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)
V Vakul

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.

Preparation Products

Knockout BITSAT 2020

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 1999/-
Buy Now
Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout BITSAT-JEE Main 2021

An exhaustive E-learning program for the complete preparation of JEE Main and Bitsat.

₹ 27999/- ₹ 16999/-
Buy Now
Knockout BITSAT-JEE Main 2020

An exhaustive E-learning program for the complete preparation of JEE Main and Bitsat.

₹ 14999/- ₹ 7999/-
Buy Now
Exams
Articles
Questions