Join NEET Medical Exam Telegram Channel  Click Here 
Join IIT JEE Engg. Maths Telegram Channel  Click Here 
Join Physics Chemistry Biology Channel  Click Here 
Join Board & Class 11th & 12th Channel  Click Here 
Join Our Quiz for All Exam Channel  Click Here 
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 Σx^{2} = n [σ^{2 }+ ($xˉ_{2}$) = 200 [152 + 402] = 365000
Correct Σx^{2 }= 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 4^{th} item + value of 5^{th} 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} = 5^{th} term.
Now, the last four observations have increased by 2.
The median is 5^{th} 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:
[x_{1} + x_{2} + ….. + x_{10}] / [10] = 4.5
[x_{1} + x_{2} + ….. + x_{10}] = 45 and [x_{11} + x_{12} +….. +x_{40}] / [30] = 3.5
x_{11 }+ x_{12 }+….. + x40 = 105
x_{1} + x_{2} + ….. + x_{40} = 150
[x_{1} + x_{2} + ….. + x_{40}] / [40] = 150 / 40 = 15 / 4.
Subjectwise Tricks Tips & Question with Solution PDFs
S.NO  Subject Name  Topicwise PDFs Download Link 
1.  Chemistry Notes PDF  Click Here to Download Now 
2.  Maths Notes PDF  Click Here to Download Now 
3.  Physics Notes PDF  Click Here to Download Now 
4.  Biology Notes PDF  Click Here to Download Now 
Syllabus and Previous Year Papers 

Chemistry Syllabus for NEET & AIIMS Exams  Click Here 
Chemistry Syllabus for JEE Mains & Advanced  Click Here 
Chapter Wise NEET Chemistry Syllabus  Click Here 
Physics Syllabus for NEET & AIIMS Exams  Click Here 
Physics Syllabus for JEE Mains & Advanced  Click Here 
Chapter Wise NEET Physics Syllabus  Click Here 
Biology Syllabus for NEET & AIIMS Exams  Click Here 
Chapter Wise NEET Biology Syllabus  Click Here 
Maths Syllabus for JEE Mains & Advanced  Click Here 
Download NEET Previous Year Question Papers with Solution  Click Here 


https://www.facebook.com/ExamsRoadOfficial  
Telegram  https://telegram.me/ExamsRoad 
https://twitter.com/ExamsRoad  
https://www.instagram.com/ExamsRoad/  
YouTube  Click Here To Subscribe Now 
Thank You.
By TEAM ExamsRoad.com