TCS Digital Elementary Statistics Quiz 1

A.M = \left ( \frac{9+64}{2} \right ) = 36.5

G.M = \sqrt{9\times 64} = (3\times 8) = 24

H.M = \left( \frac{2\times 9\times 64}{9 + 64} \right)= 15.78



Video solution:

\therefore = \frac{\Sigma (f\times x)}{\Sigma f} = \frac{1510}{40} = 37.75kg



Video solution:

we have:



A.M = A + \frac{\Sigma (f\times d)}{\Sigma f} = \left( 30+ \frac{230}{50} \right ) = 34.6



Video solution:

Weighted Mean = \left ( \frac{23\times 1 + 35\times 1 + 46\times 2 + 30\times 3 + 32\times 5}{1+1+2+3+5} \right ) = \frac{400}{12} = 33.3



Video solution:

The maximum frequency is 30. So, the Modal class is 15 - 20.

\therefore L_1 = 15, L_2 = 20, f = 30, f_1 = 25 \text{ and } f_2 = 16

Mode = L_1 + \frac{f - f_1}{2f - f_1 - f_2}\times (L_2 - L_1) = 15 + \frac{(30 - 25)}{(60 - 25 - 16)} = 16.3



Video solution:

Here n = 40 , So, \frac{n}{2} = 20

Now, 20 lies in class 25 - 30.

So, Median Class is 25 - 30

\therefore L_1 = 25, L_2 = 30, n = 40, C = 13 \& f = 9
\therefore \text{ Median } = L_1 + \frac{(L_2 - L_1)}{f}\times \left( \frac{n}{2} - C \right )
25 + \frac{(30 - 25)}{9}\times (20 - 13) = \left ( 25 + \frac{35}{9} \right ) = 28.88



Video solution:

The A.M of given Data is M, M = \frac{80}{5} = 16

\Sigma | \delta | = |10 - 16 | + | 9 - 16 | + | 21 - 16 | + | 16 - 16 | + | 24 - 16 |
=(6 + 7 + 5 + 0 + 8) = 26
\therefore \text { Mean deviation } = \frac{\Sigma|\delta|}{n} = \frac{26}{5} = 5.2



Video solution:

From the given set of numbers , we get ,
Mode = 4
Median = \frac{(3+3)}{2}=3
Mean = \frac{30}{10}=3

Now, solving the given expression, we get,
3(mean) + 2(mode) – 6(median)
= 3(3) + 2(4) - 6(3)
= -1



Video solution:

We know that ,
Mode = 3(Median)-2(Mean)
or, 7 = 3(5)-2(Mean)
or, Mean = 4



Video solution:

Firstly, we have to arrange the scores in ascending order ,
then, we get,
162, 163, 165, 172, 180, 180, 181

Therefore,
Median = [\frac{n+1}{2}]^{th} , when n = odd

Thus, Median = 172.



Video solution: