There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test: Marks 0 1 2 3 4 5 Frequency x - 2 x x 2 x + 1 2 2 x x + 1 where, x is positive integer. Determine the mean and standard deviation of the marks.

Given that sum of frequencies is 60 . ∴ x - 2 + x + x 2 + ( x + 1 ) 2 + 2 x + ( x + 1 ) = 60 ⇒ 4 x - 2 + x 2 + x 2 + 2 x + 1 + x + 1 = 60 ⇒ 2 x 2 + 7 x - 60 = 0 ⇒ 2 x 2 + 15 x - 8 x - 60 = 0 ⇒ x ( 2 x + 15 ) - 4 ( 2 x + 15 ) = 0 ⇒ ( 2 x + 15 ) ( x - 4 ) = 0 ⇒ x - 4 = 0 and 2 x + 15 = 0 ⇒ x = 4 a n d x = - 15 2 [Rejected] ∵ x ∈ I + So, x = 4 Now put x = 4 in the frequency distribution table x i f i d i = x i - 3 f i d i f i d i 2 0 2 - 3 - 6 18 1 4 - 2 - 8 16 2 16 - 1 - 16 16 3 25 0 0 0 4 8 1 8 8 5 5 2 10 20 N = 60 ∑ f i d i = - 12 ∑ f j d i 2 = 78 Let assumed mean A = 3 Mean = A + ∑ f i d i N = 3 + - 12 60 = 3 - 1 5 = 14 5 = 2 . 8 and S . D . ( σ ) = ∑ f i d i 2 N - ∑ f i d i N 2 = 78 60 - - 12 60 2 = 1 . 3 - 0 . 04 = 1 . 26 = 1 . 12

Determine the mean and standard deviation for the following distribution: Marks 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Frequency 1 6 6 8 8 2 2 3 0 2 1 0 0 0 1

x f i f i x i d i = x i - x ¯ f i d i f i d i 2 2 1 2 - 4 - 4 16 3 6 18 - 3 - 18 54 4 6 24 - 2 - 12 24 5 8 40 - 1 - 8 8 6 8 48 0 0 0 7 2 14 1 2 2 8 2 16 2 4 8 9 3 27 3 9 27 10 0 0 4 0 0 11 2 22 5 10 50 12 1 12 6 6 36 13 0 0 7 0 0 14 0 0 8 0 0 15 0 0 9 0 0 16 1 16 10 10 100 N = 40 ∑ f i x i = 239 ∑ f i d i = - 1 ∑ f i d i 2 = 325 Mean x ¯ = 239 40 = 5 . 975 ≈ 6 ∴ S D = σ = ∑ f i d i 2 N - ∑ f i d i N 2 = 325 40 - - 1 40 2 = 8 . 125 - 0 . 000625 = 8 . 124375 = 2 . 85 Hence, the required mean = 6 and MD = 2 . 85 .