Mean = (

Here

Hence mean deviation about mean =

=  =

total students = 1250

No. of students like football =

No. of students who like Cricket =

And, No. of students who like Badminton =

No. of students who like only Football =

No. of students who like only Cricket =

No. of students who like only Badminton =

Suppose k students like all sports, x only footballl and criket, y only football and badminton and z only cricket and badminton

 ...(i)

Similarly,  ...(ii)

And  ...(iii)

And,

Or ....(iv)

After solving we get the value of x= 60, y= 90, z= 80 and k = 110

We have to find number of students like at least game

Two games are

= all students – students who likes all games

= 1250 – 520 = 730

Required number of percentages =

S.D.=

Consider

GM =

GM =

GM =  =

We know that if we subtract a constant value to every term of given data then mean of data is also reduced by that constant value.

We know that if we divide every term of given data by a constant value then mean of data is also divided by that constant value.

Total number of observation = 200

Mean = 60

If 10 is subtracted from each observation then Mean = 60-10 = 50

After that each observation is divided by 5

Hence New mean =

We know that if we subtract a constant value to every term of given data then standard deviation of data remains unchanged.

Also, if we divide every term of given data by a constant value then standard deviation of data is also divided by that constant value.

Total number of observation = 200

Standard deviation = 12

If 10 is subtracted from each observation then Standard deviation = 12

After that each observation is divided by 5

Hence New mean =

Required Sum = 10 + 2.4 = 12.4