# Statistics Chapter 14 Extra Questions and Solutions For Class 10 CBSE Mathematics

You are going to go through Statistics Chapter 14 Extra Questions and Solutions For Class 10 CBSE Mathematics. This post presents to the students a clear conception of how to move with the basics of Extra Questions and answers. The expert prepared The Extra Questions and And Answers. https://cbsencertanswers.com/is very much to make things way simpler and easier for the students. Especially those who are appearing for the board exams. We took every care to make sure that the effort serves the purpose. So, let us find out Statistics Chapter 14 Extra Questions and Solutions For Class 10 CBSE MathematicsOn this page, you can find Statistics Chapter 14 Extra Questions and Solutions For Class 10 CBSE Mathematics.

## Solution.

MEAN = Σ FX / N = 2620 / 106 = 25

Therefore, Mean is 25.

## Solution.

Now,

Mean = Σ FX / N = ( 530 + 25p ) / 50

It is given that mean is 20.6 ,

20.6 = ( 530 + 25p ) / 50

( 20.6 * 50 ) = 530 + 25p

P = 500 / 25 = 20

Therefore, the value of p is 20.

## Solution.

Mean = Σ FX / N =  ( 445 + 10p) / ( p + 27)

Now, Mean = 15

Mean = (445 + 10p) / ( p + 27 )

15 = (445 + 10p)/( p+ 27)

15p + 405 = 445 + 10p

5p = 40

p = 8.

So, the value of the following missing frequency p is 8.

## Solution.

And , Mean = Σ FX / N = ( 524 + 7p) / 50

Mean = 12.58,

12.58 = ( 524 + 7p) / 50

12.58 * 50 = 524 + 7p

629 – 524 = 7p

7p = 105

p = 15

So, the value of p is 15 .

## Solution.

Mean = Σ FX / N = ( 303 + 9p) / (41 + p)

Here, mean = 7.68

7.68 = (303 + 9p)/(41 + p)

1.32p = 11.88

p= 9

Therefore, the value of p is 9 whose mean is 7.68.

## Solution.

Let the assumed mean (A)  is 3

Mean number of calls = A + (Σ FU/ N)

=  3 + ( 135 / 250 )

= 885 / 250

= 3.54

So, the mean number of calls per interval is 3.54 .

## Solution.

Let the assumed mean (A) is 4

Average no of branches per plant = A + (Σ FU/N)

= 4 – 77/200  =  (800 – 77)/200 = 3.615

So, the average number of branches per plant is 3.615.

## Solution.

Let assumed mean (A) is 275

A = 275 and h =50

Mean = A + h * (Σ FU / N)

= 275 + 50 * ( -35 /200)

=  266.25

## Solution.

From the given data we have to find the class interval we know that,

Class marks (x) = (upper class limit + lower class limit)/2

Mean = Σ FU / N

= 162 / 20 = 8.1

Thus , the mean number of plant per house is 8.1 . Here we will use the Direct Method as values of class marks X and F are very small.

## Solution.

Let the Assumed Mean (A) is 150

A = 150 and h = 20

Mean = A + h * (Σ FU /N)

= 150 + 20 * ( -12 /50 )

=  150 – 4.8

=  145.20 .

## Solution:

Let the Assumed Mean (A) is 15.

Here, A = 15 and h = 6

Mean = A + h x (Σ FU /N)

= 15 + 6 x (3/40) = 15 + 0.45 == 15.45

## Solution.

Arranging the given data in ascending order, we have

694, 696, 699, 705, 710, 712, 715, 716, 719, 721, 725, 728, 729, 734, 745.

Here the number of terms is an old number i.e., N = 15

We use the following procedure to find the median.

Median = (N + 1)/2 th term

= (15 + 1)/2 th term

= 8th term

So, the 8th term in the arranged order of the given data should be the median.

Therefore, 716 is the median of the data.

## Solution.

Here, we have N as 420. The cumulative frequency just greater than N/2 is 275 then 165.5 – 168.5 is the median class. Now,

L = 165.5, f = 142, F = 133 and h = (168.5 – 165.5) = 3

=165.5 + 1.63 = 167.13

Statistics

## Solution.

Here , N = 250  so, N/2 = 125.

The cumulative frequency just greater than N/ 2 is 127 then median class is 50 – 60 such that L = 50, f = 31, F = 96, h = 60 -50 = 10

= 50 + 9.35 = 59.35

## Solution.

Let the unknown frequency be x.

Given, median = 24.

Then, median class = 20 – 30; L = 20, h = 30 -20 = 10, f = x, F = 30

4x = 275 + 5x – 300

4x – 5x = – 25

– x = – 25

x = 25

Therefore, the Missing frequency is 25 where the median is given as 24.

## Solution.

It’s seen that, the cumulative frequency just greater than n/2 (400/2 = 200) is 216 and it belongs to the class interval 3000 – 3500 which becomes the Median class = 3000 – 3500. Frequency (F) of median class = 86 and Cumulative frequency (CF) of class preceding median class = 130 and, the Class size (h) = 500.

We use the following formula to solve it:

= 3406.98

Therefore, the median lifetime of lamp is 3406.98 hours.

Statistics

## Solution:

It’s given that, N = 200,Therefore

⇒ 46 + x + y + 25 + 10 + 5 = 200

⇒ x + y = 200 – 46 – 25 – 10 – 5

⇒ x + y = 114 —- (i)

And also given, Mean = 1.46

⇒ Sum/ N = 1.46

⇒ (x + 2y + 140)/ 200 = 1.46

⇒ x + 2y = 292 – 140

⇒ x + 2y = 152 —- (ii)

Subtract equation (i) from equation (ii), we get

x + 2y – x – y = 152 – 114

⇒ y = 38

Now, on putting the value of y in equation (i), we find x = 114 – 38 = 76

Therefore, the missing frequencies are 76 and 38 .

Statistics

## Solution.

(i) 3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4

Thus , the mode = 5 since it occurs the maximum number of times.

(ii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15

Thus, the mode = 15 since it occurs the maximum number of times.

## Solution:

From the data its observed that, Model shirt size = 40 since it was the size which occurred for the maximum number of times.

## Solution.

It’s seen that the maximum frequency is 50. So, the corresponding class i.e., 35 – 40 is the modal class. And,

l = 35, h = 40 – 35 = 5, f = 50, f1 = 34, f= 42

So, we use the following formula:

= 38.33