Important Questions For Class 10 Maths Chapter 14- Statistics - GMS - Learning Simply
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Important Questions For Class 10 Maths Chapter 14- Statistics

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Important Questions For Class 10 Maths Chapter 14- Statistics

Important questions for Class 10 Maths Chapter 14 Statistics are available, which are created as per the latest exam pattern and referring to the sample papers 2021 design. Students who are preparing for the CBSE class 10 board 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.  By solving these Important Questions of Class 10 Maths Chapter 14 Statistics, students will thoroughly revise this chapter and get well versed with the questions which are expected to be asked in the Maths paper.

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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.3Find 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:

MondayTuesdayWednesdayThursdayFridaySaturday
353027322328

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.)100-120120-140140-160160-180180-200
Number of workers12148610

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)/20fiui
100-12012110-2-24
120-14014130-1-14
140-160815000
160-180617016
180-20010190220
TotalSum fi = 50Sum 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 minute65-6868-7171-7474-7777-8080-8383-86
Number of women2438742

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 IntervalNumber of women (fi)Mid-point (xi)ui = (xi – 75.5)/hfiui
65-68266.5-3-6
68-71469.5-2-8
71-74372.5-1-3
74-77875.500
77-80778.517
80-83481.528
83-86284.536
Sum fi= 30Sum 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 :

ExpenditureNumber of families
1000-150024
1500-200040
2000-250033
2500-300028
3000-350030
3500-400022
4000-450016
4500-50007

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+(fmf12fmf1f2)×h

Substitute the values in the formula, we get;

Mode=1500+(4024802433)×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 Intervalfixidi = xi – aui = di/hfiui
1000-1500241250-1500-3-72
1500-2000401750-1000-2-80
2000-2500332250-500-1-33
2500-3000282750000
3000-3500303250500130
3500-40002237501000244
4000-45001642501500348
4500-5000747502000428
fi = 200fiui = -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 carsFrequency
0-107
10-2014
20-3013
30-4012
40-5020
50-6011
60-7015
70-808

Solution:

From the given data:

Modal class = 40 – 50, l = 40,

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

Mode=l+(fmf12fmf1f2)×h

Substitute the values

Mode=40+(2012401211)×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 seats100-104104-108108-112112-116116-200
Frequency1520321815

Determine the mean number of seats occupied over the flights.

Solution:

Class IntervalClass Marks (xi)Frequency (fi)Deviation (di = xi – a)fidi
100 – 10410215– 8– 120
104 – 10810620– 4– 80
108 – 1121103200
112 – 11611418472
116 – 120118158120
N = Σfi = 100Σfidi = – 8

∴ Assumed mean, a = 110

Class width, h = 4

And total observations, N = 100

Mean(x¯¯¯)=a+fidifi

= 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 1404
Less than 14511
Less than 15029
Less than 15540
Less than 16046
Less than 16551

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 intervalsFrequencyCumulative frequency
Below 14044
140-145711
145-1501829
150-1551140
155-160646
160-165551

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+(n2cff)×h

we have Median =145+(25.51118)×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
118-1263
127-1355
136-1449
145-15312
154-1625
163-1714
172-1802

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)FrequencyCumulative frequency
117.5-126.533
126.5-135.558
135.5-144.5917
144.5-153.51229
153.5-162.5534
162.5-171.5438
171.5-180.5240

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+(n2cff)×h

Median =144.5+(201712)×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 IntervalFrequency
0-105
10-20x
20-3020
30-4015
40-50y
50-605
Total60

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+(n2cff)×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 530
More than or equal to 1028
More than or equal to 1516
More than or equal to 2014
More than or equal to 2510
More than or equal to 307
More than or equal to 353

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.

Imp questions class 10 maths chapter 14 - 10

 

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

ClassesNumber of shopsCumulative frequency
5-1022
10-151214
15-20216
20-25420
25-30323
30-35427
35-40330

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 Yield50-5555-6060-6565-7070-7575-80
Number of farms2812243816

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 50100
More than or equal to 55100-2 = 98
More than or equal to 6098-8= 90
More than or equal to 6590-12=78
More than or equal to 7078-24=54
More than or equal to 7554-38 =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)1-33-55-77-99-1111-13
Number of families204580554012

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)85-100100-115115-130130-145
Number of players11985

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)11-1313-1515-1717-1919-2121-2323-35
Number of children76913f54

4. Consider the data :

Class65-8585-105105-125125-145145-165165-185185-205
Frequency4513201474

(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

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

6) The average score of boys in the examination of a school is 71 and that of the the girls is 73. The average score of the school in the examination is 71.8. Find the ratio of the number of boys to the number of girls who appeared in the examination.

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At the helm of GMS Learning is Principal Balkishan Agrawal, a dedicated and experienced educationist. Under his able guidance, our school has flourished academically and has achieved remarkable milestones in various fields. Principal Agrawal’s visio…

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