The mean and variance of X are 4.8 and 2.88 respectively. Consider a normally di

Question

The mean and variance of X are 4.8 and 2.88 respectively. Consider a normally distributed random variable with the same mean and variance: Y N(4.8,2.88) Using statistical tables determine: a) 1 mark] Compute the relative error in using P(Y S 2) as an approximation to b) |2 marks) Obtain a more accurate approximation to P(X s 2) using the normal c)1 mark Compute the relative error in using your answer from part b) as an ap- d)12 marks) What is P(X S2) if the sampling is done without replacement (and P(X S2) approximation to the binomial distribution with continuity correction. proximation to P(X 2). everything else is otherwise the same)?

Explanation / Answer

Here for the given binomial distribution where

np = 4.8 and np(1-p) = 2.88

dividing this two equations we get

1-p = 0.6

p = 0.4

n = 4.8/0.4 = 12

so P(Y < =2) by binomial = BIN (Y <=2 ; 12; 0.4) = 0.0834

By normal approximation

Pr(X <=2) = NORMAL (X <=2 ; 4.8; 2.88)

Z = (2 - 4.8)/ sqrt(2.88) = -1.65

Pr(X <=2) = Pr(Z < -1.65) = 0.0495

Relative error = (0.0834 - 0.0495) * 100/ 0.0834 = 40.65%

(b) Here by using continuity correction.

Pr(X <=2) = Pr(X < 2.5) = NORM(X < 2.5; 4.8; 2.88)

Z = (2.5 - 4.8)/ sqrt(2.88) = -1.3553

Pr(Z < -1.3553) = 0.08766

(c) Relative error = (0.0834 - 0.08766)* 100 / 0.0834 = 5.11%

(d) Here if sampling is done without replacement. Here the question doesn't consist full information. Here population size is not given so can't comprehend

Pr(X < =2) = Pr(X =0) + Pr(X = 1) + Pr(X = 2)

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