Section 5.7

1. Section 5.7 Suppose X1 , X2 , …, Xn is a random sample from a Bernoulli distribution with success probability p. Then Y = X1 + X2 + … + Xn has a distribution. According to the Central Limit Theorem, can be treated as having an approximate N(0, 1) distribution for sufficiently large n. This approximation is generally considered to be good if both np 5 and n(1 –p)  5. When using a normal distribution to approximate a discrete type distribution, a continuity correction can be employed for increased accuracy. b(n, p) np p Y– X – ———— = ———– np(1–p) p(1–p)/n

2. 1. Suppose Y has a b(400, 0.001) distribution. Explain why the normal distribution should not be used to approximate P(Y > 3). Since np = 400(0.001) = 0.4 < 5, the normal approximation should not be used. The Poisson approximation would be good, since p = 0.001 is so small. 2. (a) The random variable X has a b(200, 0.4) distribution. Explain why the normal distribution could be used to approximate P(X 65). Since np = 200(0.4) = 80 > 5 and n(1–p) = 200(0.6) = 120 > 5, the normal approximation would be good.

3. 2.-continued (b) Use the Central Limit Theorem to approximate P(X 65). np 80 X– 65.5 – ————  ———— = P(X 65) = P(X 65.5) = P np(1–p) 48 P(Z ) = – 2.09 (– 2.09) = 1 – (2.09) = 0.0183

4. 3. The random sample X1 , X2 , … , X600 is taken from a U(–2, 3) distribution and the following random variable is defined: Y = number of positive Xis in the random sample . Use the Central Limit Theorem to approximate P(Y > 370) . b( , ) Y has a distribution. 600 0.6 Since the normal approximation would be good. np = 600(0.6) = 360 > 5 and n(1–p) = 600(0.4) = 240 > 5, np 360 Y– 370.5 – ———— ————— = P(Y> 370) = P(Y> 370.5) = P np(1–p) 12 P(Z> ) = 0.875 1 – (0.875) = 1 – 0.8092 = 0.1908 0.8078 + 0.8106 ——————— 2

5. 4. (a) The random variable Y has a Poisson distribution with mean 85. Explain why we can think of the random variable Y as Y = X1 + X2 + … + X85 where X1 , X2 , … , X85 is a random sample from a Poisson distribution with mean 1 (one). We know from Text Exercise 5.4-4 that the sum of n independent Poisson random variables with respective means 1 , 2 , … , n, is a Poisson random variable with mean 1 + 2 + … + n .

6. (b) Use the Central Limit Theorem to approximate P(86 < Y < 92) . P(86 < Y < 92) = P(86.5 < Y < 91.5) = 86.5 – Y– 91.5 – P( ———— < ———— < ———— ) = 85 85 85 85 85 85 P(0.16 < Z < 0.71) = (0.71)– (0.16) = 0.7611 – 0.5636 = 0.1975

7. 5. A child’s game in an arcade allows the child to roll six balls (one at a time) down a slide into one of six slots labeled 1, 4, 2, 5, 3, and 6 in order from left to right. The game adds the numbers on the slots that the child rolls the six balls into. If this sum is under 19 or over 23, the child wins the game. Use the Central Limit Theorem to approximate the probability that a child who randomly rolls the six balls into the slots wins the game. Let the random variable Xi be the number of the slot that the child rolls the ith ball into for i = 1, 2, 3, 4, 5, 6. Let the random variable Y be the sum of the numbers of the slots that the child rolls the six balls into, that is, Y = X1 + X2 + X3 + X4 + X5 + X6 . For each i, the distribution of Xi is the uniform distribution on the first six positive integers. E(Xi) = 3.5 Var(Xi) = 35/12 P(losing with random rolls) = P(19 Y 23) =

8. 18.5 – Y– 23.5 – P( ———— < ———— < ———— ) = 21 21 21 P(18.5 Y 23.5) = 35 / 2 35 / 2 35 / 2 P(– 0.60 < Z < 0.60) = (0.60)–  (–0.60 ) = 0.7257 – (1 – 0.7257) = 0.4514 P(winning with random rolls) = 1 – 0.4514 = 0.5486