Given below are the **Class 10 Maths** Important and extra Questions for Statistics

(a) Fill in the blank's

(b) True and false

(c) Multiple choice questions

(d) Match the column

(e) Long answer type

(a) Fill in the blank's

(b) True and false

(c) Multiple choice questions

(d) Match the column

(e) Long answer type

(a) The mean of 50 numbers is 18, the new mean will be _____ if each observation is increased by 4. ( 22/24/20)

(b) The sum of deviations of a set of values {a,b,c,d,e,f,g.h.i ______} n items measured from 26 is -10 and the sum of deviations of the values from 20 is 50. The value of n is ______(10/12/9) And mean of the items is _______. ( 19/18/17/25)

(c) For a given data with 110 observations the ‘less than ogive’ and the ‘more then ogive’ intersect at (18, 20). The median of the data is _____. (18/20/19)

(d) The curve drawn by taking upper limits along x-axis and cumulative frequency along y-axis is _______.( less than ogive /more than ogive)

(e) The mean of five numbers is 40. If one number is excluded, their mean becomes 28. The excluded number is _________. (68 / 88)

(f) The modal class of the grouped size frequency table given below is

(a) 24

So M=22

(b) 10,25

Subtracting ,we get

-6n=-60 => n=10

So $\sum x=250$

Mean=250/10=25

(c) 18. Median is the point of intersection of ogive curves

(d) less than ogive curve

(e) 88

Sum of five number=5Xmean=200

Sum of four number=4Xmean=112

Subtracting, we get the number=88

(f) Modal class is 5-5.2

(a) The mean is 33

(b) The modal class is 35-45

(c) The mode is 34

(d) The Frequency for less than is 35 is 38

(e) The median is

Lets find the mean value first

Mean is given by

The class having highest frequency is called the Modal class, so Modal class is 35-45

Where

l = lower limit of the modal class,

h = size of the class interval (assuming all class sizes to be equal),

f

f

f

Now l=35 h=10 f

So substituting all the values we get

Now

3 Median=Mode +2 Mean

So median= (Mode +2 Mean)/3=35.853

(a) False. I

(b) True

(c) False

(d) True.

(e) False

While computing mean of grouped data, we assume that the frequencies are

(a)centred at the upper limits of the classes

(b)centred at the lower limits of the classes

(c)centred at the class-marks of the classes

(d)evenly distributed over all the classes

Which of the following is the measure of central tendency?

(a) Mean

(b)Mode

(c)Median

(d) Range

For drawing a frequency polygon of a continuous frequency distribution, we plot the Points whose ordinates are the frequencies of the respective classes and abscissae are respectively :

(a) class marks of the classes

(b) upper limits of preceding classes

(c) lower limits of the classes

(d) upper limits of the classes

Find the mean of 32 numbers given mean of ten of them is 12 and the mean of other 20 is 9. And last 2 number is 10

(a)10

(b) 12

(c) 13

(d) 14

Mean of 10 number =12

Sum of these 10 numbers =120

Mean of 20 number =9

Sum of these 20 numbers =180

mean of 2 number =10

Sum of these 2 numbers =20

Mean of 32 number= Sum of all number/32=(120+180+20)/32=10

The median and mean of the first 10 natural numbers.

(a) 5.5,5.5

(b) 5.5,6

(c) 5,6

(d) None of these

Mean =5.5

Median is mean of 5 and 6 th term, So 5

Anand says that the median of 3, 14, 19, 20, 11 is 19. What doesn’t the Anand understand about finding the median?

(a) The dataset should be ascending order

(b) Highest no in the dataset is the median

(c) Average of lowest and highest is the median

(d) None of these

The following observations are arranged in ascending order :

20, 23, 42, 53, x, x + 2, 70, 75, 82, 96

If the median is 63, find the value of

(a) 62

(b) 64

(c) 60

(d) None of these

Median is mean of 5 and 6 term

So x+1=63

x=62

The mean of 20 observations was 60. It was detected on rechecking that the value of 125 was wrongly copied as 25 for computation of mean. Find the correct mean

(a) 67

(b) 66

(c) 65

(d) None of the above

Let x be the sum of observation of 19 numbers leaving 125,

Then

$x+25=20 \times 60=1200$

Now

$x+125=20 \times y=20y$

Subtracting

$125-25=20y-1200$

$20y=1300$

y=65

Compute the Median for the given data

First calculate the Class mark and cumulative frequency of the data

We have N=266, So N/2=133,Cummulative frequency first greater than 133,lies in class 130-140

Now

Median is calculated as

Where

l = lower limit of median class,

n = number of observations,

cf = cumulative frequency of class preceding the median class,

f = frequency of median class,

h = class size (assuming class size to be equal)

Here l=130 n=266 cf=89 f=72 h=10

Substituting these

M= 130+(133-89)x10/72=136.11

**Notes**-
**Assignments** -
**NCERT Solutions**

Class 10 Maths Class 10 Science