# Important Questions for CBSE Class 11 Maths Chapter 15 - Statistics

## CBSE Class 11 Maths Chapter-15 Important Questions - Free PDF Download

Free PDF download of Important Questions with solutions for CBSE Class 11 Maths Chapter 15 - Statistics prepared by expert Maths teachers from latest edition of CBSE(NCERT) books. CoolGyan.Org to score more marks in your Examination.

1 Marks Questions

1. In a test with a maximum marks 25, eleven students scored 3,9,5,3,12,10,17,4,7,19,21 marks respectively. Calculate the range.

Ans. The marks can be arranged in ascending order as 3,3,4,5,7,9,10,12,17,19,21.

Range = maximum value – minimum value

=21-3

= 18

2. Coefficient of variation of two distributions is 70 and 75, and their standard deviations are 28 and 27 respectively what are their arithmetic mean?

Ans. Given C.V (first distribution) = 70

Standard deviation = = 28

C.V

=

Similarly for second distribution

C.V

3. Write the formula for mean deviation.

Ans.MD

4. Write the formula for variance

Ans. Variance

5. Find the median for the following data.

579101215

862226

Ans.

 5 7 9 10 12 15 8 6 2 2 2 6 8 14 16 18 20 26

Median is the average of 13th and 14th item, both of which lie in the c.f 14

6. Write the formula of mean deviation about the median

Ans.

7. Find the rang of the following series 6,7,10,12,13,4,8,12

Ans. Range = maximum value – minimum value

= 113-4

=9

8. Find the mean of the following data 3,6,11,12,18

Ans. Mean =

9. Express in the form of a + ib (3i-7) + (7-4i) – (6+3i) + i23

Ans. Let

Z =

10. Find the conjugate of

Ans.

11. Solve for x and y, 3x + (2x-y) i= 6 – 3i

Ans.3x = 6

x = 2

2x – y = - 3

2 × 2 – y = - 3

- y = - 3 – 4

y = 7

12. Find the value of 1+i2 + i4 + i6 + i8 + ---- + i20

Ans.

13. Multiply 3-2i by its conjugate.

Ans. Let z = 3 – 2i

14. Find the multiplicative inverse 4 – 3i.

Ans. Let z = 4 – 3i

15. Express in term of a + ib

Ans.

16. Evaluate

Ans.

17. If 1, w, w2 are three cube root of unity, show that (1 – w + w2) (1 + w – w2) = 4

Ans. (1 – w + w2) (1 + w – w2)

(1 + w2 - w) (1 + w – w2)

18. Find that sum product of the complex number

Ans.

19. Write the real and imaginary part 1 – 2i2

Ans. Let z = 1 – 2i2

=1 – 2 (-1)

= 1 + 2

= 3

= 3 + 0.i

Re (z) = 3, Im (z) = 0

20. If two complex number z1, z2 are such that |z1| = |z2|, is it then necessary that z1 = z2

Ans. Let z1 = a + ib

21. Find the conjugate and modulus of

Ans. Let

22. Find the number of non zero integral solution of the equation |1-i|x = 2x

Ans.

Which is false no value of x satisfies.

23. If (a + ib) (c + id) (e + if) (g + ih) = A + iB then show that

Ans.

4 Marks Questions

1.The mean of 2,7,4,6,8 and p is 7. Find the mean deviation about the median of these observations.

Ans.Observations are 2, 7, 4, 6, 8 and p which are 6 in numbers

The near of these observations is 7

Arrange the observations in ascending order 2,4,6,7,8,15

Medias (M) =

Calculation of mean deviation about Median.

 xi xi-M |xi-M| 2 -4.5 4.5 4 -2.5 2.5 6 -0.5 0.5 7 0.5 0.5 8 1.5 1.5 15 8.5 8.5 Total 18

2.Find the mean deviation about the mean for the following data!

1030507090

42428168

Ans. To calculate mean, we requirevalues then for mean deviation, we require || values and values.

 10 4 4 40 160 30 24 720 20 480 50 28 1400 0 0 70 16 1120 20 320 90 8 720 40 320 80 4000 1280

MD

3.Find the mean, standard deviation and variance of the first natural numbers.

Ans. The given numbers are 1, 2, 3, ……, n

Mean

Variance

Standard deviation

4.Find the mean variance and standard deviation for following data

Ans.

 4 8 11 17 20 24 32 3 5 9 5 4 3 1

Note: - 4th, 5th and 6th columns are filled in after calculating the mean.

 4 3 12 -10 100 300 8 5 40 -6 36 180 11 9 99 -3 9 81 17 5 85 3 9 45 20 4 80 6 36 144 24 3 72 10 100 300 32 1 32 18 324 324 Total 30 402 1374

Here

Mean

Variance

Standard deviation

= 6.77

5.The mean and standard deviation of 6 observations are 8 and 4 respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Ans. Let be the six given observations

Then

Also

As each observation is multiplied by 3, new observations are

New near =

Let be the new standard deviation, then

6.Prove that the standard deviation is independent of any change of origin, but is dependent on the change of scale.

Ans. Let us use the transformation to change the scale and origin

Now

Also

Both,are positive which shows that standard deviation is independent of choice of origin, but depends on the scale.

7.Calculate the mean deviation about the mean for the following data

Expenditure0-100100-200200-300300-400400-500500-600600-700700-800

persons 489107543

Ans.

