Section 5.6

Important Theorem in the Text:

The Central Limit Theorem Theorem 5.6-1

1.

(a)

Let X1 , X2 , … , Xn be a random sample from a U(–2, 3) distribution.

Define the random variable Y = X1 + X2 + … + Xn = n Xi . i = 1

Use the Central Limit Theorem to find a and b, both depending on

n, so that the limiting distribution of is N(0, 1).Y – a—— b

Y is the sum of n independent and identically distributed random variables each of which has mean = and variance 2 =1/2 25/12.

Therefore, the Central Limit Theorem tells us that the limiting

distribution of is N(0, 1). Y –———–n/2

5n—–23

Use the Central Limit Theorem to find a and b, with only b

depending on n, so that the limiting distribution of

is N(0, 1).

X – a—— b

(b)

X is the mean of n independent and identically distributed random variables each of which has mean = and variance 2 =1/2 25/12.

Therefore, the Central Limit Theorem tells us that the limiting

distribution of is N(0, 1). X –———–

1/2

5—–—23n

Suppose n = 25. Use the Central Limit Theorem to approximate P(Y 12).

1.-continued

(c)

P(Y 12) = PY – 12 –——— ———— =

25/2

25—–23

25/2

25—–23

P(Z ) =– 0.07

(– 0.07) = 1 – (0.07) = 0.4721

2.

(a)

A random sample X1 , X2 , … , Xn is taken from a N(100, 64) distribution. Find each of the following:

P(96 < Xi < 104) for each i = 1, 2, …, n .

P(96 < Xi < 104) = 96 – Xi – 104 –P( ———— < ———— < ———— ) =

100 100 100

8 8 8

P(– 0.50 < Z < 0.50) = (0.50) – (– 0.50) =

(0.50) – (1 – (0.50)) = 0.6915 – (1 – 0.6915) = 0.3830

P(96 < X < 104) = 96 – X – 104 –P( ———— < ———— < ———— ) =

100 100 100

4 4 4

P(– 1.00 < Z < 1.00) = (1.00) – (– 1.00) =

(1.00) – (1 – (1.00)) = 0.8413 – (1 – 0.8413) = 0.6826

2.-continued

(b) P(96 < X < 104) when n = 4.

(c) P(96 < X < 104) when n = 16.

P(96 < X < 104) = 96 – X – 104 –P( ———— < ———— < ———— ) =

100 100 100

2 2 2

P(– 2.00 < Z < 2.00) = (2.00) – (– 2.00) =

(2.00) – (1 – (2.00)) = 0.9772 – (1 – 0.9772) = 0.9544

3.

(a)

A random sample X1 , X2 , … , X25 is taken from a distribution defined by the p.d.f. f(x) = x / 50 if 0 < x < 10 .

P(Xi < 6) for each i = 1, 2, …, 25 .

P(Xi < 6) = x— dx =50 9 / 25

0

6

x2

—— = 100

0

6

20 / 3 = 2 = 50 – 400 / 9 = 50 / 9

P(X < 6) = X – 6 –P( ———— < ———— ) =

20/3 20/3

2 / 3 2 / 3P(Z < – 1.41) =

(– 1.41) = 1 – (1.41) = 1 – 0.9207 = 0.0793

(b) Use the Central Limit Theorem to approximate P(X < 6).