Question 1194474
### Example of a Binomial Event Approximated by a Normal Distribution

**Scenario**: A factory produces light bulbs, and the probability of a bulb being defective is \( p = 0.02 \). The factory produces \( n = 500 \) bulbs in a day. Let \( X \) represent the number of defective bulbs in a batch of 500.

We want to find the probability that exactly 15 bulbs are defective (\( P(X = 15) \)).

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### Part A: Calculating the Probability Using Both Approaches

#### 1. Binomial Approach
The probability mass function for a binomial distribution is given by:

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

For \( n = 500 \), \( p = 0.02 \), and \( k = 15 \), the exact probability is:

\[
P(X = 15) = \binom{500}{15} (0.02)^{15} (0.98)^{485}
\]

We'll calculate this value explicitly.

#### 2. Normal Approximation
The binomial distribution can be approximated by a normal distribution if \( np \geq 5 \) and \( n(1-p) \geq 5 \). Here:

\[
np = 500 \times 0.02 = 10, \quad n(1-p) = 500 \times 0.98 = 490
\]

Both conditions are satisfied.

The approximating normal distribution has mean and standard deviation:

\[
\mu = np = 10, \quad \sigma = \sqrt{np(1-p)} = \sqrt{500 \times 0.02 \times 0.98} \approx 3.13
\]

To approximate \( P(X = 15) \), we use the continuity correction, finding \( P(14.5 \leq X \leq 15.5) \):

\[
Z_1 = \frac{14.5 - 10}{3.13}, \quad Z_2 = \frac{15.5 - 10}{3.13}
\]

The corresponding probabilities are found using the standard normal table or software.

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### Part B: Conditions for Less Accurate Approximation

The normal approximation becomes less accurate when:
1. \( p \) is close to 0 or 1, making the distribution highly skewed.
2. \( n \) is small, violating the \( np \geq 5 \) and \( n(1-p) \geq 5 \) conditions.
3. The event involves a discrete probability at the tails of the distribution where the binomial and normal curves differ significantly.

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### Part C: When Not to Use the Normal Approximation Despite Meeting the Criteria

Even if \( np \geq 5 \) and \( n(1-p) \geq 5 \), the normal approximation may not be ideal when:
1. **Precision is critical**: The binomial distribution provides exact probabilities, whereas the normal approximation involves rounding and continuity corrections.
2. **Events near the tails**: For probabilities of rare events (e.g., \( P(X = n) \) when \( n \) is far from the mean), the normal approximation may misestimate the probabilities.
3. **Computational ease**: With modern software, binomial probabilities are easy to compute, making the exact approach preferable.

For example, if \( n = 20 \) and \( p = 0.25 \), both criteria are satisfied, but the distribution is still not symmetric, making the normal approximation less accurate. Using the exact binomial distribution would yield better results.