Directions: Read the following frequency distribution for two series of observations and answer the items that follows: Class interval Frequency Series-I Series-II 10-20 20 4 20-30 15 8 30-40 10 4 40-50 x 2x 50-60 y y Total 100 100 What is the mean of frequency distribution of Series-I?

First, find the missing frequencies for Series-I by using the total frequency: 20 + 15 + 10 + x + y = 100, which means x + y = 55. The cumulative frequency up to the 40-50 class is 45, so using the step-deviation or direct mean method, the midpoints are 15, 25, 35, 45, and 55. The sum of the product of frequencies and midpoints is (20*15) + (15*25) + (10*35) + (x*45) + (y*55) = 300 + 375 + 350 + 45x + 55y = 1025 + 45x + 55y. Substituting y = 55 - x into the sum gives 1025 + 45x + 55(55 - x) = 1025 + 45x + 3025 - 55x = 4050 - 10x. Using the fact that the mean is total sum over total frequency, (4050 - 10x) / 100 = 37.6, which means 4050 - 10x = 3760, resulting in 10x = 290 and x = 29. The mean of the frequency distribution is 37.6.

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