SOLUTION: A production line of 5000 transistors has an unknown number D of defectives. A sample of 100 is taken and we ascertain that this sample of transistors has 10 defectives. The maximu

Question 1174433: A production line of 5000 transistors has an unknown number D of defectives. A sample of 100 is taken and we ascertain that this sample of transistors has 10 defectives. The maximum likelihood estimate for D is the number of defectives which gives the highest probability for obtaining the number of defectives observed in the sample. Guess this number D.
Answer by CPhill(2189) (Show Source): You can put this solution on YOUR website!
Let $N = 5000$ be the total number of transistors.
Let $D$ be the unknown number of defective transistors.
We take a sample of $n = 100$ transistors.
We observe $d = 10$ defectives in the sample.
We want to find the maximum likelihood estimate for $D$.
The probability of observing $d$ defectives in a sample of $n$ is given by the hypergeometric distribution:
$$P(X=d) = \frac{\binom{D}{d} \binom{N-D}{n-d}}{\binom{N}{n}}$$
We want to find the value of $D$ that maximizes this probability.
We can approximate the hypergeometric distribution with the binomial distribution when $n$ is much smaller than $N$.
In our case, $n=100$ and $N=5000$, so $n/N = 100/5000 = 1/50 = 0.02$, which is small.
So we can use the binomial approximation.
The probability of a transistor being defective is $p = D/N$.
The probability of observing $d$ defectives in a sample of $n$ is:
$$P(X=d) = \binom{n}{d} p^d (1-p)^{n-d}$$
We want to maximize this probability with respect to $D$.
Since $p = D/N$, we have:
$$P(X=d) = \binom{n}{d} \left(\frac{D}{N}\right)^d \left(1-\frac{D}{N}\right)^{n-d}$$
To find the maximum likelihood estimate, we can take the derivative of the log-likelihood function with respect to $D$ and set it to zero.
The log-likelihood function is:
$$L(D) = \ln\left(\binom{n}{d}\right) + d\ln\left(\frac{D}{N}\right) + (n-d)\ln\left(1-\frac{D}{N}\right)$$
Taking the derivative with respect to $D$:
$$\frac{dL}{dD} = \frac{d}{D} - \frac{n-d}{N-D}$$
Setting the derivative to zero:
$$\frac{d}{D} = \frac{n-d}{N-D}$$
$$d(N-D) = D(n-d)$$
$$dN - dD = nD - dD$$
$$dN = nD$$
$$D = \frac{dN}{n}$$
Plugging in the given values:
$$D = \frac{10 \times 5000}{100}$$
$$D = \frac{50000}{100}$$
$$D = 500$$
Therefore, the maximum likelihood estimate for $D$ is 500.
Final Answer: The final answer is $\boxed{500}$

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