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:
𝑥𝑛 + 𝑥0
𝑀𝑒𝑎𝑛(𝜇) =
;
2
(𝑥𝑛 − 𝑥0 + 1)2 − 1
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒(𝜎 2 ) =
12
𝑥𝑛 = 0.19; 𝑥0 = 0.15
0.34
𝑀𝑒𝑎𝑛(𝜇) =
= 0.17
2
(1.4)2 − 1
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒(𝜎 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
𝑃(< 272 𝐾) =
=
= 0.54
48 + 60 + 92 200
𝑛
𝑃(𝑋 = 𝑥) = ( ) (1 − 𝑝)𝑛−𝑥 𝑝 𝑥
𝑥
(a) What is the probability that exactly 12 reactions result in a final temperature less than 272
Solution:
K?
𝑥 = 12; 𝑛 = 20;
20
𝑃(𝑋 = 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?
𝑃(𝑋 = 19) + 𝑃(𝑋 = 20)
20
𝑃(𝑋 = 19) = ( ) (0.46)1 ∗ (0.54)19 = 0.0000757
19
20
𝑃(𝑋 = 20) = ( ) (0.46)0 ∗ (0.54)20 = 0.00000823
20
𝑃(𝑋 ≥ 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
𝑃(𝑋 = 18) = ( ) (0.46)2 ∗ (0.54)18 = 0.0006129
18
𝑃(𝑋 ≥ 18) = 𝑃(𝑋 ≥ 19) + 𝑃(𝑋 = 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?
𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑉𝑎𝑙𝑢𝑒 = 𝑀𝑒𝑎𝑛(𝜇) = 𝑛𝑝 = 20 ∗ 0.54 = 10.8 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛𝑠
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 … ∴ 𝑃 =
= 0.4
𝑝
2.5
𝑥−1
𝑃(𝑋 = 𝑥) = (1 − 𝑝)
∗𝑝
𝑀𝑒𝑎𝑛(𝜇) =
(a) 𝑃(𝑋 = 1)
Solution:
𝑃(𝑋 = 1) = (0.6)0 ∗ (0.4)1 = 0.4
(b) 𝑃(𝑋 = 4)
Solution:
𝑃(𝑋 = 4) = (0.6)3 ∗ (0.4)1 = 0.0864
(c) 𝑃(𝑋 = 5)
Solution:
𝑃(𝑋 = 5) = (0.6)4 ∗ (0.4)1 = 0.05184
𝑃(𝑋 ≤ 3)
(d)
Solution:
𝑃(𝑋
𝑃(𝑋
𝑃(𝑋
𝑃(𝑋
𝑃(𝑋
𝑃(𝑋
≤ 3) = 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) + 𝑃(𝑋 = 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:
𝑃(𝑋 > 3) = 1 − 𝑃(𝑋 ≤ 3) = 1 − 0.784 = 0.216
Page 1 of 3
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.
𝑝 = 0.1
𝑃(𝑋 = 𝑥) = (
𝑥 − 1 (1
) − 𝑝)𝑥−𝑟 𝑝𝑟
𝑟−1
(a) What is the probability four or more people will have to be tested before two with the gene are detected?
Solution:
𝑃(𝑋 ≥ 4) = 1 − 𝑃(𝑋 < 4) = 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) + 𝑃(𝑋 = 3)
𝑠𝑖𝑛𝑐𝑒 𝑤𝑒 𝑛𝑒𝑒𝑑 𝑡𝑤𝑜 𝑠𝑢𝑐𝑐𝑒𝑠𝑠;
𝑃(𝑋 ≥ 4) = 1 − 𝑃(𝑋 < 4) = 𝑃(𝑋 = 2) + 𝑃(𝑋 = 3)
1
𝑃(𝑋 = 2) = ( ) ∗ (0.9)0 (0.1)2 = 0.01
1
2
𝑃(𝑋 = 3) = ( ) (0.9)1 (0.1)2 = 0.018
1
𝑃(𝑋 ≥ 4) = 1 − 0.028 = 0.972
(b) How many people are expected to be tested before two with the gene are detected?
