A box contains tickets labeled with the numbers {4,2,0,3,5}. The next two questi

Question

A box contains tickets labeled with the numbers {4,2,0,3,5}. The next two questions have to do with the results of 100 random draws with replacement from this box.

In 100 random draws with replacement from the box, the SE of the sumof just the positive numbers on the ticketsdrawn is closest to

(a) 10×2.191

(b) 10×1.959

(c) 10×3.286

(d)10×1.743

(e) 10×3.67414.

The normal approximation to the probability that the sum of all 100 numbers drawn with replacement from this boxis between -32.8 and 65.7 is closest to

(a) 30%

(b) 50%

(c) 65%

(d)80%

(e) 97%

Explanation / Answer

Back-up Theory

Let p1, p2, p3, p4 and p5 be respectively the probability that tickets with number 4, - 2, 0, 3 and – 5 are drawn. Since the draw is random and also with replacement, p1 = p2 = p3 = p4 = p5 = 1/5 = 0.2 = p, say.

Let X1, X2, X3, X4 and X5 be respectively the number of times tickets with number 4, - 2, 0, 3 and – 5 are drawn in 100 draws. Then, Xi ~ B(100, 0.2), E(Xi) = np = 100 x 0.2 = 20, V(Xi) = np(1 - p) = 100 x 0.2 x 0.8 = 16 and hence SD(Xi) = 4 for all i = 1 to 5.

Part (1)

Let S1 = sum of just the positive numbers drawn in 100 draws. Then, S1 = 4X1 + 0X3 + 3X4 and hence

SD(S1) = 4SD(X1) + 0SD(X3) + 3SD(X4) = (4 + 3)SD(Xi) = 7 x 4 = 28 ANSWER

Part (2)

Let S = sum of all the numbers drawn in 100 draws. Then, S = 4X1 – 2X2 + 0X3 + 3X4 – 5X5 and hence E(S) = 4E(X1) – 2E(X2) + 0E(X3) + 3E(X4) – 5E(X5) = 20(4 – 2 + 0 + 3 - 5) = 0 and

SD(S) = 4SD(X1) + 2SD(X2) + 0SD(X3) + 3SD(X4) + 5SD(X5) = 4(4 + 2 + 0 + 3 + 5) = 56.

By Normal approximation, S ~ N(0, 56). Hence,

P(- 32.8 < S < 65.7)

= P[{(- 32.8 - 0)/56} < Z < {(65.7 - 0)/56}]

= P(- 0.586 < Z < 1.173)

= P(Z < 1.173) - P(Z < - 0.586)

= 0.8796 – 0.2789

= 0.6007 ANSWER

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