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Probabilities Essay

4895 words - 20 pages

Statistics 100A Homework 5 Solutions
Ryan Rosario
Chapter 5 1. Let X be a random variable with probability density function c(1 − x2 ) −1 < x < 1 0 otherwise

f (x) = (a) What is the value of c?

We know that for f (x) to be a probability distribution −∞ f (x)dx = 1. We integrate f (x) with respect to x, set the result equal to 1 and solve for c.
1

1 =
−1

c(1 − x2 )dx cx − c x3 3
1 −1

= = = = c = Thus, c =
3 4

c c − −c + c− 3 3 2c −2c − 3 3 4c 3 3 4 .

(b) What is the cumulative distribution function of X? We want to find F (x). To do that, integrate f (x) from the lower bound of the domain on which f (x) = 0 to x so we will get an expression in terms of x.
x

F ...view middle of the document...

Break the problem up into two steps: finding the probability that 1 device functions for at least 15 hours, and finding the probability that 3 of the 6 devices function for at least 15 hours. Step 1: One Device First we must find the probability that one randomly selected device will function for at least 15 hours. It is given that each device follows a probability distribution f (x) given earlier. Let X be the number of hours that the device functions. Then, 10 dx 2 15 x 10 ∞ = − x 15 10 10 = lim − − − x→∞ x 15 10 2 = − lim + x→∞ x 3 =
2 3 ∞

P (X ≥ 15) =

Step 2: 3 of the 6 Devices We are interested in n = 6 devices. Note that each device constitutes a trial with one of two outcomes: device functions at least 15 hours or it doesn’t. Each of the 6 trials are Bernoulli trials and we assume the trials are independent. We also have a 2 constant probability of success p = 3 . We use the binomial model. Let X be the number of devices that function at least 15 hours. Then,
3 3 4 2 5 1

P (X ≥ 3) =

6 3 +

2 3 6 6

1 3 2 3
6

+ 1 3
0

6 4

2 3

1 3

+

6 5

2 3

1 3

= ≈

656 729 0.9

3

6. Compute E(X) if X has a density function given by (a) f (x) =
−x 1 2 4 xe

0

x>0 otherwise

Recall that by definition,
b

E(X) =
a

xf (x)f x

Then, for the PDF given in this problem,

E(X) =
0

x

1 −x xe 2 4

dx =

1 4

∞ 0

x2 e− 2 dx

x

We use what my Calculus II professor called a “tricky u-substitution.” First, let y = x . 2 1 Then x = 2y and dx = 2dy. The 4 from x2 = (2y)2 = 4y 2 cancels the 4 in front of the integral, and we must carry the 2 from dx outside the integral to get, E(X) = 1 4
∞ 0

x2 e− 2 dx =

x

1 4

∞ 0

(2y)2 e−y (2dy) = 2
0

y 2 e−y dy

Recall that when we have an integral of the form h(x) = f (x)g(x)dx we use integration by parts. We choose values for u and dv. Then we compute du and v and substitute into h(x) = uv − vdu. Now use integration by parts, with u = y 2 , dv = e−y so du = 2ydy, v = −e−y . Then,
Z E(X) = = = 2
0 ∞

y 2 e−y dy

– Z 2 −y 2 e−y − −e−y (2y) dy » – Z 2 −y 2 e−y + 2 ye−y dy integrate by parts again. Use: u = y, dv = e−y thus du = dy, v = −e−y »  ff– Z 2 −y 2 e−y + 2 −ye−y − −e−y dy ˆ ˘ ¯˜ 2 −y 2 e−y + 2 −ye−y − e−y −2y 2 e−y − 4ye−y − 4e−y ˆ ` ´˜∞ −2e−y y 2 + 2y + 2 0 ˆ ` 2 ´˜ ˆ ` ´˜ lim −2e−y y + 2y + 2 − −2e−y y 2 + 2y + 2 x=0
x→∞

»

= = = = =

=

=== ==

L H

=== ==

L H

The limit evaluates to an indeterminate form 0 · ∞. – » ´ ˆ ` ´˜ 1 ` 2 y + 2y + 2 − −2e−y y 2 + 2y + 2 x=0 −2 lim y x→∞ e The limit evaluates to ∞ so use l’Hˆpital’s Rule. o ∞ ˆ −y ` 2 ´˜ 2y + 2 −2 lim +2 e y + 2y + 2 x=0 x→∞ ey ∞ The limit evaluates to ∞ still so use l’Hˆpital’s Rule again. o ˆ −y ` 2 ´˜ 2 −2 lim y +2 e y + 2y + 2 x=0 x→∞ e | {z }
=0

=

4

4

(b) f (x) =

c(1 − x2 ) −1 < x < 1 0 otherwise
1

E(X) =
−1

x c 1 − x2
1

dx

= c
−1 1

x 1 − x2 dx x − x3 dx
−1 x2

= c = c = c = 0

2

...

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