# Maths Quiz on Statistics for IIT JEE & Engineering Exam

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## JEE Maths Quiz on Statistics

JEE Maths Quiz on Statistics : In this article you will get to Online test for JEE Main, JEE Advanced, UPSEE, WBJEE and other engineering entrance examinations that will help the students in their preparation. These tests are free of cost and will useful in performance and inculcating knowledge. In this post we are providing you Maths Quiz on Statistics.

## JEE Maths Quiz on Statistics

Q1. Mean of 100 observations is 45. It was later found that two observations 19 and 31 were incorrectly recorded as 91 and 13. The correct mean is _________.
a) 45.49
b) 45.44
c) 44.46
d) 44.84

c) 44.46

Solution:

Sum of 100 items = 45 × 100 = 4500

Sum of items added = 19 + 31 = 50

Sum of items replaced = 91 + 13 = 104

New sum = 4500 − 104 + 50 = 4446

New mean= 4446 / 100 = 44.46

Q2. The mean and S.D. of the marks of 200 candidates were found to be 40 and 15 respectively. Later, it was discovered that a score of 40 was wrongly read as 50. The correct mean and S.D. respectively are ______________.
a) 14.98
b) 15.45
c) 13.48
d) 14.59

a) 14.98

Solution:

Corrected Σx = 40 × 200 − 50 + 40 = 7990

Corrected x bar = 7990 / 200 = 39.95

Incorrect Σx2 = n [σ+ (\bar{x}^2) = 200 [152 + 402] = 365000

Correct Σx= 365000 − 2500 + 1600 = 364100

Corrected σ =√(364100/200) − (39.95)2

= √(1820.5 − 1596)

= √224.5

= 14.98

Q3. If a variable takes the discrete values α − 4, α − 7 / 2, α − 5 / 2, α − 3, α − 2, α + 1 / 2, α − 1 / 2, α + 5 (α > 0), then the median is ____________.
a) α − 4 / 5
b) α + 5 / 4
c) α − 5 / 4
d) α + 4 / 5

c) α − 5 / 4

Solution:

Arrange the data as α − 7/2, α − 3, α − 5/2, α − 2, α − 1/2, α + 1/2, α – 4, α + 5

Median = [1 / 2] [value of 4th item + value of 5th item ]

Median = [(α − 2) + (α − 1 / 2)] / 2

= [2α − 5 / 2] / 2

= α − 5 / 4

Q4.  The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observations of the set is increased by 2, then the median of the new set is _______.
a) 21.5
b) 20.25
c) 20
d) No Change

d) No Change

Solution:

Since n = 9, then median term = ([9 + 1] / 2)th = 5th term.

Now, the last four observations have increased by 2.

The median is 5th observation, which is remaining unchanged.

There will be no change in the median.

Q5. The number of observations in a group is 40. If the average of the first 10 is 4.5 and that of the remaining 30 is 3.5, then the average of the whole group is _________.
a) 12 / 5
b) 17 / 6
c) 15 / 4
d) 16 /5

c) 15 / 4

Solution:

[x1 + x2 + ….. + x10] /  = 4.5

[x1 + x2 + ….. + x10] = 45 and [x11 + x12 +….. +x40] /  = 3.5

x11 + x12 +….. + x40 = 105

x1 + x2 + ….. + x40 = 150

[x1 + x2 + ….. + x40] /  = 150 / 40 = 15 / 4.

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