Question

mean of four observations is 12 .if first,second, and third,and fourth aberration are decreased by 1,2,3and4 respectively. then find the new mean

Answer 1

A1 +a2+ a3 +a4/4=12
A1 a2 a3 a4 -1 -2-3-4=48-10/4=38/4
=9 approx

Answer 2

Answer:

9.5

Step-by-step explanation:

Let the four observations be a,b,c and d.

Now we are given that mean of four observations is 12

$Mean =\frac{\text{Sum of observations}}{\text{no. of observations}}$

So,  $Mean =\frac{\text{Sum of 4 observations}}{4}$

$12 =\frac{a+b+c+d}{4}$

$48=a+b+c+d$

Now we are given that first,second, and third,and fourth aberration are decreased by 1,2,3 and 4 respectively.

So, a-1+b-2+c-3+d-4 = a+b+c+d-10 =48-10=38

So, new sum =38

So,  $\text{New Mean} =\frac{38}{4}$

$\text{New Mean} =9.5$

Hence the new mean is 9.5.