 Expenditure No. of persons Mid point 0-100 4 50 200 308 1232 100-200 8 150 1200 208 1664 200-300 9 250 2250 108 972 300-400 10 350 3500 8 80 400-500 7 450 3150 92 644 500-600 5 550 2750 192 960 600-700 4 650 2600 292 1168 700-800 3 750 2250 392 1176 50 17900 7896

mean =

8.Find the mean deviation about the median for the following data

Marks 0-1010-2020-3030-4040-5050-60

No. of boys 810101642

Ans.

 Marks No. of boys Cumulative Frequency Mid points 0-10 8 8 5 22 176 10-20 10 18 15 12 120 20-30 10 28 25 2 20 30-40 16 44 35 8 128 40-50 4 48 45 18 72 50-60 2 50 55 28 56 total 50 572

which is the median class.

Median

= 27

9.An analysis of monthly wages point to workers in two firms A and B, belonging to the same industry, given the following result. Find mean deviation about median.

Firm AFirm B

No of wages earns586648

Average monthly wagesRs 5253Rs 5253

Ans.For firm A, number of workers = 586

Average monthly wage is Rs 5253

Total wages = Rs 5253586

= Rs 3078258

For firm B, total wages = Rs 253648

=Rs 3403944

Hence firm B pays out amount of monthly wages.

10.Find the mean deviation about the median of the following frequency distribution

Class 0-66-1212-1818-2424-30

Frequency8101295

Ans.

 Class Mid value Frequency 0-6 3 8 8 11 88 6-12 9 10 18 5 50 12-18 15 12 30 1 12 18-24 21 9 39 7 63 21-30 27 5 44 13 65

12-18 is the medias class

Medias =

Medias

11.Calculate the mean deviation from the median from the following data

Salary per week(in Rs) 10-2020-3030-4040-5050-6060-70

no. of workers 461020106

Ans.

 Salary per Week (in Rs) Mid value Frequency 10-20 15 4 4 30 120 20-30 25 6 10 20 120 30-40 35 10 20 10 100 40-50 45 20 40 0 0 50-60 55 10 50 10 100 60-70 65 6 56 20 120 70-80 75 4 60 30 120

40-50 is the median class

Medias =

Mean deviation = =

12.Let values of a variable Y and let ‘a’ be a non zero real number. Then prove that the variance of the observations is also, find their standard deviation.

Ans.Let value of variables such that then

13.If

Ans.

Taking conjugate both side

14.If

Ans.

15.Solve

Ans.

16.Find the modulus

Ans.i25 + (1+3i)3

17.If

Ans. (i) (Given)

(ii) [taking conjugate both side

(i) × (ii)

18.Evaluate

Ans.

19.Find that modulus and argument

Ans.

20.For what real value of x and y are numbers equal (1+i) y2 + (6+i) and (2+i) x

Ans.(1+i) y2 + (6 + i) = (2 + i) x

y2 + iy2 + 6 + i = 2x + xi

(y2 + 6) + (y2 + 1) i = 2x + xi

y2 + 6 = 2x

y2 + 1 = x

y 2 = x – 1

x – 1 + 6 = 2x

5 = x

21.If x + iy =

Ans.

taking conjugate both side

x2 + y2 = 1

Proved.

22.Convert in the polar form

Ans.

23.Find the real values of x and y if (x - iy) (3 + 5i) is the conjugate of – 6 – 24i

Ans.

(x – iy) (3 + 5i) = - 6 + 24i

3x + 5xi – 3yi – 5yi2 = - 6 + 24i

24.If

Ans. If

6 Marks Questions

1.Calculate the mean, variance and standard deviation of the following data:

 Classes 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Frequency 3 7 12 15 8 3 2

Ans.

 Classes Frequency Mid Point 30-40 3 35 105 729 2187 40-50 7 45 315 289 2023 50-60 12 55 660 49 588 60-70 15 65 975 9 135 70-80 8 75 600 169 1352 80-90 3 85 255 529 1587 90-100 2 95 190 1089 2178 Total 50 3100 10050

Here

Mean

Variance

Standard deviation

2.The mean and the standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who mistook one observation as 50 instead of 40. What are the correct mean and standard deviation?

Ans. Given that

Incorrect mean

Incorrect S.D

As

= incorrect sum of observation =4000

= correct sum of observations = 4000-50+40

= 3990

So correct mean =

Also

Using incorrect values,

= 162601

= incorrect

= correct

= 162601-2500+1600=161701

Correct

Hence, correct mean is 39.9 and correct standard deviation is 5.

3.200 candidates the mean and standard deviation was found to be 10 and 15 respectively. After that if was found that the scale 43 was misread as 34. Find the correct mean and correct S.D

Ans.

Corrected = Incorrect (sum of incorrect +sum of correct value)

= 8000-34+43= 8009

Corrected mean =

Incorrect

Corrected (incorrect) – (sum of squares of incorrect values) + (sum of square of correct values)

=

Corrected =

4.Find the mean deviation from the mean 6,7,10,12,13,4,8,20

Ans.Let be the mean

 6 4 7 3 10 0 12 2 13 3 4 6 8 2 20 10 Total = 30

= 30 and n = 8

5.Find two numbers such that their sum is 6 and the product is 14.

Ans.Let x and y be the no.

x + y = 6

xy = 14

6.Convert into polar form

Ans.

7.If α and β are different complex number with |β| = 1 Then find

Ans.