NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE

JEE: NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE

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SHORT ANSWER TYPE QUESTION

Q.1. Find the mean deviation about the mean of the distribution:
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Ans.
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
MeanNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Mean deviation MD =NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Here, the required MD = 1.25

Q.2. Find the mean deviation about the median of the following distribution:
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Ans.
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
HereNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Median = NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
∴ Median = 12
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the required MD = 1.25

Q.3. Calculate the mean deviation about the mean of the set of first n natural numbers when n is an odd number.
Ans.
First n natural numbers are 1, 2, 3, ..., n. Here, n is odd.
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
The deviations of numbers from mean NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEEare
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
The absolute values of deviation from the mean i.e.NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEEare
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
The sum of absolute values of deviations from the mean i.e.
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
∴ Mean deviation about the mean
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE

Q.4. Calculate the mean deviation about the mean of the set of first n natural numbers when n is an even number.
Ans. First n natural numbers are 1, 2, 3, 4, 5, 6, …, n (even)
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
[∴ Sum of first odd n natural numbers = n2]
Hence, the required MD = n/4.

Q.5. Find the standard deviation of the first n natural numbers.
Ans.
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the required SD =NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE

Q.6. The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results: 
Number of observations = 25, mean = 18.2 seconds, standard deviation = 3.25 seconds.
Further, another set of 15 observations x1, x2, ..., x15, also in seconds, is now available and we have NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEECalculate the standard derivation based on all 40  observations.
Ans. Given that n1 = 25, NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
andNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
For the first set, we have
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
⇒ 10.5625 + 331.24 =NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
For the combined standard deviation of the 40 observation, n = 40
andNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the required SD = 3.87

Q.7. The mean and standard deviation of a set of n1 observations are NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE and s1, respectively while the mean and standard deviation of another set of n2 observations areNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEEand s2, respectively. Show that the standard deviation of the combined set of (n1 + n2) observations is given by
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Ans.
Let xi, i = 1, 2, 3, 4, …, n1 
and yj, j = 1, 2, 3, 4, …, n2
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Now mean of the combined series is given by
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Therefore, σ2 of the combined series is
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
[∴ The algebraic sum of the deviation of values of first series from their mean is zero]
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Similarly, we have
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Now combined Standard Deviation (SD)
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE

Q.8. Two sets each of 20 observations, have the same standard derivation 5. The first set has a mean 17 and the second a mean 22. Determine the standard deviation of the set obtained by combining the given two sets.

Ans.
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Now we know for combined two series that
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the required SD = 5.59

Q.9. The frequency distribution:
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
where A is a positive integer, has a variance of 160. Determine the value of A.
Ans.
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the value of A is 7.

Q.10. For the frequency distribution:
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Find the standard distribution.
Ans.
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the required SD = 1.37

Q.11. There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test:

NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
where x is a positive integer. Determine the mean and standard deviation of the marks.
Ans. Given thatNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
∴ x – 2 + x + x2 + (x + 1)+ 2x + (x + 1) = 60
⇒ 4x – 2 + x2 + x2 + 2x + 1 + x + 1 = 60
⇒ 2x2 + 7x – 60 = 0
⇒ 2x2 + 15x – 8x – 60 = 0
⇒ x(2x + 15) – 4(2x + 15) = 0
⇒ (2x + 15) (x  4)    =    0     
⇒ 2x + 15 = 0
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
∴ x = 4 [∴ x ∈ I+]
Now put x = 4 in the frequency distribution table
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Let assumed mean A = 3
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
andNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the required mean = 2.8 and SD = 1.12

Q.12. The mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours. Find the overall standard deviation.

Ans.
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
we know that for a combined series.
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the required SD = 8.9

Q.13. Mean and standard deviation of 100 items are 50 and 4, respectively. Find the sum of all the item and the sum of the squares of the items.

Ans.
Given thatNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
and varianceNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the required sum are 5000 and 251600.

Q.14. If for a distribution ∑(x − 5) = 3, ∑(x − 5)2 = 43 and the total number of item is 18, find the mean and standard deviation.
Ans.
Given that NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the required mean is 5.17 and SD = 1.54

Q.15. Find the mean and variance of the frequency distribution given below:
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Ans.
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Mean =NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
VarianceNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the required mean = 4.15 and variance = 4.04

LONG ANSWER TYPE QUESTION

Q.16. Calculate the mean deviation about the mean for the following frequency distribution:
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Ans.
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Mean =NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
and Mean deviation =NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the required MD = 3.84

Q.17. Calculate the mean deviation from the median of the following data:
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Ans.
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Median class =NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE th term 10th term i.e.12 – 18
∴ Median =NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
andNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the required MD = 7

Q.18. Determine the mean and standard deviation for the following distribution:
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Ans.
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
MeanNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Here, the required mean = 6 and MD = 2.85

Q.19. The weights of coffee in 70 jars is shown in the following table:
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Determine variance and standard deviation of the above distribution.
Ans.
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
∴ Variance =NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the required variance = 1.165 and SD = 1.08 g

Q.20. Determine mean and standard deviation of first n terms of an A.P. whose first term is a and common difference is d.
Ans.
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
We know thatNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the required SD = NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE

Q.21. Following are the marks obtained, out of 100, by two students Ravi and Hashina in 10 tests.

NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
who is more intelligent and who  is more consistent?
Ans.
Case I: For Ravi
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
andNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Now for Hashina
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Assumed mean A = 55
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, Hashina is more consistent and intelligent.

Q.22. Mean and standard deviation of 100 observations were found to be 40 and 10, respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.
Ans.
Given that n = 100,NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Corrected NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
and Corrected mean =NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NowNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the required SD = 10.24.

Q.23. While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find  the correct mean and the variance.
Ans. Given that n = 10,NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
CorrectedNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
= 423
∴ Correct MeanNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
andNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
and corrected variance
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
= 1833.1 – 1789.3 = 43.8
Hence the required mean = 42.3 and variance = 43.8

OBJECTIVE ANSWERS TYPE QUESTIONS

Q.24. The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is 
(a) 2 
(b) 2.57 
(c) 3 
(d) 3.75
Ans. (b)
Solution.
Observations are given by 3, 10, 10, 4, 7, 10 and 5
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the correct option is (b).

Q.25. Mean deviation for n observations x1, x2, ..., xn from their meanNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
(a) NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
(b) NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
(c) NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
(d) NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Ans. (b)
Solution.
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the correct option is (b).

Q.26. When tested the lines (in hours) of 5 bulbs were noted as follows: 
1357, 1090, 1666, 1494, 1623 
The mean deviation (in hours) from their mean is 
(a) 178 
(b) 179 
(c) 220 
(d) 356
Ans. (a)
Solution.
The lines of 5 bulbs are given by
1357, 1090, 1666, 1494, 1623
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the correct option is (a).

Q.27. Following are the marks obtained by 9 students in a Mathematics test
50, 69, 20, 33, 53, 39, 40, 65, 59.

The mean deviation from the median is
(a) 9
(b) 10.5
(c) 12.67
(d) 14.76

Ans. (c)
Solution.
Marks obtained are 50, 69, 20, 33, 53, 39, 40, 65 and 59
Let us write in ascending order
20, 33, 39, 40, 50, 53, 59, 65, 69.
Here n = 9
∴ Median =NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEEth term = 5th term i.e. 50
∴ Median = 50
Now
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the correct option is (c).

Q.28. The standard deviation of the data 6, 5, 9, 13, 12, 8, 10 is
(a)NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
(b) 52/7
(c) √6 
(d) 6
Ans. (a)
Solution.

Given data are 6, 5, 9, 13, 12, 8 and 10
∴ n = 7
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the correct option is (a).

Q.29. Let x1, x2, ..., xn be n observations andNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEEbe their arithmetic mean. The formula for the standard deviation is given by

(a)NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
(b)NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
(c)NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
(d)NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Ans. (c)
Solution.

The formula for S.D =NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the correct option is (c).

Q.30. If the mean of 100 observations is 50 and their standard deviation is 5, then the sum of all the squares of all the observations is 

(a) 50,000 
(b) 250000 
(c) 252500 
(d) 255000
Ans. (c)
Solution.

HereNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the correct option is (c).

Q.31. 
If a, b, c, d and e be the observations with mean m and standard deviations S, then find the standard deviation of the observations a + K, b + K, c + K, d + K and e + K is
(a) S
(b) KS
(c) S + K
(d) S/K

Ans. (a)
Solution.

Given observation are a, b, c, d and e
∴ Mean =NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Now mean of a + K, b + K, c + K, d + K and e + K is
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the correct option is (a).

Q.32. If x1, x2, x3, x4 and x5 be the observations with mean m and standard deviations S then, the standard deviation of the observations Kx1, Kx2, Kx3, Kx4 and Kx5 is 
(a) K + S 
(b) S/K 
(c) KS 
(d) S
Ans. (c)
Solution.

HereNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Here, the correct option is (c).

Q.33. Let x1, x2, x3, …, xn be n observations. Let wi = lxi + k for i = 1, 2, …, n where l and k are constants. If the mean of xi’s is 48 and their standard deviation is 12, the mean of wi’s is 55 and standard deviations of wi’s is 15, then the values of l and k should be 
(a) l = 1.25, k = – 5 
(b) l = – 1.25, k = 5 
(c) l = 2.5, k = – 5 
(d) l = 2.5, k = 5
Ans. (a)
Solution.

