Q: Consider the following frequency distribution : Value 4 5 8 9 6 12 11 Frequency 5 f_1 f_2 2 1 1 3 Suppose that the sum of the frequencies is 19 and the median of this frequency distribution is 6. For

$ array |c|c|c|c|c|c|c| X_i & f_ i & array l x =7 \\ d_i = |x_i- x | array & array l M =6 \\ ei = | xi - M | array & f_ i d _ i & f_ i e _ i & f_ i d _ i ^2 \\ 4 & 5 & 3 & 2 & 15 & 10 & 45 \\ 5 & 4 & 2 & 1 & 8 & 4 & 16 \\ 6 & 1 & 1 & 0 & 1 & 0 & 1 \\ 8 & 3 & 1 & 2 & 3 & 6 & 3 \\ 9 & 2 & 2 & 3 & 4 & 6 & 8 \\ 11 & 3 & 4 & 5 & 12 & 15 & 48 \\ 12 & 1 & 5 & 6 & 5 & 6 & 25 \\ & & & & 48 & 47 & 146 \\ array $ aligned & f_1=4 \\ & f_2=3 \\ & = 48 19 , = 47 19 , ^2= 146 19 aligned

Question 1

Consider a data consisting of 10 observations x_1, x_2, , x_ 10 , whose mean is 5 and variance is 7. If the mean and the variance of the first 8 observations x_1, x_2, , x_8 are 4 and 3.5, respectivel

Accepted Answer

Given _ i=1 ^ 10 x_i 10 = 5 _ i=1 ^ 10 x_i = 50 _ i=1 ^ 10 x_i^2 10 - 5^2 = 7 _ i=1 ^ 10 x_i^2 = 320 For the first 8 observations: _ i=1 ^ 8 x_i 8 = 4 _ i=1 ^ 8 x_i = 32 _ i=1 ^ 8 x_i^2 8 - 4^2 = 3.5 _ i=1 ^ 8 x_i^2 = 8(3.5 + 16) = 156 Subtracting the sums, we get: x_9 + x_ 10 = 50 - 32 = 18 x_9^2 + x_ 10 ^2 = 320 - 156 = 164 Using (x_9 + x_ 10 )^2 = x_9^2 + x_ 10 ^2 + 2x_9 x_ 10 : 18^2 = 164 + 2x_9 x_ 10 2x_9 x_ 10 = 324 - 164 = 160 x_9 x_ 10 = 80 The numbers x_9 and x_ 10 are roots of the equa

Question 2

Consider the following frequency distribution : Value 4 5 8 9 6 12 11 Frequency 5 f_1 f_2 2 1 1 3 Suppose that the sum of the frequencies is 19 and the median of this frequency distribution is 6. For

Accepted Answer

$ array |c|c|c|c|c|c|c| X_i & f_ i & array l x =7 \ d_i = |x_i- x | array & array l M =6 \ ei = | xi - M | array & f_ i d _ i & f_ i e _ i & f_ i d _ i ^2 \ 4 & 5 & 3 & 2 & 15 & 10 & 45 \ 5 & 4 & 2 & 1 & 8 & 4 & 16 \ 6 & 1 & 1 & 0 & 1 & 0 & 1 \ 8 & 3 & 1 & 2 & 3 & 6 & 3 \ 9 & 2 & 2 & 3 & 4 & 6 & 8 \ 11 & 3 & 4 & 5 & 12 & 15 & 48 \ 12 & 1 & 5 & 6 & 5 & 6 & 25 \ & & & & 48 & 47 & 146 \ array $ aligned & f_1=4 \ & f_2=3 \ & = 48 19 , = 47 19 , ^2= 146 19 aligned

Question 3

Let X be a random variable, and let P(X=x) denote the probability that X takes the value x. Suppose that the points (x, P(X=x)), x=0,1,2,3,4, lie on a fixed straight line in the x y-plane, and P(X=x)=

Accepted Answer

Let equation of line is y = mx + c array |c|c|c|c|c|c|c| x & 0 & l & 2 & 3 & 4 & R -\ 0,1,2,3,4\ \ P ( x ) & c & m + c & 2 ~m + c & 3 ~m + c & 4 ~m + c & 0 \ array _ x=0 ^4(m x+c)=1 10 m+5 c=1 2 m+c= 1 5 ...(1) aligned & mean = x_i P_i= _ i =0 ^4 ( mx _ i + c ) x _ i =30 ~m +10 c = 5 2 \ & 3 ~m + c = 1 4 (2) aligned aligned & P _ i x _ i ^2= _ i =0 ^4 ( mx _ i + c ) x _1^2 \ & = _ i =0 ^4 ( mx _ i ^3+ cx _ i ^2 ) 100 ~m +30 c (Now putting m and c ) \ & P _ i x _ i ^2=5+3=8 \ & Variance = P

Question 4

Consider the given data with frequency distribution x i 3 8 11 10 5 4 f i 5 2 3 2 4 4 Match each entry in List-I to the correct entries in List-II.   List-I   List-II P The mean of the above

Accepted Answer

Arranging the given data in ascending order we get, x i 3 4 5 8 10 11 f i 5 4 4 2 2 3 Now finding the mean of the above data we get, Mean = 3 &#215; 5 + 8 &#215; 2 + 11 &#215; 3 + 10 &#215; 2 + 5 &#215; 4 + 4 &#215; 4 5 + 2 + 3 + 2 + 4 + 4 = 15 + 16 + 33 + 20 + 20 + 16 20 = 120 20 = 6 Now, median = 1 2 10 th + 11 th &#160; observation = 1 2 5 + 5 = 5 Hence, Mean deviation about mean will be, = 3 &#215; 5 + 2 &#215; 4 + 1 &#215; 4 + 2 &#215; 2 + 4 &#215; 2 + 5 &#215; 3 20 = 54 20 = 2 . 7 And Mean

Statistics — Previous Year Questions & Solutions