sta_5325_hw_2_ramin_shamshiri

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Ramin Shamshiri STA 5325, HW #2 Due 06/01/09

STA 5325, Homework #2

Due June 01, 2009

Ramin Shamshiri

UFID # 90213353 Note: Problem numbers are according to the 6

th text edition. Selected problem are highlighted.

Assignment for Tuesday, May 26th

, 2009

4.5. Suppose that Y posses the density function

𝑓 𝑦 = 𝑐𝑦, 0 ≤ 𝑦 ≤ 20, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒.

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

b. Find F(y)

c. Graph f(y) and F(y)

d. Use F(y) to find 𝑃(1 ≤ 𝑌 ≤ 2).

e. Use f(y) and geometry to find 𝑃(1 ≤ 𝑌 ≤ 2).

Solution a: According to the properties of density function,

𝑓 𝑦 ≥ 0

𝑓(𝑦)∞

−∞𝑑𝑦 = 1

Therefore:

𝐹 ∞ = 𝑓(𝑦)∞

−∞

𝑑𝑦 = 𝑐𝑦2

0

𝑑𝑦 = 𝑐𝑦2

2

0

2

= 1 => 𝑐 =1

2

𝑓 𝑦 =

𝑦

2, 0 ≤ 𝑦 ≤ 2

0, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒.

Solution b:

According to definition 4.3,

𝑓 𝑦 =𝑑𝐹(𝑦)

𝑑𝑦= 𝐹′(𝑦)

So, F(y) can be written as:

𝐹 𝑦 = 𝑓(𝑡)𝑦

−∞

𝑑𝑡 = 𝑡

2

𝑦

−∞

𝑑𝑡 =𝑦2

4

Solution c:

Plots are shown in Figures 4.5a and 4.5b.

Fig 4.5.a Fig 4.5.b

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Solution d:

𝑃 1 ≤ 𝑌 ≤ 2 = 𝐹(𝑦) 12 = 𝑦

2

4 1

2

= 1 −1

4=

𝟑

𝟒

Solution e:

According to the figure below, the area corresponding to (1 ≤ 𝑌 ≤ 2) can be calculated by calculating the area

corresponding to (0 ≤ 𝑌 ≤ 2) minus (0 ≤ 𝑌 ≤ 1).

Area under 0 ≤ 𝑌 ≤ 2 = 2 1

2= 1

Area under 0 ≤ 𝑌 ≤ 1 = 1 0.5

2=

1

4

1 −1

4=

𝟑

𝟒

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4.7. A supplier of kerosene has a 150-gallon tank that is filled at the begging of each week. His weekly demand

shows a relative frequently behavior that increases steadily up to 100 gallons and then levels off between 100 and 150 gallons. If Y denotes weekly demand in hundreds of gallons, the relative frequency of demand can be

modeled by

𝑓 𝑦 = 𝑦, 0 ≤ 𝑦 ≤ 1

1, 0 < 𝑦 ≤ 1.50, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒.

a. Find F(y)

b. Find 𝑃(1 ≤ 𝑌 ≤ 0.5).

c. Find 𝑃(0.5 ≤ 𝑌 ≤ 1.2).

Solution a:

𝐹 𝑦 = 𝑓 𝑡 𝑦

−∞

𝑑𝑡

For (𝑦 < 0), 𝐹 𝑦 = 𝑃(𝑌 ≤ 𝑦) = 𝟎

For (0 ≤ 𝑦 ≤ 1), 𝐹 𝑦 = 𝑡𝑦

0𝑑𝑡 = 𝑡

2

2

0

𝑦

=𝒚𝟐

𝟐

For 1 ≤ 𝑦 ≤ 1.5 , 𝐹 𝑦 = 𝑓 𝑡 1

0𝑑𝑡 + 𝑓 𝑡

𝑦

1𝑑𝑡

= 𝑡1

0

𝑑𝑡 + 1𝑦

1

𝑑𝑡 =1

2+ 𝑦 − 1 = 𝒚 −

𝟏

𝟐

For 𝑦 ≥ 1.5 , 𝐹 𝑦 = 1

𝐹 𝑦 =

0, 𝑦 < 0

𝒚𝟐

𝟐, 0 ≤ 𝑦 ≤ 1

𝒚 −𝟏

𝟐, 1 ≤ 𝑦 ≤ 1.5

1, 𝑦 > 1.5

Solution b:

𝑃 1 ≤ 𝑌 ≤ 0.5 = 𝐹 0.5 =1

8

Solution c:

𝑃 0.5 ≤ 𝑌 ≤ 1.2 = 𝐹 1.2 − 𝐹 0.5 = 1.2 − 0.5 − 1

8 = 0.575

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4.14. If, as in Exercise 4.10, Y has density function

𝑓 𝑥 = 1/2 2 − 𝑦 , 0 ≤ 𝑦 ≤ 20, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒.

find the mean and variance of Y.

