which are given as per the new CBSE(NCERT) guidelines for 2019-2020. Students who are preparing for the CBSE-2020 exams can practice these questions of Statistics to score full marks for the questions from this chapter.

This chapter is important for students from the examination perspective. Most of the long answer questions will come for the exam from this chapter. Students can also refer to the solutions prepared by BYJU’S expert teachers for all the chapters of Maths.

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Important Questions & Answers For Class 10 Maths Chapter 14 Statistics

The important questions of statistics chapter for class 10 is given here both short answer type and long answer type.

Short Answer Type Questions

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.

So, the mode is = 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. So, The mode is 30.

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

Monday Tuesday Wednesday Thursday Friday Saturday
35 30 27 32 23 28

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.

Question 9. The cumulative frequency table is useful in determining the ____________?

Solution: Median

Long Answer Type Questions

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

Daily wages (in Rs.) 100-120 120-140 140-160 160-180 180-200
Number of workers 12 14 8 6 10

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 (xi) = (upper limit + lower limit)/2

In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 150 and class interval is h = 20.
So, ui = (xi – A)/h = ui = (xi – 150)/20

Substitute and find the values as follows:

Daily wages
(Class interval)
Number of workers
frequency (fi)
Mid-point (xi) ui = (xi – 150)/20 fiui
100-120 12 110 -2 -24
120-140 14 130 -1 -14
140-160 8 150 0 0
160-180 6 170 1 6
180-200 10 190 2 20
Total Sum fi = 50 Sum fiui = -12

So, the formula to find out the mean is:

Mean = x̄ = A + h∑fiui /∑fi =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.

Number of heart beats per minute 65-68 68-71 71-74 74-77 77-80 80-83 83-86
Number of women 2 4 3 8 7 4 2

Solution:

From the given data, let us assume the mean as A = 75.5

xi = (Upper limit + Lower limit)/2
Class size (h) = 3
Now, find the ui and fi ui as follows:

Class Interval Number of women (fi) Mid-point (xi) ui = (xi – 75.5)/h fiui
65-68 2 66.5 -3 -6
68-71 4 69.5 -2 -8
71-74 3 72.5 -1 -3
74-77 8 75.5 0 0
77-80 7 78.5 1 7
80-83 4 81.5 2 8
83-86 2 84.5 3 6
Sum fi= 30 Sum fiui = 4

Mean = x̄ = A + h∑fiui /∑fi

= 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 median of an ungrouped data and the median calculated when the same data is grouped are always the same. Do you think that this is a correct statement? Give reason.

Solution:

In order to calculate the median of a grouped data, the formula used is based on the assumption that the observations in the classes are uniformly distributed or equally spaced. Hence, we cannot say that the statement “the median of an ungrouped data and the median calculated when the same data is grouped are always the same” is always correct.

Q. 4: The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :

Expenditure Number of families
1000-1500 24
1500-2000 40
2000-2500 33
2500-3000 28
3000-3500 30
3500-4000 22
4000-4500 16
4500-5000 7

Solution:

From the given data:

Modal class = 1500-2000

l = 1500

Frequencies:

fm = 40

f1 = 24, f< = 33 and

h = 500

Mode formula:

\(Mode= l+(\frac{f_m -f_1}{2f_m -f_1-f_2})\times h\)

Substitute the values in the formula, we get;

\(Mode= 1500+(\frac{40 -24}{80 – 24 – 33})\times 500\)

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, xi =(upper limit +lower limit)/2

Let us assume a mean, A be 2750

Class Interval fi xi di = xi – a ui = di/h fiui
1000-1500 24 1250 -1500 -3 -72
1500-2000 40 1750 -1000 -2 -80
2000-2500 33 2250 -500 -1 -33
2500-3000 28 2750 0 0 0
3000-3500 30 3250 500 1 30
3500-4000 22 3750 1000 2 44
4000-4500 16 4250 1500 3 48
4500-5000 7 4750 2000 4 28
fi = 200 fiui = -35

The formula to calculate the mean,

Mean = x̄ = a + (∑fiui /∑fi) 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. 5: 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:

Number of cars Frequency
0-10 7
10-20 14
20-30 13
30-40 12
40-50 20
50-60 11
60-70 15
70-80 8

Solution:

From the given data:

