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 Find Correlation coefficient for X and Y values given below : X= (1,2,3,4,5) Y= {11,22,34,43,56} a. 0.8899 b. 0.9989 c. 1.0899 d. 1.0989 Ans - b Explanation : Step 1: Find Mean for X and Y X=15/5=3 Y=166/5=33.2 Step 2: Calculate Standard Deviation for Y inputs: σx= √(1/(N-1)*((x1-xm)2+(x2-xm)2+..+(xn-xm)2)) =√(1/(5-1)((11-33.2)2+(22-33.2)2+(34-33.2)2+(43-33.2)2+(56-33.2)2)) =√(1/4((-22.2)2+(-11.2)2+(0.8)2+(9.8)2+(22.8)2)) =√(1/4((492.84)+(125.44)+(0.64)+(96.04)+(519.84))) =√(308.7) =17.5699 Step 3: Standard Deviation for X Inputs: σx= √(1/(N-1)*((x1-xm)2+(x2-xm)2+..+(xn-xm)2)) =√(1/(5-1)((1-3)2+(2-3)2+(3-3)2+(4-3)2+(5-3)2)) =√(1/4((-2)2+(-1)2+(0)2+(1)2+(2)2)) =√(1/4((4)+(1)+(0)+(1)+(4))) =√(2.5) =1.5811 Σ((X - μx) (Y - μy)) =(1-3)(11-33.2)+(2-3)(22-33.2)+(3-3)(34-33.2)+(4-3)(43-33.2)+(5-3)(56-33.2) =(-2*-22.2) + (-1*-11.2) + (0* 0.8) + (1 *9.8) + (2* 22.8) =44.4 + 11.2 + 0 + 9.8 + 45.6 =111 Correlation Coefficient = 111/((5-1)*1.5811*17.5699) Correlation Coefficient (r) = 0.9989 Hence the correlation coefficient between the two given data set is 0.9989 ............................................. Calculate Standard Error from the given data : X = 10, 20,30,40,50 a. 6.1071 b. 6.0711 c. 7.1071 d. 7.0711 Ans - d Explanation : Total Inputs (N) = (10,20,30,40,50) Total Inputs (N) =5 First find Mean: Mean (xm) = (x1+x2+x3...xn)/N Mean (xm) = 150/5 Mean (xm) = 30 Then find SD: SD = √(1/(N-1)*((x1-xm)2+(x2-xm)2+..+(xn-xm)2)) = √(1/(5-1)((10-30)2+(20-30)2+(30-30)2+(40-30)2+(50-30)2)) = √(1/4((-20)2+(-10)2+(0)2+(10)2+(20)2)) = √(1/4((400)+(100)+(0)+(100)+(400))) = √(250) = 15.811 Then Find Standard Error: Standard Error=SD / √(N) = 15.8114/√(5) = 15.8114/2.2361 = 7.0711 ............................................. A sack contains 4 black balls 5 red balls. What is probability to draw 1 black ball and 2 red balls in one draw ? a. 12/21 b. 9/20 c. 10/21 d. 11/20 Ans – c Solution : Out of 9, 3 (1 black & 2 red. are expected to be drawn) Hence sample space n(S) = 9c3 = 9!/(6!×3!) = 362880/4320 = 84 Now out of 4 black ball 1 is expected to be drawn hence nb. = 4c1 = 4 Same way out of 5 red balls 2 are expected be drawn hence n(R) = 5c2 = 5!/(3!×2!) = 120/12 = 10 Then P(B U R) = n(B)×n(R)/n(S) i.e 4×10/84 = 10/21 ……………………………………………………………………………………………………………………………………………

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