ENGG 319
Assignment #2
Question 1:
Thickness measurements of a coating process are made to the nearest hundredth of a millimeter.
The thickness measurements are uniformly distributed with values 0.15, 0.16, 0.17, 0.18, and 0.19.
Determine the mean and variance of the coating thickness for this process.
Solution:
Equation:
π₯π + π₯0
ππππ(π) =
;
2
(π₯π − π₯0 + 1)2 − 1
ππππππππ(π 2 ) =
12
π₯π = 0.19; π₯0 = 0.15
0.34
ππππ(π) =
= 0.17
2
(1.4)2 − 1
ππππππππ(π 2 ) =
= 0.08
12
Question 2:
Consider the endothermic reactions in Exercise 3-28. A total of 20 independent reactions
are to be conducted.
48 + 60
108
π(< 272 πΎ) =
=
= 0.54
48 + 60 + 92 200
π
π(π = π₯) = ( ) (1 − π)π−π₯ π π₯
π₯
(a) What is the probability that exactly 12 reactions result in a final temperature less than 272
Solution:
K?
π₯ = 12; π = 20;
20
π(π = 12) = ( ) (0.46)8 ∗ (0.54)12 = 0.155
12
(b) What is the probability that at least 19 reactions result in a final temperature less than 272
Solution:
K?
π(π = 19) + π(π = 20)
20
π(π = 19) = ( ) (0.46)1 ∗ (0.54)19 = 0.0000757
19
20
π(π = 20) = ( ) (0.46)0 ∗ (0.54)20 = 0.00000823
20
π(π ≥ 19) = 0.0000757 + 0.00000823 = 0.00008393
(c) What is the probability that at least 18 reactions result in a final temperature less than 272
Solution:
K?
20
π(π = 18) = ( ) (0.46)2 ∗ (0.54)18 = 0.0006129
18
π(π ≥ 18) = π(π ≥ 19) + π(π = 18) = 0.00008393 + 0.0006129 = 0.0006968
(d) What is the expected number of reactions result in a final temperature less than 272
Solution:
K?
πΈπ₯ππππ‘ππ ππππ’π = ππππ(π) = ππ = 20 ∗ 0.54 = 10.8 πππππ‘ππππ
Question 3:
Suppose the random variable X has a geometric distribution with a mean of 2.5. Determine the
following probabilities:
1
1
= 2.5 … ∴ π =
= 0.4
π
2.5
π₯−1
π(π = π₯) = (1 − π)
∗π
ππππ(π) =
(a) π(π = 1)
Solution:
π(π = 1) = (0.6)0 ∗ (0.4)1 = 0.4
(b) π(π = 4)
Solution:
π(π = 4) = (0.6)3 ∗ (0.4)1 = 0.0864
(c) π(π = 5)
Solution:
π(π = 5) = (0.6)4 ∗ (0.4)1 = 0.05184
π(π ≤ 3)
(d)
Solution:
π(π
π(π
π(π
π(π
π(π
π(π
≤ 3) = π(π = 1) + π(π = 2) + π(π = 3)
= 1) = 0.4
= 2) = (0.6)1 (0.4)1 = 0.24
= 3) = (0.6)2 (0.4)1 = 0.144
≤ 3) = 0.4 + 0.24 + 0.144 = 0.784
> 3)
(e)
Solution:
π(π > 3) = 1 − π(π ≤ 3) = 1 − 0.784 = 0.216
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ENGG 319
Assignment #2
Question 4:
In a clinical study, volunteers are tested for a gene that has been found to increase the risk for a
disease. The probability that a person carries the gene is 0.1.
π = 0.1
π(π = π₯) = (
π₯ − 1 (1
) − π)π₯−π ππ
π−1
(a) What is the probability four or more people will have to be tested before two with the gene are detected?
Solution:
π(π ≥ 4) = 1 − π(π < 4) = π(π = 1) + π(π = 2) + π(π = 3)
π ππππ π€π ππππ π‘π€π π π’ππππ π ;
π(π ≥ 4) = 1 − π(π < 4) = π(π = 2) + π(π = 3)
1
π(π = 2) = ( ) ∗ (0.9)0 (0.1)2 = 0.01
1
2
π(π = 3) = ( ) (0.9)1 (0.1)2 = 0.018
1
π(π ≥ 4) = 1 − 0.028 = 0.972
(b) How many people are expected to be tested before two with the gene are detected?
Solution:
πΈπ₯ππππ‘ππ ππ’ππππ = ππππ(π) =
π
2
=
= 20 ππππππ
π 0.1
Question 5:
The analysis of results from a leaf transmutation experiment (turning a leaf into a petal) is summarized
by type of transformation completed:
Total Color
Transformation
Yes
No
Total Textural Transformation
Yes
No
243
26
13
18
A naturalist randomly selects three leaves from this set, without replacement. Determine the following probabilities.
π−πΎ πΎ
)( )
π(π = π₯) = π − π₯ π₯
π
( )
π
π = 300; π = 3
(
(a) Exactly one has undergone both types of transformations.
Solution:
πΎ = 243; π₯ = 1
57 243
( )(
)
1 = 0.0871
π(π = 1) = 2
300
(
)
3
(b) At least one has undergone both transformations.
Solution:
π(π ≥ 1) = 1 − π(π < 1) = 1 − π(π = 0)
57 243
( )(
)
0 = 0.006568
π(π = 0) = 3
300
(
)
3
π(π ≥ 1) = 1 − 0.006568 = 0.99342
(c) Exactly one has undergone one but not both transformations.
Solution:
πΎ = 26 + 13 = 39; π₯ = 1
261 39
(
)( )
1 = 0.297
π(π = 1) = 2
300
(
)
3
(d) At least one has undergone at least one transformation.
Solution:
πΎ = 26 + 13 + 243 = 282; π₯ ≥ 1
π(π ≥ 1) = 1 − π(π < 1) = 1 − π(π = 0)
18 282
( )(
)
0 = 0.0001832
π(π = 0) = 3
300
(
)
3
π(π ≥ 1) = 1 − 0.0001832 = 0.9998
Question 6:
Assume the number of errors along a magnetic recording surface is a Poisson random variable with a
mean of one error every 105 bits. A sector of data consists of 4096 eight-bit bytes.
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ENGG 319
Assignment #2
(a) What is the probability of more than one error in a sector?
Solution:
π −ππ‘ (ππ‘)π₯
π₯!
4096 ∗ 8
π = ππ =
= 0.32768; π₯ > 1;
105
π(π > 1) = 1 − π(π ≤ 1) = 1 − π(π = 0) − π(π = 1)
π −0.32768 ∗ 0.327680
π(π = 0) =
= 0.72059
0!
π −0.32768 ∗ 0.327680
π(π = 1) =
= 0.23612
0!
π(π > 1) = 1 − 0.72059 − 0.23612 = 0.04328
π(π₯) ==
(b) What is the mean number of sectors until an error is found?
Solution:
π(π ≥ 1) = 1 − π(π < 1) = 1 − π(π = 0)
π −0.32768 ∗ 0.327680
π(π = 0) =
= 0.72059
0!
π(π ≥ 1) = 1 − 0.72059 = 0.27941
πππππ π€π πππ πππππππ πππ π‘βπ ππππ ππ π’ππ‘ππ 1π π‘ π π’ππππ π
1
1
ππππ(π) =
=
= 3.57897 ππππ‘πππ
π(π ≥ 1) 0.27941
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