Given that wi = xi + k,NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
wi = 55 and SD(wi) = 15
thenNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
⇒ 55 = 48 + k (i)
SD of wi = SD of xi 
15 = l x 12
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE(ii)
from eq. (i) and (ii) we have
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE= 55 – 1.25 x 48 = 55 - 60 = 5
Here, the correct option is (a).

Q.34. The standard deviations for first 10 natural numbers is 

(a) 5.5 
(b) 3.87 
(c) 2.97 
(d) 2.87
Ans. (d)
Solution.

We know that SD of first n natural numbersNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Here n = 10
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the correct option is (d).

Q.35. Consider the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. If 1 is added to each numbers, the variance of the numbers, so obtained is 
(a) 6.5 
(b) 2.87 
(c) 3.87 
(d) 8.25
Ans. (d)
Solution.

Given numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Numbers obtained when 1 is added to the above numbers is
2, 3, 4, 5, 6, 7, 8, 9, 10 and 11.
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE= 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
= 5[4 + 9] = 5 x 13 = 65
NowNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE= 22 + 32 + 42 + ... + 112 
= (12 + 22 + 32 + 42 + ... + 112) - (1)2
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
= 22 x 23 - 1 = 506 - 1 = 505
∴ VarianceNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
= 50.5 – 42.25 = 8.25
Hence, the correct option is (d).

Q.36. Consider the first 10 positive integers. If we multiply each number by – 1 and then add 1 to each number, the variance of the numbers, so obtained is (a) 8.25 

(b) 6.5 
(c) 3.85 
(d) 2.87
Ans. (a)
Solution.

First 10 positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
on multiplying each number by – 1, we get
– 1, – 2, – 3, – 4, – 5, – 6, – 7, – 8, – 9, – 10

on adding 1 to each of the number, we get
0, – 1, – 2, – 3, – 4, – 5, – 6, – 7, – 8, – 9
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE= 0 – 1 – 2 – 3 – 4 – 5 – 6 – 7 – 8 – 9
= – 45
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
∴ Variance = (SD)2 =NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the correct option is (a).

Q.37. The following information relates to a sample of size 60 NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEEthen the variance is 
(a) 6.63 
(b) 16 
(c) 22 
(d) 44
Ans. (d)
Solution.

We know that varianceNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the correct option is (d).

Q.38. Coefficient of variation of two distributions are 50 and 60, and their arithmetic means are 30 and 25 respectively. Difference of their standard deviation is 
(a) 0 
(b) 1 
(c) 1.5 
(d) 2.5
Ans. (a)
Solution.

Here, we have CV1 = 50, CV2 = 60
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the correct option is (a).

Q.39. The standard deviation of some temperature data in °C is 5. If the data were converted into ºF, the variance would be 
(a) 81 
(b) 57 
(c) 36 
(d) 25
Ans. (a)
Solution.

Given that σC = 5
We know thatNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the correct option is (a)

FILL IN THE BLANKS

Q.40. Coefficient of variation =NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Ans.
NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the value of the filler is SD. 

Q.41. IfNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEEis the mean of n values of x, then NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEEis always equal to _______. If a has any value other than NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE _________ than ∑(xi - a)2.
Ans. IfNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEEis the mean of n observations of x, thenNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE and if ‘a’ has the value other thanNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE then NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEEis less thanNCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the value of the fillers are 0 and less.      

Q.42. If the variance of a data is 121, then the standard deviation of the data 
is _______.
Ans. We know that SD =NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE
Hence, the value of the filler is 11.

Q.43. The standard deviation of a data is ___________ of any change in origin, but is _____ on the change of scale.
Ans. Since the standard deviation of any data is independent of any change in origin but is dependent of any change of scale.     
Hence, the value of the fillers are independent and dependent.

Q.44. The sum of the squares of the deviations of the values of the variable is _______ when taken about their arithmetic mean.
Ans. The sum of the squares of the deviations of the value of variable is minimum when taken about their arithmetic mean.     
Hence, the value of the filler is minimum.

Q.45. The mean deviation of the data is _______ when measured from the median.
Ans. The mean deviation of the data is least when measured from the median.      Hence, the value of the filler is least.

Q.46. The standard deviation is _______ to the mean deviation taken from the arithmetic mean.
Ans. The standard deviations is greater than or equal to the mean deviation taken from the arithmetic mean.
Hence, the value of the filler is greater than or equal.    

The document NCERT Exemplar: Statistics Notes | Study Mathematics (Maths) Class 11 - JEE is a part of the JEE Course Mathematics (Maths) Class 11.
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