Solution:
𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑁𝑢𝑚𝑏𝑒𝑟 = 𝑀𝑒𝑎𝑛(𝜇) =
𝑟
2
=
= 20 𝑝𝑒𝑜𝑝𝑙𝑒
𝑝 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.
𝑁−𝐾 𝐾
)( )
𝑃(𝑋 = 𝑥) = 𝑛 − 𝑥 𝑥
𝑁
( )
𝑛
𝑁 = 300; 𝑛 = 3
(
(a) Exactly one has undergone both types of transformations.
Solution:
𝐾 = 243; 𝑥 = 1
57 243
( )(
)
1 = 0.0871
𝑃(𝑋 = 1) = 2
300
(
)
3
(b) At least one has undergone both transformations.
Solution:
𝑃(𝑋 ≥ 1) = 1 − 𝑃(𝑋 < 1) = 1 − 𝑃(𝑋 = 0)
57 243
( )(
)
0 = 0.006568
𝑃(𝑋 = 0) = 3
300
(
)
3
𝑃(𝑋 ≥ 1) = 1 − 0.006568 = 0.99342
(c) Exactly one has undergone one but not both transformations.
Solution:
𝐾 = 26 + 13 = 39; 𝑥 = 1
261 39
(
)( )
1 = 0.297
𝑃(𝑋 = 1) = 2
300
(
)
3
(d) At least one has undergone at least one transformation.
Solution:
𝐾 = 26 + 13 + 243 = 282; 𝑥 ≥ 1
𝑃(𝑋 ≥ 1) = 1 − 𝑃(𝑋 < 1) = 1 − 𝑃(𝑋 = 0)
18 282
( )(
)
0 = 0.0001832
𝑃(𝑋 = 0) = 3
300
(
)
3
𝑃(𝑋 ≥ 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.
Page 2 of 3
ENGG 319
Assignment #2
(a) What is the probability of more than one error in a sector?
Solution:
𝑒 −𝜆𝑡 (𝜆𝑡)𝑥
𝑥!
4096 ∗ 8
𝜆 = 𝑛𝑝 =
= 0.32768; 𝑥 > 1;
105
𝑃(𝑋 > 1) = 1 − 𝑃(𝑋 ≤ 1) = 1 − 𝑃(𝑋 = 0) − 𝑃(𝑋 = 1)
𝑒 −0.32768 ∗ 0.327680
𝑃(𝑋 = 0) =
= 0.72059
0!
𝑒 −0.32768 ∗ 0.327680
𝑃(𝑋 = 1) =
= 0.23612
0!
𝑃(𝑋 > 1) = 1 − 0.72059 − 0.23612 = 0.04328
𝑓(𝑥) ==
(b) What is the mean number of sectors until an error is found?
Solution:
𝑃(𝑋 ≥ 1) = 1 − 𝑃(𝑋 < 1) = 1 − 𝑃(𝑋 = 0)
𝑒 −0.32768 ∗ 0.327680
𝑃(𝑋 = 0) =
= 0.72059
0!
𝑃(𝑋 ≥ 1) = 1 − 0.72059 = 0.27941
𝑆𝑖𝑛𝑐𝑒 𝑤𝑒 𝑎𝑟𝑒 𝑙𝑜𝑜𝑘𝑖𝑛𝑔 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑚𝑒𝑎𝑛 𝑜𝑓 𝑢𝑛𝑡𝑖𝑙 1𝑠𝑡 𝑠𝑢𝑐𝑐𝑒𝑠𝑠
1
1
𝑚𝑒𝑎𝑛(𝜇) =
=
= 3.57897 𝑆𝑒𝑐𝑡𝑜𝑟𝑠
𝑃(𝑋 ≥ 1) 0.27941
Page 3 of 3