2 6 15 30 45 43.5 22.5

hoblem -3 x 68.5 40 30 x 10 -40 685 30 40 22.83 62.83 Average marks is 62.83 lem the average marks from the following distribution iculate 10-15 15-20 20-25 30 30-35 rks 5-10 No. of students tion, Marks (CH) No. student (f) Midvalue(x) d Fd 2.5 5-10 10-15 125 15-20 20-25 225 +1 25-30 27.5 2 +15 30-35 +3 32.5 25 Using step deviation formula Assumed Mean (A) 17.5 Common factor (i is 5 30 x 5 17.5 17.5 17.5 t 1.2 18.7 llinnon four An were 5's, 30 were 6's and the 540

Apply step-deviation method to find the A M of the following frequency distribution Variate(x) 5 10 15 20 25 30 35 40 45 50 Frequency(f) 20 43 75 67 72 45 39 9 8 6

The value of (4.35 xx 4.35 xx 4.35 + 3.25 xx 3.25 xx 3.25)/(43.5 xx 43.5 + 32.5 xx 32.5 -43.5 xx 32.5) is : (4.35 xx 4.35 xx 4.35 + 3.25 xx 3.25 xx 3.25)/(43.5 xx 43.5 + 32.5 xx 32.5 -43.5 xx 32.5) का मान ज्ञात करें:

The average of 20, 25, 35, 15, a, 30 is 22.5. The value of a is :

24 xx (15 + 45) div 6 + (45 + 5) - 5

Find the median for the following distribution: Class 5−10 10−15 15−20 20−25 25−30 30−35 35−40 40−45 Frequency 5 6 15 10 5 4 2 2

Calculate the median from the following distribution: Class: 5-10 10-15 15-20 20-25 25-30 30-35 35-40 40-45 Frequency: 5 6 15 10 5 4 2 2

Calculate the median from the following distribution: Class: 5-10 10-15 15-20 20-25 25-30 30-35 35-40 40-45 Frequency: 5 6 15 10 5 4 2 2

Find the fourth proportional to : (i) 1.5 , 4.5 and 3.5 (ii) 3a, 6a^(2) and 2ab^(2)

Simplify ( 4.5 a + 1.5 b ) ^(2) + ( 4. 5b + 1. 5a ) ^(2)