. The distribution of marks in a certain exami- nation was found to be normal with 23% of the candidates scoring above 60 and 21% below 40. Find the mean and the standard devia- tion of the distribution of marks.

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A: Farmula used  P(X/Y=y)=P(X,Y=y)/P(Y=y)

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A:

Q: f(x, y) 2 4 1 0.10 0.15 Y 3 0.20 0.30 5 0.10 0.15 a. Determine the marginal distribution g(x), x =…

A: Farmula used  P(X/Y=y)=P(X,Y=y)/P(Y=y)

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Q: Let F be the co

A: Here,for G∈F and △∈D={cG1-G2:CER,Gj∈F, j=1,2}. Then, TG+t△-TG=∫G+t△z-ydG+t△ y-∫Gx-ydGy…

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Q: Determine the range, mean, and standard deviation of the data set. 9,5,9,17,17,16,3,8

A:

Q: 56 56 44 54 42 35 6 26 43 17 43   7 30 28 58 19 27 37 44 79 56 57

A: The data is given below 56 56 44 54 42 35 6 26 43 17 43 7 30 28 58 19 27 37 44 79 56 57

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A: Given: From the given table

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Q: Given that P(A) = 0.7, P (A/B) = 0.7, P(A∩B) notation same as P (A and B) = 0.42, find  the…

A: Since you have posted a question with multiple sub-parts, we will solve first three subparts for…

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A: Given, x y Rx Ry d=Rx-Ry d2 8 4 2 7 -5 25 9 3 1 9.5 -8.5 72.25 5 3 8 9.5 -1.5 2.25 7 4…

Q: 4. x = 81.72 μ = 79.83 s = 15 n = 29 5. X = 70 μ = 71.5 s = 8 n = 10

A: Formula : Test statistic

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A: Helpful formulas (i)VaX=a2VX(ii)VC=0,where c is the constant(iii)VaX+ bY=a2VX+ b2VY

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A: Given that,

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