Solution:

The mean of Y is the expected value of Y:

𝜇 = 𝐸 𝑌 = 𝑦.∞

−∞

𝑓 𝑦 𝑑𝑦 =1

2 𝑦.

2

0

2 − 𝑦 𝑑𝑦 =𝟐

𝟑

The variance can be calculated as:

𝜎2 = 𝐸 𝑌2 − 𝜇2

𝐸 𝑌2 = 𝑦2.∞

−∞

𝑓 𝑦 𝑑𝑦 =1

2 𝑦2.

2

0

2 − 𝑦 𝑑𝑦 =2

3

𝜎2 = 𝐸 𝑌2 − 𝜇2 =2

3−

2

3

2

=𝟐

𝟗

4.16. If, as in Exercise 4.12, Y has density function

𝑓 𝑥 = 0.2, − 1 ≤ 𝑦 ≤ 0

0.2 + 1.2 𝑦, 0 < 𝑦 ≤ 10, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒.

find the mean and variance of Y.

Solution:

𝜇 = 𝐸 𝑌 = 𝑦.∞

−∞

𝑓 𝑦 𝑑𝑦 = 𝑦.0

−1

0.2 𝑑𝑦 + 𝑦.1

0

0.2 + 1.2 𝑦 𝑑𝑦 = 𝟎. 𝟒

𝜎2 = 𝐸 𝑌2 − 𝜇2

𝐸 𝑌2 = 𝑦2.0

−1

0.2 𝑑𝑦 + 𝑦2.1

0

0.2 + 1.2 𝑦 𝑑𝑦 = 0.4333

𝜎2 = 𝐸 𝑌2 − 𝜇2 = 0.4333 − 0. 42 = 𝟎. 𝟐𝟕𝟑𝟑

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Assignment for Wednesday, May 27th

, 2009

4.36. If a point is randomly located in an interval (a,b) and if Y denotes the location of the point, then Y is

assumed to have a uniform distribution over (a,b). A plant efficiency expert randomly selects a location along a

500-foot assembly line from which to observe the work habits of the workers on the line. What is the probability that the point she selects

a. is whiting 25 feet of the end of the line?

b. is within 25 feet of the beginning of the line? c. is closer to the beginning of the line that to the end of the line?

Solution: According to the definition of uniform probability distribution, a random variable Y is said to have a

continuous uniform probability distribution on the interval (𝜃1, 𝜃2) if and only if the density function of Y is:

𝑓 𝑦 =1

𝜃2−𝜃1 𝜃1<y<𝜃2

Here we have: 𝜃1 = 0 and 𝜃2 = 500, therefore 𝑓 𝑦 =1

500

Solution a: Probability that the selected point is within 25 of the end of the line means P(475<y<500), so we have:

= 𝑓 𝑦 500

475

𝑑𝑦 = 1

500

500

475

𝑑𝑦 = 1 −475

500= 0.05

Solution b:

Probability that the selected point is within 25 of the beginning of the line means P(0<y<25), so we have:

= 𝑓 𝑦 25

0

𝑑𝑦 = 1

500

25

0

𝑑𝑦 =25

500= 0.05

Solution c: P(0<y<250),

= 𝑓 𝑦 250

0

𝑑𝑦 = 1

500

250

0

𝑑𝑦 = 250/500 = 0.5

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4.38. Beginning at 12:00 midnight, a computer center is up for 1 hour and then down for 2 hours on a regular

cycle. A person who is unaware of this schedule dials the center at a random time between 12:00 midnight and 5:00 AM. What is the probability that the center is up when the person’s call comes in?

