# R S Aggarwal Solutions for Class 11 Maths Chapter 30 Statistics Exercise 30C

Class 11 Chapter 30 – Statistics exercise 30C solutions are provided here. These solutions are outlined keeping in mind the current CBSE syllabus, hence possessing a great chance of scoring good marks in the exams. In this exercise, students will get a chance to solve miscellaneous word problems on the concept of statistics.

Class 11 Maths Chapter 30 Statistics Exercise 30C is based on the topic- Miscellaneous word problems. R S Aggarwal chapter 30 solutions for Class 11 are solved by the experts at BYJUâ€™S and they have provided correct answers to all the questions. Students can download their pdf now and start practising.

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### Access Answers to Maths R S Aggarwal Class 11 Chapter 30 Statistics Exercise 30C Page number 941

Question 1: If the standard deviation of the numbers 2, 3, 2x, 11 is 3.5, calculate the possible values of x.

Solution:

Standard Deviation (Ïƒ) = 3.5

Sum of observations: 2 + 3 + 2x + 11 = 16 + 2x

Total number of observations = 4

Mean = (16+2x)/4 = (8+x)/2

12.25 Ã— 16 = 280 â€“ 64x + 12x2

196 = 280 â€“ 64x + 12x2

12x2 â€“ 64x + 84 = 0

or 3x2 â€“ 16x + 21 = 0

or (3x â€“ 7)(x â€“ 3) = 0

=> Either 3x â€“ 7 = 0 or x â€“ 3 = 0

=>x = 7/3 or x = 3

Therefore, possible values of x are 3 and 7/3.

Question 2: The variance of 15 observations is 6. If each observation is increased by 8, find the variance of the resulting observations.

Solution: Let x1, x2, x3, x4,….., x15 are any random 15 observations.

Variance = 6 and n = 15 (Given)

We know that,

If each observation is increased by 8, then let new observations be y1, y2, y3, y4,â€¦â€¦â€¦, y15;

where yi = xi + 8 â€¦.â€¦(2)

Now, find the variance for new observations:

Mean of new observations,

Using equation (2) and (3) in equation (1), we get

Question 3: The variance of 20 observations is 5. If each observation is multiplied by 2. Find the variance of the resulting observations.

Solution:

Given: Variance = 5 and n = 20

Let x1, x2, x3, x4,….., x20 are any random 20 observations.

If each observation is multiplied by 2, then let new observations be y1, y2, y3, y4,â€¦â€¦â€¦, y20.

where yi = 2xi ……..(2)

Now, find the variance for new observations:

Mean of new observations,

Using equation (2) and (3) in equation (1), we get

Question 4: The mean and variance of five observations are 6 and 4 respectively. If three of these are 5, 7 and 9, find the other two observations.

Solution:

Mean of five observations = 6 and

Variance of five observations = 4

Let the other two observations be x and y, then new set of observations be 5, 7, 9, x and y

Total number of observations = 5

Sum of all the observations = 5 + 7 + 9 + x + y = 21 + x + y

We know, Mean = (Sum of all the observations) / (Total number of observations)

=> 6 = (21 + x + y)/5

=> 9 = x + y â€¦â€¦(1)

Also,

20 = 11 + (x2 + 36 â€“ 12x) + (y2 + 36 â€“ 12y)

9 = x2 + y2 + 72 â€“ 12(x + y)

x2 + y2 + 72 â€“ 12(9) â€“ 9 = 0

(using equation (1))

x2 + y2 + 63 â€“ 108 = 0

x2 + y2 â€“ 45 = 0

or x2 + y2 = 45 ….(2)

Form (1); x + y = 9

Squaring both sides,

(x + y) 2 = (9) 2

(x2 + y2) + 2xy = 81

45 + 2xy = 81 (using equation (2))

2xy = 81 â€“ 45

or xy = 18

or x = 18/y

(1)=> 18/y + y = 9

y2 â€“ 9y +18 = 0

(y â€“ 3)(y â€“ 6) = 0

Either (y â€“ 3) = 0 or (y â€“ 6) = 0

=> y = 3, 6

For y = 3

x = 18/3 = 6

and for y = 6

x = 18/6 = 3

Thus, remaining two observations are 3 and 6.

Question 5: The mean and variance of five observations are 4.4 and 8.24 respectively. If three of these are 1, 2 and 6, find the other two observations.

Solution:

Mean of five observations = 4.4 and

Variance of five observations = 8.24

Let the other two observations be x and y, then new set of observations be 1, 2, 6, x and y.

Total number of observations = 5

Sum of all the observations = 1 + 2 + 6 + x + y = 9 + x + y

We know, Mean = (Sum of all the observations) / (Total number of observations)

=> 4.4 = (9 + x + y)/5

=> 13 = x + y â€¦â€¦(1)

Also,

41.2 = 19.88 + (x2 + 19.36 â€“ 8.8x) + (y2 + 19.36 â€“ 8.8y)

21.32 = x2 + y2 + 38.72 â€“ 8.8(x + y)

x2 + y2 + 38.72 â€“ 8.8(13) â€“ 21.32 = 0

(using equation (1))

x2 + y2 â€“ 97 = 0 …(2)

Squaring equation (1) both the sides, we get

(x + y)2 = (13)2

x2 + y2 + 2xy = 169

97 + 2xy = 169

(using equation (2))

xy = 36

or x = 36/y

(1)=> 36/y + y = 13

y2 + 36 = 13y

y2 â€“ 13y + 36 = 0

(y â€“ 4)(y â€“ 9) = 0

Either (y â€“ 4) = 0 or (y â€“ 9) = 0

=> y = 4 or y = 9

For y = 4

x = 36/y = 36/4 = 12

For y = 9

x = 36/9 = 4

Thus, remaining two observations are 4 and 9.

## Access other exercise solutions of Class 11 Maths Chapter 30 Statistics

Exercise 30A Solutions

Exercise 30B Solutions

Exercise 30D Solutions