Modal class = 40 – 50, l = 40,

class width (h) = 10, fm = 20, f1 = 12 and f2 = 11

\(Mode= l+(\frac{f_m -f_1}{2f_m -f_1-f_2})\times h\)

Substitute the values

\(Mode= 40+(\frac{20-12}{40-12-11})\times 10\)

Mode = 40 + (80/17) = 40 + 4.7 = 44.7

Thus, the mode of the given data is 44.7 cars

Q. 6: An aircraft has 120 passenger seats. The number of seats occupied during 100 flights are given in the following table :

Number of seats 100-104 104-108 108-112 112-116 116-200
Frequency 15 20 32 18 15

Determine the mean number of seats occupied over the flights.

Solution:

Class Interval Class Marks (xi) Frequency (fi) Deviation (di = xi – a) fidi
100 – 104 102 15 – 8 – 120
104 – 108 106 20 – 4 – 80
108 – 112 110 32 0 0
112 – 116 114 18 4 72
116 – 120 118 15 8 120
N = Σfi = 100 Σfidi = – 8

∴ Assumed mean, a = 110

Class width, h = 4

And total observations, N = 100

\(Mean (\overline{x})=a+\frac{\sum f_id_i}{\sum f_i}\)

= 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).

Q. 7: A survey regarding the heights (in cm) of 51 girls of Class X of a school was conducted and the following data were obtained:

Height (in cm) Number of girls
Less than 140 4
Less than 145 11
Less than 150 29
Less than 155 40
Less than 160 46
Less than 165 51

Find the median height.

Solution:

To calculate the median height, we need to find the class intervals and their corresponding frequencies.

The given distribution being of the less than type, 140, 145, 150,… ., 165 give the upper limits of the corresponding class intervals.

So, the classes should be below 140, 140 – 145, 145 – 150,… ., 160 – 165.

Observe that from the given distribution, we find that there are 4 girls with height less than 140, i.e., the frequency of class interval below 140 is 4.

Now, there are 11 girls with heights less than 145 and 4 girls with height less than 140. Therefore, the number of girls with height in the interval of 140 – 145 is 11 – 4 = 7.

Similarly, the frequency of 145 – 150 is 29 – 11 = 18, for 150 – 155, it is 40 – 29 = 11, and so on. So, our frequency distribution table with the given cumulative frequencies becomes:

Class intervals Frequency Cumulative frequency
Below 140 4 4
140-145 7 11
145-150 18 29
150-155 11 40
155-160 6 46
160-165 5 51

Now n = 51. So, n/2 = 51/2 =25.5

This observation lies in class 145 – 150.

Then, l (the lower limit) = 145,

cf (the cumulative frequency of the class preceding 145 – 150) = 11,

f (the frequency of the median class 145 – 150) = 18,

h (the class size) = 5.

Using the formula,

Median \(=l+(\frac{\frac{n}{2}-cf}{f})\times h\)

we have Median \(=145+(\frac{25.5-11}{18})\times 5\)

= 145 + 72.5/18 = 149.03.

So, the median height of the girls is 149.03 cm.

This means that the height of about 50% of the girls is less than this height, and 50% are taller than this height.

Q. 8: The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :

Length(in mm) Number of leaves
118-126 3
127-135 5
136-144 9
145-153 12
154-162 5
163-171 4
172-180 2

Find the median length of leaves.  

(Hint: The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, . . ., 171.5 – 180.5.)

Solution:

Since the data are not continuous reduce 0.5 in the lower limit and add 0.5 in the upper limit.

Class Interval (CI) Frequency Cumulative frequency
117.5-126.5 3 3
126.5-135.5 5 8
135.5-144.5 9 17
144.5-153.5 12 29
153.5-162.5 5 34
162.5-171.5 4 38
171.5-180.5 2 40

From the above tab

n = 40 and n/2 = 20

Median class = 144.5-153.5

then, l = 144.5,

cf = 17, f = 12 & h = 9

Median \(=l+(\frac{\frac{n}{2}-cf}{f})\times h\)

Median \(=144.5+(\frac{20-17}{12})\times 9\)

=144.5+(9/4)

=146.75 mm

Therefore, the median length of the leaves = 146.75 mm.

