STAT 400 Discussion 05 Solutions Spring 2018

1. 2.3-11 2.5-3

If the moment-generating function of X is

M X ( t ) = , Find the mean, variance, and pmf of X.

f ( 1 ) = , f ( 2 ) = , f ( 3 ) = .

M X' ( t ) = . E ( X ) = M X' ( 0 ) = 2.

OR

E ( X ) = ( 1 ) + ( 2 ) + ( 3 ) = 2.

M X'' ( t ) = . E ( X 2 ) = M X'' ( 0 ) = = 4.8.

OR

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2 + ( 2 ) 2 + ( 3 )

2 = = 4.8.

Var ( X ) = E ( X 2

) – [ E ( X ) ] 2 = 4.8 – 2

2 = 0.8.

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2. Suppose a discrete random variable X has the following probability distribution:

f ( 0 ) = , f ( k ) = , k = 2, 4, 6, 8, … .

( possible values of X are even non-negative integers: 0, 2, 4, 6, 8, … ). Recall Discussion #2 Problem 1 (a): this is a valid probability distribution. a) Find the moment-generating function of X, M X ( t ). For which values of t does it exist?

M X ( t ) = E ( e t X ) = =

= = = .

Must have < 1 for geometric series to converge. Þ t < ln 3.

b) Use M X ( t ) to find E ( X ).

Recall Week 04 Discussion Problem 1 (a): E ( X ) = .

M 'X ( t ) = = , t < ln 3.

E ( X ) = M 'X ( 0 ) = = .

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3. Suppose a discrete random variable X has the following probability distribution:

f ( 1 ) = ln 3 – 1, f ( k ) = , k = 2, 3, 4, … . ( possible values of X are positive integers: 1, 2, 3, 4, … ). Recall Discussion #2 Problem 1 (b): this is a valid probability distribution.

“Hint”: Recall that .

a) Find the moment-generating function of X, M X ( t ). For which values of t does it exist?

M X ( t ) = = e t × ( ln 3 – 1 ) +

= e t × ( ln 3 – 1 ) + = e t ln 3 – e t + – 1 – e t ln 3

= , t Î R.

b) Use M X ( t ) to find E ( X ).

Recall Week 04 Discussion Problem 1 (b): E ( X ) = 3 ln 3 – 1 » 2.2958.

, E ( X ) = = 3 ln 3 – 1.

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4. Suppose the moment-generating function of X is M X ( t ) = 0.1 e

2 t + 0.3 e 4 t + 0.6 e

7 t. a) Find µ = E ( X ).

M X' ( t ) = 0.2 e 2 t + 1.2 e

4 t + 4.2 e 7 t.

E ( X ) = M X' ( 0 ) = 0.2 + 1.2 + 4.2 = 5.6.

b) Find s = SD ( X ).

M X" ( t ) = 0.4 e

2 t + 4.8 e 4 t + 29.4 e

7 t. E ( X

2 ) = M X" ( 0 ) = 0.4 + 4.8 + 29.4 = 34.6.

Var ( X ) = E ( X

2 ) – [ E ( X ) ]

2 = 34.6 – 5.6 2 = 3.24.

SD ( X ) = = 1.8.

OR M X ( t ) = 0.1 e

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1.296 0.768 1.176

5.6 34.6 3.240 µ = E ( X ) = S x × f ( x ) = 5.6. Var ( X ) = S ( x – µ ) 2 × f ( x ) = 3.24. OR Var ( X ) = S x 2 × f ( x ) – µ 2 = 34.6 – 5.6

2 = 3.24. SD ( X ) = = 1.8.

24.3

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5. Suppose a discrete random variable X has the following probability distribution: f ( k ) = P ( X = k ) = , k = 2, 3, 4, 5, 6, … , zero otherwise.

a) Find the value of a that makes this is a valid probability distribution. Must have = 1.

Þ 1 = = = .

Þ a 2 + a – 1 = 0. Þ a = .

< 0. Þ a = » 0.618034.

Note: a = = j – 1, where j is the golden ratio.

b) Find P ( X is even ). P ( X is even ) = f ( 2 ) + f ( 4 ) + f ( 6 ) + f ( 8 ) + …

= a 2 + a

4 + a 6 + a

8 + … =

= = = a » 0.618034.

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c) Find the moment-generating function of X, M X ( t ). For which values of t does it exist?

M X ( t ) = E ( e t X

) = = = = ,

< 1 Û t < ln = ln j » 0.48121.

d) Find E ( X ).

M 'X ( t ) = = ,

t < ln .

