Name : Jericho Hyansalem Wicaksono NRP : 5019201004
Course : Ship Maintenance Lecturer : Ir. Dwi Priyanta, MSE
Assignment : Workbook 3 – Probability Distribution
1. Let us consider a power plant with three generators supplying 250-MW load. Data on the individual units, the possible states and their probabilities, the MW capacities in and out, and the corresponding losses of load are listed in the following table.
Determine the expected power loss of the above generators configuration.
Answer:
Based on the calculation in the table above, it can be said that for unit 1, the expected load loss is 10 MW. For unit 2, the expected load loss is 3 MW. For unit 3, the expected load loss is 6 MW. Hence, the expected total power loss above the generators is 10 + 3 + 6 (MW)
= 19 MW. However, it must be known that the loss is 19 MW out of 450 MW.
2. The tensile strength of paper used to make grocery bags is an important quality characteristics. It is known that the strength, say X, is normally distributed with mean 𝝁 = 𝟒𝟎 𝒍𝒃/𝒊𝒏𝟐 and standard deviation 𝝈 = 𝟐 𝒍𝒃/𝒊𝒏𝟐, denoted as X ~ N (40.2). The purchaser of the bags requires them to have strength of at least 𝟑𝟓 𝒍𝒃/𝒊𝒏𝟐. Determine the probability that a bag produced from this paper will meet or exceed this specification.
Answer:
This problem is related to “Standard Normal Random Variable” concept. Hence, we can solve this question as follows:
0 0.99 0 -
100 0.1 100 10
10 MW
0 0.98 0 -
150 0.02 150 3
3 MW
0 0.97 0 -
200 0.03 200 6 MW
6 MW Load Loss Expected
Load Loss Unit 1
(100 MW) Unit 2 (150 MW)
Unit 3 (200 MW)
No. Capacity
Out Probability
𝜇 = 40 𝑙𝑏/𝑖𝑛2 𝜎 = 2 𝑙𝑏/𝑖𝑛2 𝑋 = 35 𝑙𝑏/𝑖𝑛2
Let us say that X is the strength of the paper bags. Hence. The probability that the paper bags will meet or exceed the specification is:
𝑃(𝑋 ≥ 35) = 1 − (𝑃(𝑋 < 35))
If X is a normal random variable with mean 𝜇 and variance 𝜎2, then Z is a standard normal random variable and it is obtained by standardizing X. Hence, the value of Z is:
𝑍 =(𝑋 − 𝜇) 𝜎 𝑍 =(35 − 40)
2 = −2.5
𝑃(𝑋 < 35) = 𝑃(𝑍 < −2.5) = 𝜙(−2.5) And, based on standard normal cumulative probability table,
We take 𝑍 = −2.51, because the probability value we want to find is less than -2.5. Hence, 𝑃(𝑋 < 35) = 𝑃(𝑍 < −2.5) = 0.0060
So,
𝑃(𝑋 ≥ 35) = 1 − (0.0060) 𝑃(𝑋 ≥ 35) = 0.994
Therefore, the probability that the paper bags will meet or exceed the specification is 0.994 or 99.4% chance.
3. The diameter of metal shaft used in a disk drive unit is normally distributed with mean 0.2508 in and standard deviation 0.0005 in. The specifications on the shaft have been established as 0.2500 ± 0.0015 in. We wish to determine what fraction of the shafts produced conform to specifications.
Answer:
Let P(S) the probability of the metal shaft conform to specifications. The specifications on the shaft have been established as 0.2500 ± 0.0015 in. Hence, it can be written as:
𝑃(𝑆) = 𝑃(0.2485 ≤ 𝑋 ≤ 0.2515) 𝑃(𝑆) = 𝑃(𝑋 ≤ 0.2515) − 𝑃(𝑋 ≥ 0.2485) We know that,
𝑧 = (𝑥 − 𝜇) 𝜎 𝑃(𝑍 < 𝑧) = 𝜙(𝑧) Hence,
𝑃(𝑋 ≤ 0.2515) = 𝑃 (𝑍 ≤0.2515 − 0.2508 0.0005 ) 𝑃(𝑋 ≤ 0.2515) = 𝑃(𝑍 ≤ 1.4)
𝑃(𝑋 ≤ 0.2515) = 𝑃(𝑍 ≤ 1.4) = 𝜙(1.4) And
𝑃(𝑋 ≥ 0.2485) = 𝑃 (𝑍 ≥0.2485 − 0.2508 0.0005 ) 𝑃(𝑋 ≥ 0.2485) = 𝑃(𝑍 ≥ −4.6) 𝑃(𝑋 ≥ 0.2485) = 𝑃(𝑍 ≥ −4.6) = 𝜙(−4.6)
Therefore, based on the standard normal cumulative probability table for both positive and negative z-values, the P(S) is:
𝑃(𝑆) = 𝜙(1.4) − 𝜙(−4.6) 𝑃(𝑆) = 0.9192 − 0.0000
𝑃(𝑆) = 0.9192
In conclusion, the fraction of the shafts produced conform to specifications is 0.9192 in probability or 91.92% chance.