Solution:

Let 12:00 midnight be 𝜃1 = 0, then we will have 𝜃2 = 5, therefore 𝑓 𝑦 =1

5

Beginning at 𝜃1 = 0, the computer is up for 1 hour, so: P(0<Y<1)

The computer is down for 2 hours and then up for another one hour, so P(3<Y<4)

The probability that the center is up when the person call is: P(0<Y<1) + P(3<Y<4)

= 𝑓 𝑦 1

0

𝑑𝑦 + 𝑓 𝑦 4

3

𝑑𝑦

= 1

5

1

0

𝑑𝑦 + 1

5

4

3

𝑑𝑦 =1

5+

4

5−

3

5=

𝟐

𝟓

4.49. A company that manufactures and bottles apple juice uses a machine that automatically fills 16-ounces bottles. There is some variation, however, in the amounts of liquid dispensed into the bottles that are filled. The

amount dispensed has been observed to be approximately normally distributed with mean 16 ounces and

standard deviation 1 ounce. What portion of bottles will have more than 17 ounces dispenses into them?

Solution:

Mean = 16

Standard deviation=1

Transforming normal random variable Y to a standard normal random variable Z:

𝑧 =𝑦 − 𝜇

𝜎=

17 − 16

1= 1

Using Table 4, page 792, P(Z>1)=0.1587

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4.57. Wires manufacture for use in a computer system are specified to have resistance of between 0.12 and 0.14 ohms. The actual measured resistances of the wires produced by company A have a normal probability

distribution with mean 0.13 ohm and standard deviation 0.005 ohm.

a. What is the probability that a randomly selected wire from company A’s production will meet the

specifications?

b. If four of these wires are used in each computer system and all are selected from company A, what is

the probability that all four in a randomly selected system will meet the specifications?

Solution a:

Mean=0.13 SD=0.005

Probability to meet the specification, P(0.12<Z<0.14)

Transforming

𝑧1 =𝑦 − 𝜇

𝜎=

0.12 − 0.13

0.005= −2

𝑧2 =𝑦 − 𝜇

𝜎=

0.14 − 0.13

0.005= 2

Using table 4;

P(-2<Z<2)=2[0.5-P(Z>2)]=2[0.5-0.0228]=0.9544

Solution b:

Probability that all four meet the specifications is (0.9544)4=0.829

4.61. A soft-drink machine can be regulated so that it discharges and average of 𝜇 ounces per cup. If the ounces

of fill are normally distributed with standard deviation 0.3 ounce, give the setting for 𝜇 so that 8-ounce cups will

overflow only 1% of the time.

Solution:

Y is a normal random variable with (𝜇, 𝜎 = 0.3)

From table 4.1, P(Z>z)=0.01, so z should be 2.33.

𝑧 =𝑦 − 𝜇

𝜎=

8 − 𝜇

0.3= 2.33

Therefore 𝝁 = 𝟕. 𝟑𝟎𝟏

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Assignment for Thursday, May 28th

, 2009

4.74. One-hour carbon monoxide concentrations in air samples from a large city have an approximately

exponential distribution with mean 3.6 parts per million.

a. Find the probability that the carbon monoxide concentration exceeds 9 parts per million during a

randomly selected 1-hour period.

b. A traffic control strategy reduced the mean to 2.5 parts per million. Now find the probability that the

concentration exceeds 9 parts per million.

Solution

This is an exponential distribution with Mean= 𝛽= 3.6. The density function of random variable Y in

exponential distribution is:

𝑓 𝑦 =

1

𝛽𝑒

−𝑦𝛽 , 0 ≤ 𝑦 < ∞

0, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒

Solution a:

Probability of exceeding 9 means P(Y>9), therefore we will have:

𝑃 𝑌 > 9 = 1

𝛽𝑒

−𝑦𝛽

∞

9

𝑑𝑦 = 1

3.6𝑒−

𝑦3.6

∞

9

𝑑𝑦 = −𝑒−𝑦

3.6 9

∞

= 0.0821

Solution b:

𝛽= 2.5

𝑃 𝑌 > 9 = 1

𝛽𝑒

−𝑦𝛽

∞

9

𝑑𝑦 = 1

2.5𝑒−

𝑦2.5

∞

9

𝑑𝑦 = −𝑒−𝑦

2.5 9

∞

= 0.0273

4.76. Suppose that a random variable Y has a probability density function given by

𝑓 𝑦 = 𝑘𝑦3𝑒−𝑦2 , 𝑦 > 0


Find the value of k that makes f(y) a density function.

