Question

FREE 100 POINTS❤

stats ch15 pg15.18 eg15.1.18
plz explain in detail step by step.(plz. send ✍hand written ans.)​

Answer 1

Step-by-step explanation:

x                w

1                 1²

2                 2²

3                 3²

.                   .

.                    .

n                   n²

$\therefore Weighted \ Mean = \frac{w_{1}x_{1} + w_{2}x_{2} + .. +w_{n}x_{n}}{w_{1} + w_{2}+..+w_{n}}$

$\Longrightarrow \frac{1.1^2 + 2.2^2+...+n.n^2}{1^2+2^2+..+n^2}$

$\Longrightarrow \frac{1^2+2^3+..+n^3}{1^2+2^2+...+n^2}$

$\Longrightarrow \frac{\frac{n^2(n+1)^2}{4}}{\frac{n(n+1)(2n+1)}{6}}$

$\Longrightarrow \frac{\frac{n.n.(n+1).(n+1)}{2}}{\frac{n(n+1)(2n+1)}{3}}$

$\Longrightarrow \frac{\frac{n(n+1)}{2}}{\frac{2n+1}{3}}$

$\Longrightarrow {\boxed{\frac{3n(n+1)}{2(2n+1)}}$

Hope it helps

Answer 2

Answer:

We have to find the weighted A.M and H.M where weights are equal to the square of the first natural numbers .

The formula of weighted mean is :

$\mathtt{\dfrac{Sum\:of\:products\:of\:weights\:and\:original\:set\:of\:numbers}{Sum\:of\;weights}}$

OR :

$\mathsf{\dfrac{w_1x_1+w_2x_2+...}{w_1+w_2+...}}$

$x_1 ,x_2,x_3=1,2,3...and\:so\:on$

$w_1,w_2,w_3=1,4,9....and\:so\:on$

The given weights are :

1² , 2² , 3² .......... n

$\mathsf{Weighted\:mean=\dfrac{1.1^2+2.2^2+...n.n^2}{1^2+2^2..n^2}}\\\\\implies \dfrac{\Sigma n^3}{\Sigma n^2}$

Sum of cubes of number is given by [ n² ( n + 1 )² ] / 4

Sum of square of natural number is n ( n + 1 )( 2 n + 1 ) / 6

$\mathsf{Weighted\:A.M=\dfrac{\dfrac{n^2(n+1)^2}{4}}{\dfrac{n(n+1)(2n+1)}{6}}}\\\\\implies \mathsf{\dfrac{6n(n+1)}{4(2n+1)}}\\\\\implies \boxed{\mathsf{\dfrac{3n(n+1)}{2(2n+1)}}}$