Question

The sum of deviations of a certain number of item measured from 2.5 is 50 and 3.5 is -50 find N and X​

Answer 1

Answer: No. Of items = 100 and arithmetic mean = 3

Explanation:

2.5 and 3.5 given in question are assumed means & 50 and -50 are their respective sum of deviations

Answer 2

Answer:

The value of number of items (N) will be equal to 100.

The value of arithmetic mean (X) is 3.

Explanation:

We have given

$A_{1}=2.5\\A_{2}=3.5$             $\sum dX_{1}=50\\\sum dX_{2}=-50$

We know  $NX-NA=\sum dX$

So,  $NX-N(2.5)=50$                                     ...............(1)

$NX-N(3.5)=-50$                                          ................(2)

Subtract eq.(1) from eq.(2),

$NX-N(2.5)-NX+N(3.5)=50-(-50)$

$3.5N-2.5N=100$

$N=100$

Therefore the number of items, $N=100$

Put the value of n in eq.(1),

$X(100)-(100)(2.5)=50\\100X=50+250\\100X=300\\X=3$

Therefore arithmetic mean, $X=3$