Solution:

From definition of gamma distribution:

A random variable Y is said to have gamma distribution with parameters 𝛼 > 0 and 𝛽 > 0 if and only if the

density function of Y is:

𝑓 𝑦 = 𝑦𝛼−1𝑒−𝑦/𝛽

𝛽𝛼Γ(𝛼) 0 ≤ 𝑦 < ∞


Comparing this with the given probability density function;

𝛼 − 1 = 3 => 𝛼 = 4

𝛽 = 2

Γ 𝑛 = 𝑛 − 1 ! => Γ 4 = 6

𝛽𝛼 = 24 = 16 Therefore:

𝑘 =1

𝛽𝛼Γ(𝛼)=

1

6 × 16=

𝟏

𝟗𝟔

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4.78. Consider the plant of exercise 4.77. How much of the bulk product should be stocked so that the plant’s

chance of running out of the product is only 0.05?

Solution:

From problem 4.77, the amount of product used in one day can be modeled by an exponential distribution with

𝛽 = 4.

Plant’s chance of running out of the product is 0.05, which means P(Y>x)=0.05

P Y > 𝑥 = 1

𝛽𝑒

−𝑦𝛽

∞

𝑥

𝑑𝑦 = 1

4𝑒−

𝑦4

∞

𝑥

𝑑𝑦 = −𝑒−𝑦4

𝑥

∞

= 𝑒−𝑥4 = 0.05

𝐿𝑛(𝑒−𝑥4) = 𝐿𝑛(0.05)

−𝑥

4= −2.995

=> 𝒙 = 𝟏𝟏. 𝟗𝟖𝟐

4.88. If Y has a probability density function given by

𝑓 𝑦 = 4𝑦2𝑒−2𝑦 , 𝑦 > 0


Obtain E(Y) and V(Y) by inspection.

Solution:

Comparing the given density function with gamma distribution function, 𝑦𝛼−1𝑒−𝑦 /𝛽

𝛽𝛼Γ(𝛼)

𝛼 − 1 = 2 => 𝜶 = 𝟑

𝜷 =𝟏

𝟐

According to Theorem 4.8, the mean and variance of gamma distribution is:

𝜇 = 𝐸 𝑌 = 𝛼𝛽

𝜎2 = 𝑉 𝑌 = 𝛼𝛽2

Therefore:

𝜇 = (3) 1

2 =

𝟑

𝟐

𝑉 𝑌 = 3 . 1

2

2

=𝟑

𝟒

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4.101. The proportion of time per day that all checkout counters in a supermarket are busy is a random variable

Y with a density function given by

𝑓 𝑦 = 𝑐𝑦2 1 − 𝑦 4, 0 ≤ 𝑦 ≤ 10, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒.

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

b. Find E(Y).

Solution a:

A random variable Y is said to have a beta probability distribution with parameters 𝛼 > 0 and 𝛽 > 0 if and only if the density function of Y is:

𝑓 𝑦 = 𝑦𝛼−1 1 − 𝑦 𝛽−1

𝐵(𝛼, 𝛽), 0 ≤ 𝑦 ≤ 1


𝐵 𝛼, 𝛽 =Γ α Γ(β)

Γ(α + β)

Comparing the given density function with beta probability distribution function:

𝛼 − 1 = 2 => 𝛼 = 3

𝛽 − 1 = 4 => 𝛽 = 5 and

𝑐 =1

𝐵 𝛼, 𝛽 =

Γ α + β

Γ α Γ β

=> 𝑐 =1

𝐵 3,5 =

Γ(3 + 5)

Γ 3 Γ(5)=

7.6.5

2=

210

2= 𝟏𝟎𝟓

Knowing that Γ 𝑛 = 𝑛 − 1 !

Solution b: According to Theorem 4.11, the mean of beta distribution is:

𝜇 = 𝐸 𝑌 =𝛼

𝛼 + 𝛽=

3

8= 𝟎. 𝟑𝟕𝟓

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Assignment for Friday, May 29th

, 2009

4.104. Suppose that the waiting time for the first customer to enter a retail shop after 9:00 AM is a random

variable Y with an exponential density function given by

𝑓 𝑦 = 1

𝜃 𝑒−𝑦/𝜃 , 𝑦 > 0


a. Find the moment-generating function for Y.

b. Use the answer from (a) to find E(Y) and V(Y).

