SOLUTION: We have been using the normal distribution to approximate situations that are, in fact, binomial events. Create an example of a binomial event that can be approximated by a normal

Question 1194474: We have been using the normal distribution to approximate situations that are, in fact, binomial events. Create an example of a binomial event that can be approximated by a normal distribution and:
a)Demonstrate how accurate the approximation is by using both approaches to find the probability of the same event. Hint: Calculate the probability of the event as a binomial (sum of all binomial events) and calculate the probability of the approximated event using a normal distribution, and compare them to see how close the approximation is.
b)Describe the conditions under which the normal distribution would give a less accurate approximation.
c)Explain a situation in which the criteria for using the normal approximation would be met, i.e.np≥5 and n(1-p)≥5, and yet you would decide not to use the normal distribution.
Answer by yurtman(42) (Show Source): You can put this solution on YOUR website!
### 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 (\( P(X = 15) \)).
---
### 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.
---
### 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.
---
### 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.
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