Important Questions for CBSE Class 11 Maths Chapter 15  Statistics
CBSE Class 11 Maths Chapter15 Important Questions  Free PDF Download
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
=213
= 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 13^{th} and 14^{th} 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
= 1134
=9
8. Find the mean of the following data 3,6,11,12,18
Ans. Mean =
9. Express in the form of a + ib (3i7) + (74i) – (6+3i) + i^{23}
Ans. Let
Z =
10. Find the conjugate of
Ans.
11. Solve for x and y, 3x + (2xy) 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+i^{2} + i^{4} + i^{6} + i^{8} +  + i^{20}
Ans.
13. Multiply 32i 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, w^{2} are three cube root of unity, show that (1 – w + w^{2}) (1 + w – w^{2}) = 4
Ans. (1 – w + w^{2}) (1 + w – w^{2})
(1 + w^{2}  w) (1 + w – w^{2})
18. Find that sum product of the complex number
Ans.
19. Write the real and imaginary part 1 – 2i^{2}
Ans. Let z = 1 – 2i^{2}
=1 – 2 (1)
= 1 + 2
= 3
= 3 + 0.i
Re (z) = 3, Im (z) = 0
20. If two complex number z_{1}, z_{2 }are such that z_{1} = z_{2}, is it then necessary that z_{1} = z_{2 }
Ans. Let z_{1} = a + ib
21. Find the conjugate and modulus of
Ans. Let
22. Find the number of non zero integral solution of the equation 1i^{x} = 2^{x}
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  xiM  xiM 
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 
Media’s deviation about median
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 
Mean deviation about the mean
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:  4^{th}, 5^{th} and 6^{th} 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
Expenditure0100100200200300300400400500500600600700700800
persons 489107543
Ans.
Expenditure  No. of persons  Mid point  
0100  4  50  200  308  1232 
100200  8  150  1200  208  1664 
200300  9  250  2250  108  972 
300400  10  350  3500  8  80 
400500  7  450  3150  92  644 
500600  5  550  2750  192  960 
600700  4  650  2600  292  1168 
700800  3  750  2250  392  1176 
 50  17900  7896 
mean =
8.Find the mean deviation about the median for the following data
Marks 01010202030304040505060
No. of boys 810101642
Ans.
Marks  No. of boys  Cumulative Frequency  Mid points  
010  8  8  5  22  176 
1020  10  18  15  12  120 
2030  10  28  25  2  20 
3040  16  44  35  8  128 
4050  4  48  45  18  72 
5060  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 06612121818242430
Frequency8101295
Ans.
Class  Mid value  Frequency  
06  3  8  8  11  88 
612  9  10  18  5  50 
1218  15  12  30  1  12 
1824  21  9  39  7  63 
2130  27  5  44  13  65 

1218 is the medias class
Medias =
Medias
Mean deviation about median =
11.Calculate the mean deviation from the median from the following data
Salary per week(in Rs) 102020303040405050606070
no. of workers 461020106
Ans.
Salary per Week (in Rs)  Mid value  Frequency  
1020  15  4  4  30  120 
2030  25  6  10  20  120 
3040  35  10  20  10  100 
4050  45  20  40  0  0 
5060  55  10  50  10  100 
6070  65  6  56  20  120 
7080  75  4  60  30  120 

4050 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.i^{25} + (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) y^{2} + (6+i) and (2+i) x
Ans.(1+i) y^{2} + (6 + i) = (2 + i) x
y^{2} + iy^{2} + 6 + i = 2x + xi
(y^{2} + 6) + (y^{2} + 1) i = 2x + xi
y^{2} + 6 = 2x
y^{2} + 1 = x
y ^{2} = x – 1
x – 1 + 6 = 2x
5 = x
21.If x + iy =
Ans.
taking conjugate both side
x^{2} + y^{2} = 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 – 5yi^{2} =  6 + 24i
24.If
Ans. If
6 Marks Questions
1.Calculate the mean, variance and standard deviation of the following data:
Classes  3040  4050  5060  6070  7080  8090  90100 
Frequency  3  7  12  15  8  3  2 
Ans.
Classes  Frequency  Mid Point  
3040  3  35  105  729  2187 
4050  7  45  315  289  2023 
5060  12  55  660  49  588 
6070  15  65  975  9  135 
7080  8  75  600  169  1352 
8090  3  85  255  529  1587 
90100  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 = 400050+40
= 3990
So correct mean =
Also
Using incorrect values,
= 162601
= incorrect
= correct
= 1626012500+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)
= 800034+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.