A manufacturer of liquid crystal displays (LCDs) is studying their production lines. The probability of sampling a defective LCD is 0.1. A sample of 5 LCDs is taken. You may assume that an LCD being defective is independent of any of the others being defective.
What is the expected value for the number of defective LCDs?
The expected value for the number of defective LCDs can be calculated using the formula:
Expected Value (E) = n * p
Where:
- n is the number of LCDs sampled (in this case, 5)
- p is the probability of sampling a defective LCD (in this case, 0.1)
Using the formula, we can calculate the expected value:
E = 5 * 0.1
E = 0.5
Therefore, the expected value for the number of defective LCDs is 0.5.
To find the expected value for the number of defective LCDs, we will multiply each possible number of defective LCDs by its corresponding probability and then sum them up.
Let's denote X as the number of defective LCDs. Since each LCD being defective is independent of any of the others being defective, we can use the Bernoulli distribution.
Given that the probability of sampling a defective LCD is 0.1 (or 10%), we can say that the probability of an LCD being non-defective is 1 - 0.1 = 0.9 (or 90%).
The probability mass function (PMF) for the number of defective LCDs is given by the binomial distribution:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
where n is the total number of LCDs sampled (in this case, 5), k is the number of defective LCDs, and p is the probability of sampling a defective LCD (0.1 or 10%).
Let's calculate the expected value:
E(X) = sum(k * P(X = k)) for k = 0 to 5
E(X) = (0 * P(X = 0)) + (1 * P(X = 1)) + (2 * P(X = 2)) + (3 * P(X = 3)) + (4 * P(X = 4)) + (5 * P(X = 5))
Now let's calculate each term:
P(X = 0) = (5 choose 0) * (0.1)^0 * (0.9)^(5-0) = 1 * 1 * 0.9^5 = 0.59049
P(X = 1) = (5 choose 1) * (0.1)^1 * (0.9)^(5-1) = 5 * 0.1 * 0.9^4 = 0.32805
P(X = 2) = (5 choose 2) * (0.1)^2 * (0.9)^(5-2) = 10 * 0.01 * 0.9^3 = 0.0729
P(X = 3) = (5 choose 3) * (0.1)^3 * (0.9)^(5-3) = 10 * 0.001 * 0.9^2 = 0.0081
P(X = 4) = (5 choose 4) * (0.1)^4 * (0.9)^(5-4) = 5 * 0.0001 * 0.9^1 = 0.00045
P(X = 5) = (5 choose 5) * (0.1)^5 * (0.9)^(5-5) = 1 * 0.00001 * 0.9^0 = 0.00001
Now let's substitute the values into the formula for expected value:
E(X) = (0 * 0.59049) + (1 * 0.32805) + (2 * 0.0729) + (3 * 0.0081) + (4 * 0.00045) + (5 * 0.00001)
E(X) = 1.64466
Therefore, the expected value for the number of defective LCDs is approximately 1.64.
To find the expected value for the number of defective LCDs, we need to multiply the probability of each possible outcome (number of defective LCDs) by its corresponding value, and then sum up all these products.
In this case, the possible outcomes are: 0 defective LCDs, 1 defective LCD, 2 defective LCDs, 3 defective LCDs, 4 defective LCDs, and 5 defective LCDs.
Let's calculate the expected value step by step:
1. Probability of 0 defective LCDs:
The probability of not sampling a defective LCD is (1 - 0.1) = 0.9. Since sampling is independent, the probability of all 5 LCDs being non-defective is 0.9^5 = 0.59049. The expected value for 0 defective LCDs is 0 * 0.59049 = 0.
2. Probability of 1 defective LCD:
The probability of sampling 1 defective LCD is calculated as follows:
Choose 1 out of the 5 LCDs to be defective: 5C1 = 5
Multiply it by the probability of getting 1 defective LCD and 4 non-defective LCDs: 0.1^1 * 0.9^4 = 0.0371
So, the expected value for 1 defective LCD is 1 * 0.0371 = 0.0371.
3. Probability of 2 defective LCDs:
The probability of sampling 2 defective LCDs is calculated as follows:
Choose 2 out of the 5 LCDs to be defective: 5C2 = 10
Multiply it by the probability of getting 2 defective LCDs and 3 non-defective LCDs: 0.1^2 * 0.9^3 = 0.0081
So, the expected value for 2 defective LCDs is 2 * 0.0081 = 0.0162.
4. Probability of 3 defective LCDs:
Following the same logic, the probability of sampling 3 defective LCDs is calculated as follows:
Choose 3 out of the 5 LCDs to be defective: 5C3 = 10
Multiply it by the probability of getting 3 defective LCDs and 2 non-defective LCDs: 0.1^3 * 0.9^2 = 0.000729
So, the expected value for 3 defective LCDs is 3 * 0.000729 = 0.002187.
5. Probability of 4 defective LCDs:
Using the same method, the probability of sampling 4 defective LCDs is:
Choose 4 out of the 5 LCDs to be defective: 5C4 = 5
Multiply it by the probability of getting 4 defective LCDs and 1 non-defective LCD: 0.1^4 * 0.9^1 = 0.0006561
So, the expected value for 4 defective LCDs is 4 * 0.0006561 = 0.002624.
6. Probability of 5 defective LCDs:
Finally, the probability of sampling 5 defective LCDs is:
Choose all 5 LCDs to be defective: 5C5 = 1
Multiply it by the probability of getting 5 defective LCDs and 0 non-defective LCD: 0.1^5 * 0.9^0 = 0.00001
So, the expected value for 5 defective LCDs is 5 * 0.00001 = 0.00005.
To find the overall expected value, we sum up the expected values for each possible outcome:
Expected value = 0 + 0.0371 + 0.0162 + 0.002187 + 0.002624 + 0.00005 = 0.057161.
Therefore, the expected value for the number of defective LCDs is approximately 0.057.