A quality control inspector has drawn a sample of 2020 light bulbs from a recent production lot. If the number of defective bulbs is 11 or less, the lot passes inspection. Suppose 20%20% of the bulbs in the lot are defective. What is the probability that the lot will pass inspection? Round your answer to four decimal places.




This problem can be modeled using a binomial distribution, where each light bulb is either defective or not defective. The probability of a defective bulb is given as 20% (or 0.2), and the probability of a non-defective bulb is 80% (or 0.8).

The number of bulbs sampled is \( n = 2020 \), and the probability of finding a defective bulb is \( p = 0.2 \).

We are asked to find the probability that the number of defective bulbs, \( X \), is 11 or fewer. This is the cumulative probability \( P(X \leq 11) \).

The probability of exactly \( k \) defective bulbs in a sample of \( n \) can be calculated using the binomial probability formula:

\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]

To compute \( P(X \leq 11) \), we sum the probabilities from \( X = 0 \) to \( X = 11 \). However, since \( n = 2020 \) is large, using the binomial distribution directly can be computationally intensive. Instead, we can approximate the binomial distribution with a normal distribution.

### Step 1: Normal Approximation to the Binomial Distribution
The binomial distribution can be approximated by a normal distribution with mean \( \mu \) and standard deviation \( \sigma \), where:

\[
\mu = n \cdot p = 2020 \cdot 0.2 = 404
\]
\[
\sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{2020 \cdot 0.2 \cdot 0.8} = \sqrt{323.2} \approx 17.98
\]

### Step 2: Applying the Continuity Correction
To approximate the probability \( P(X \leq 11) \), we apply the continuity correction. We want to calculate the probability:

\[
P(X \leq 11.5)
\]

This can be translated into the corresponding z-score for the normal distribution:

\[
z = \frac{11.5 - \mu}{\sigma} = \frac{11.5 - 404}{17.98} \approx -21.84
\]

### Step 3: Calculating the Probability
For a z-score of -21.84, the cumulative probability is effectively 0. This means that the probability of having 11 or fewer defective bulbs is extremely small.

Thus, the probability that the lot will pass inspection is approximately:

\[
\boxed{0.0000}
\]

Would you like further details or have any questions?

---

Here are 5 related questions:
1. How is the continuity correction applied in binomial to normal approximations?
2. What are the key conditions for using the normal approximation to the binomial distribution?
3. How would the probability change if the number of bulbs sampled was smaller?
4. How can we use the exact binomial formula for smaller values of \(n\)?
5. How does changing the defect rate affect the probability of passing inspection?

**Tip:** When using the normal approximation to the binomial distribution, make sure the sample size \( n \) and probabilities \( p \) and \( 1 - p \) are large enough to satisfy the rule of thumb \( np > 5 \) and \( n(1 - p) > 5 \).

A quality control inspector has drawn a sample of 2020 light bulbs from a recent production lot. If the number of defective bulbs is 11 or less, the lot passes inspection. Suppose 20% of the bulbs in the lot are defective. What is the probability that the lot will pass inspection? Round your answer to four decimal places.

Determine the probability that a lot of 2020 light bulbs will pass inspection, given 20% defective bulbs. Learn how to use binomial distribution and its normal approximation with a step-by-step solution, including key formulas and concepts.

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