E ( X ) = M 'X ( 0 ) = = = 3 + a » 3.618034.

OR

E ( X ) = a × E ( X ) = Þ × E ( X ) = = .

Therefore, E ( X ) = = = = 3 + a » 3.618034.

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6. Let X be a continuous random variable with the probability density function

f ( x ) = , x > 5, zero otherwise.

a) Find the value of C that would make f ( x ) a valid probability density function.

Must have = 1.

1 = = – = . C = 375.

b) Find the cumulative distribution function of X, F ( x ) = P ( X £ x ). “Hint”: Should be F ( 5 ) = 0, F ( ¥ ) = 1.

F ( x ) = P ( X £ x ) = = – = , x ³ 5.

c) Find the probability P ( 6 < X < 10 ).

P ( 6 < X < 10 ) = = – =

» 0.5787 – 0.1250 = 0.4537.

OR

P ( 6 < X < 10 ) = F ( 10- ) – F ( 6 ) =

» 0.8750 – 0.4213 = 0.4537.

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f) Find the 80th percentile of the distribution of X, p 0.80 .

P ( X £ p 0.80 ) = F ( p 0.80 ) = = 0.80.

p 0.80 = » 8.55.

OR

P ( X ³ p 0.80 ) = 0.20.

= – = = 0.20.

p 0.80 = » 8.55. For fun:

Median = » 6.3.

g) Find the expected value of X, E ( X ).

E ( X ) = = = = – = 7.5.

h) Find the standard deviation of X, SD ( X ).

E ( X ) = = = = – = 75.

Var ( X ) = E ( X

2 ) – [ E ( X ) ]

2 = 75 – 7.5 2 = 18.75.

SD ( X ) = » 4.33.

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7. Let X be a continuous random variable with the probability density function f ( x ) = C x

2, 3 £ x £ 9, zero otherwise. a) Find the value of C that would make f ( x ) a valid probability density function.

1 = . Þ C = .

b) Find the probability P ( X < 5 ).

P ( X < 5 ) = » 0.1396.

c) Find the probability P ( X > 7 ).

P ( X > 7 ) = » 0.5499.

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d) Find the mean of the probability distribution of X.

E( X ) = » 6.923.

e) Find the median of the probability distribution of X.

F ( x ) = P ( X £ x ) = ,

3 ≤ x < 9. F ( x ) = P ( X £ x ) = 0, x < 3.

F ( x ) = P ( X £ x ) = 1, x ³ 9.

F ( m ) = . = .

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8. Suppose a random variable X has the following probability density function:

f ( x ) = cos x, 0 < x < , zero otherwise.

a) Find P ( X < ).

P ( X < ) = = = sin – 0 = » 0.7071.

b) Find µ = E ( X ).

µ = E ( X ) = = = » 0.5708.

c) Find the median of the probability distribution of X.

F ( x ) = P ( X ≤ x ) = = sin x, 0 < x < .

Median: F ( m ) = P ( X ≤ m ) = .

Þ sin m = . m = » 0.5236.

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9. Let X be a continuous random variable with the probability density function f ( x ) = 6 x ( 1 – x ), 0 < x < 1, zero elsewhere. Compute P ( µ – 2 s < X < µ + 2 s ).

µ = E ( X ) = = = = = 0.50.

E ( X 2 ) = = = = 0.30.

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s = » 0.2236.

µ – 2 s » 0.0528, µ + 2 s » 0.9472.

P ( µ – 2 s < X < µ + 2 s ) » =

» 0.992 – 0.008 = 0.984.

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10. Suppose a random variable X has the following probability density function: f ( x ) = x e x, 0 < x < 1, zero otherwise.

a) Find P ( X < ).

P ( X < ) = = = 1 – » 0.175639.

b) Find µ = E ( X ).

µ = E ( X ) = = = e – 2.

c) Find the moment-generating function of X, M X ( t ).

M X ( t ) = =

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M X ( – 1 ) = = .

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11. Let X be a continuous random variable with the probability density function

a) Find the value of c that makes f ( x ) a valid probability density function.

Must have = 1.

1 = = = = 17 c.

Þ 1 = 17 c. Þ c = .

b) Find the probability P ( X < 5 ).

P ( X < 5 ) = = = = .

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F X ( x ) = = , 0 ≤ x < 3,

F X ( x ) = = , 3 ≤ x < 8,

F X ( x ) = 1, x ³ 8.

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e) Find the variance of the probability distribution of X.

E ( X 2 ) = = =

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Var ( X ) = E ( X 2 ) – [ E ( X ) ]

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