Q. 9: If the median of a distribution given below is 28.5 then, find the value of an x &y.

Class Interval Frequency
0-10 5
10-20 x
20-30 20
30-40 15
40-50 y
50-60 5
Total 60

Solution:

From the given data,

n = 60

Median of the given data = 28.5

Where, n/2 = 30

Median class is 20 – 30 with a cumulative frequency = 25 + x

Lower limit of median class = 20,

Cf = 5 + x ,

f = 20 & h = 10

Median = \(=l+(\frac{\frac{n}{2}-cf}{f})\times h\)

Substitute the values

28.5=20+10(30−5−x)/20)

8.5 =(25-x)/2

17 = 25-x

Therefore, x =8

Now, from cumulative frequency, we can identify the value of x + y as follows:

Since,

60=5+20+15+5+x+y

Now, substitute the value of x, to find y

60 = 5+20+15+5+8+y

y = 60-53

y = 7

Therefore, the value of x = 8 and y = 7

Q. 10: The annual profits earned by 30 shops of a shopping complex in a locality give rise to the following distribution :

Profit (Rs in lakhs) Number of shops (frequency)
More than or equal to 5 30
More than or equal to 10 28
More than or equal to 15 16
More than or equal to 20 14
More than or equal to 25 10
More than or equal to 30 7
More than or equal to 35 3

Hence obtain the median profit.

Solution :

We first draw the coordinate axes, with lower limits of the profit along the horizontal axis, and the cumulative frequency along the vertical axes.

Then, we plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7) and (35, 3). We

join these points with a smooth curve to get the ‘more than’ ogive, as shown in the below figure.

Important Question Class 10 maths chapter 14

Now, let us obtain the classes, their frequencies and the cumulative frequency from the given table.

Classes Number of shops Cumulative frequency
5-10 2 2
10-15 12 14
15-20 2 16
20-25 4 20
25-30 3 23
30-35 4 27
35-40 3 30

Using these values, we plot the points (10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27), (40, 30) on the same axes as in the above figure, to get the ‘less than’ ogive, as shown below.

Imp questions class 10 maths chapter 14

The abscissa of their point of intersection is nearly 17.5, which is the median. This can also be verified by using the formula.

Hence, the median profit (in lakhs) is Rs. 17.5.

Q. 11: The following tables give the production yield per hectare of wheat of 100 farms of a village.

Production Yield 50-55 55-60 60-65 65-70 70-75 75-80
Number of farms 2 8 12 24 38 16

Change the distribution to a more than type distribution and draw its ogive.

Solution:

Converting the given distribution to a more than type distribution, we get

Production Yield (kg/ha) Number of farms
More than or equal to 50 100
More than or equal to 55 100-2 = 98
More than or equal to 60 98-8= 90
More than or equal to 65 90-12=78
More than or equal to 70 78-24=54
More than or equal to 75 54-38 =16

From the table obtained draw the ogive by plotting the corresponding points where the upper limits in x-axis and the frequencies obtained in the y-axis are (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on this graph paper.

The graph obtained is known as more than type ogive curve.

Class 10 Chapter 14 Imp Ques. 11 Solution

Practice Questions for Class 10 Maths Chapter 14 Statistics

  1. The frequency distribution table of agricultural holdings in a village is given below :
Area of land (in hectares) 1-3 3-5 5-7 7-9 9-11 11-13
Number of families 20 45 80 55 40 12

Find the modal agricultural holdings of the village.

2. The maximum bowling speeds, in km per hour, of 33 players at a cricket coaching centre are given as follows :

Speed (km/hr) 85-100 100-115 115-130 130-145
Number of players 11 9 8 5

Calculate the median bowling speed.

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

Daily Pocket Allowance(in c) 11-13 13-15 15-17 17-19 19-21 21-23 23-35
Number of children 7 6 9 13 f 5 4

4. Consider the data :

Class 65-85 85-105 105-125 125-145 145-165 165-185 185-205
Frequency 4 5 13 20 14 7 4

(1) The difference of the upper limit of the median class and the lower limit of the modal class is

(A) 0 (B) 19 (C) 20 (D) 38

(2) In the formula x = a + h(fiui/fi), for finding the mean of grouped frequency distribution, ui =

(A) (xi+a)/h

(B) h (xi – a)

(C) (xi –a)/h

(D) (a – xi)/h

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