Solution a:

As an special case of gamma-distributed random variables, the gamma density function in which 𝛼 = 1 is

called the exponential density function. A random variable is said to have an exponential distribution with

parameter 𝛽 > 0 if and only if the density function Y is:

𝑓 𝑦 =

1

𝛽𝑒

−𝑦𝛽 , 0 ≤ 𝑦 < ∞

0, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒

Comparing the given density function with exponential density function: 𝛽 = 𝜃

The moment-generating function for a gamma distributed random variable, according to example 4.13, page

190 of the text book is:

𝑚 𝑡 =1

1 − 𝛽𝑡 𝛼

Replacing 𝛽 = 𝜃 and 𝛼 = 1, we will have:

𝒎 𝒕 =𝟏

𝟏 − 𝜽𝒕

Solution b:

𝐸 𝑌 = 𝜇 = 𝑑𝑚(𝑡)

𝑑𝑡 𝑡=0

= 𝜃

1 − 𝜃𝑡 2 𝑡=0

= 𝜽

𝐸 𝑌2 = 𝑑2𝑚(𝑡)

𝑑𝑡2 𝑡=0

= 2𝜃2

1 − 𝜃𝑡 3 𝑡=0

= 2𝜃2

𝑉 𝑌 = 𝐸 𝑌2 − 𝜇2 = 2𝜃2 − 𝜃2 = 𝜽𝟐

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4.107. The moment generating function of a normally distributed random variable, Y, with mean 𝜇 and variance

𝜎2 was shown in Exercise 4.106 to be 𝑚 𝑡 = 𝑒𝜇𝑡 +1

2𝑡2𝜎2

. Use the result in Exercise 4.105 to derive the moment

generating function of = −3𝑌 + 4 . What is the distribution of X? Why?

Solution:

Solving exercise 4.105 for 𝑈 = −3𝑌 + 4, (𝑎 = −3, 𝑏 = 4)

According to the theorem 4.12, the moment generating function of g(Y) is given by:

𝑚 𝑡 = 𝐸 𝑒𝑡𝑔 𝑌 Let 𝑔 𝑦 = 𝑋 = −3𝑌 + 4, then:

𝑚𝑋 𝑡 = 𝐸 𝑒(−3𝑌+4)𝑡 = 𝐸 𝑒−3𝑡𝑌+4𝑡 = 𝑒4𝑡𝐸 𝑒−3𝑡𝑌

=> 𝒎𝑿 𝒕 = 𝒆𝟒𝒕𝒎𝒀(−𝟑𝒕)

Solving exercise 4.106,

According to Example 4.16, if we let 𝑔 𝑦 = 𝑌 − 𝜇, where Y is a normally distributed random variable with

mean 𝜇 and variance 𝜎2, the moment generating function for g(Y) is:

𝑚𝑌−𝜇 𝑡 = 𝑒 𝑡2

2 𝜎2

In the other words,

𝑚𝑌−𝜇 𝑡 = 𝐸 𝑒(𝑌−𝜇)𝑡 = 𝐸 𝑒𝑌𝑡−𝜇𝑡 = 𝑒−𝜇𝑡 𝐸 𝑒𝑡𝑌 = 𝒆−𝜇𝒕𝒎𝒀(𝒕)

Therefore we have:

𝑒 𝑡2

2 𝜎2

= 𝑒−𝜇𝑡 𝑚𝑌(𝑡)

Which leads to:

𝒎𝒀 𝒕 = 𝒆𝝁𝒕. 𝒆 𝒕𝟐

𝟐 𝝈𝟐

Or

𝒎𝒀 𝒕 = 𝒆𝝁𝒕+

𝒕𝟐

𝟐 𝝈𝟐

Now, having 𝑚𝑋 𝑡 = 𝑒4𝑡𝑚𝑌(−3𝑡)

𝑚𝑋 𝑡 = 𝑒4𝑡𝑒𝜇 (−3𝑡)+

(−3𝑡)2

2 𝜎2

=> 𝒎𝑿 𝒕 = 𝒆(𝟒−𝟑𝝁)𝒕𝒆 𝒕𝟐

𝟐 𝟗𝝈𝟐

X has a normal distribution with mean equal to 4 − 3𝜇 and variance of 9𝜎2.

This is because of the uniqueness of moment generating function.

sta_5325_hw_2_ramin_shamshiri

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