Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Try this! The mean and SD of the marks of 200 candidates were found to be 40 and 15.Laterit was discoveredthat a scoreof 40 was wrongread as 50.The Correct mean and SD are:

The mean and SD of the marks of 200 candidates were found to be 40 and 15.Later it was discovered that a score of 40 was wrong read as 50.The Correct mean and SD are:

  • Option 1)

    14.98,39.95

  • Option 2)

    39.95,14.98

  • Option 3)

    39.95,224.5

  • Option 4)

    None of these

 

Answers (1)

As we discussed in concept

Standard Deviation -

If x1, x2...xn are n observations then square root of the arithmetic mean of 

\sigma = \sqrt{\frac{\sum \left ( x_{i}-\bar{x} \right )^{2}}{n}}

\bar{}

- wherein

where \bar{x} is mean

 

 Total Sum:=:40x200=8000

Now new sum:=:8000-50+40=8000-10=7990

\therefore\:new mean\:=\:\frac{7990}{200}=\frac{79.9}{2}=39.95

So S.D = 14.98

as S.D = \sqrt{\frac{1}{n}\varepsilon x^{2}_{s}-(\bar{X})^{2}}


Option 1)

14.98,39.95

This option is incorrect.

Option 2)

39.95,14.98

This option is correct.

Option 3)

39.95,224.5

This option is incorrect.

Option 4)

None of these

This option is incorrect.

Posted by

Vakul

View full answer

Similar Questions

Trending Articles/News

anam-khan

BITSAT 2025 syllabus out; registration link active at bitsadmission.com
anam-khan

BITSAT 2025 registration link active at bitsadmission.com; exam schedule, eligibility criteria
anam-khan

BITSAT 2025 registration begins today for 10 BE, 6 MSc, BPharm courses; eligibility, fee

Latest Question