 |
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 _________. 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
a) 45.49
b) 45.44
c) 44.46
d) 44.84
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 Solution: Corrected Σx = 40 × 200 − 50 + 40 = 7990 Corrected x bar = 7990 / 200 = 39.95 Incorrect Σx2 = n [σ2 + (\bar{x}^2) = 200 [152 + 402] = 365000 Correct Σx2 = 365000 − 2500 + 1600 = 364100 Corrected σ =√(364100/200) − (39.95)2 = √(1820.5 − 1596) = √224.5 = 14.98
a) 14.98
b) 15.45
c) 13.48
d) 14.59
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 ____________. 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
a) α − 4 / 5
b) α + 5 / 4
c) α − 5 / 4
d) α + 4 / 5
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 _______. 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.
a) 21.5
b) 20.25
c) 20
d) No Change
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 _________. c) 15 / 4 Solution: [x1 + x2 + ….. + x10] / [10] = 4.5 [x1 + x2 + ….. + x10] = 45 and [x11 + x12 +….. +x40] / [30] = 3.5 x11 + x12 +….. + x40 = 105 x1 + x2 + ….. + x40 = 150 [x1 + x2 + ….. + x40] / [40] = 150 / 40 = 15 / 4.
a) 12 / 5
b) 17 / 6
c) 15 / 4
d) 16 /5
Subject-wise Tricks Tips & Question with Solution PDFs
| S.NO | Subject Name | Topic-wise 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
