Question
A theater owner has found that 5% of patrons do not show up for the performance that they purchased tickets for. If the theater has 100 seats, find the probability that 6 or more patrons will not show up for the sold-out performance. Use the normal approximation and express your answer as a decimal. Please round to the nearest thousandth.
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Answered by
GPT-4o mini
To solve this problem, we first need to define the random variable \( X \), which is the number of patrons who do not show up for the performance. Given that 5% of patrons do not show up, we can model \( X \) using the binomial distribution \( B(n, p) \), where \( n = 100 \) (the number of seats) and \( p = 0.05 \) (the probability that a patron does not show up).
### Step 1: Calculate the mean and variance
The mean \( \mu \) and variance \( \sigma^2 \) of a binomial distribution are calculated as follows:
\[
\mu = n \cdot p = 100 \cdot 0.05 = 5
\]
\[
\sigma^2 = n \cdot p \cdot (1-p) = 100 \cdot 0.05 \cdot (1 - 0.05) = 100 \cdot 0.05 \cdot 0.95 = 4.75
\]
Thus, the standard deviation \( \sigma \) is:
\[
\sigma = \sqrt{4.75} \approx 2.179
\]
### Step 2: Normal approximation to the binomial distribution
Since \( n \) is large, we can use the normal approximation to the binomial distribution. The random variable \( X \) can be approximated by a normal distribution with mean \( \mu = 5 \) and standard deviation \( \sigma \approx 2.179 \).
### Step 3: Find the probability that 6 or more patrons do not show up
We are interested in the probability \( P(X \geq 6) \). To use the normal approximation, we apply a continuity correction. Thus, we want to find:
\[
P(X \geq 6) = P(X > 5.5)
\]
We standardize this to a standard normal variable \( Z \):
\[
Z = \frac{X - \mu}{\sigma} = \frac{5.5 - 5}{2.179} \approx \frac{0.5}{2.179} \approx 0.229
\]
### Step 4: Find the corresponding probability from z-table
Now, we need to find \( P(Z > 0.229) \). We can look this value up in the standard normal distribution table or use a calculator.
Using a standard normal distribution table:
\[
P(Z < 0.229) \approx 0.5910
\]
Thus,
\[
P(Z > 0.229) = 1 - P(Z < 0.229) \approx 1 - 0.5910 = 0.4090
\]
### Step 5: Round to the nearest thousandth
Finally, rounding this probability to the nearest thousandth gives:
\[
\boxed{0.409}
\]
This means the probability that 6 or more patrons will not show up is approximately 0.409.
### Step 1: Calculate the mean and variance
The mean \( \mu \) and variance \( \sigma^2 \) of a binomial distribution are calculated as follows:
\[
\mu = n \cdot p = 100 \cdot 0.05 = 5
\]
\[
\sigma^2 = n \cdot p \cdot (1-p) = 100 \cdot 0.05 \cdot (1 - 0.05) = 100 \cdot 0.05 \cdot 0.95 = 4.75
\]
Thus, the standard deviation \( \sigma \) is:
\[
\sigma = \sqrt{4.75} \approx 2.179
\]
### Step 2: Normal approximation to the binomial distribution
Since \( n \) is large, we can use the normal approximation to the binomial distribution. The random variable \( X \) can be approximated by a normal distribution with mean \( \mu = 5 \) and standard deviation \( \sigma \approx 2.179 \).
### Step 3: Find the probability that 6 or more patrons do not show up
We are interested in the probability \( P(X \geq 6) \). To use the normal approximation, we apply a continuity correction. Thus, we want to find:
\[
P(X \geq 6) = P(X > 5.5)
\]
We standardize this to a standard normal variable \( Z \):
\[
Z = \frac{X - \mu}{\sigma} = \frac{5.5 - 5}{2.179} \approx \frac{0.5}{2.179} \approx 0.229
\]
### Step 4: Find the corresponding probability from z-table
Now, we need to find \( P(Z > 0.229) \). We can look this value up in the standard normal distribution table or use a calculator.
Using a standard normal distribution table:
\[
P(Z < 0.229) \approx 0.5910
\]
Thus,
\[
P(Z > 0.229) = 1 - P(Z < 0.229) \approx 1 - 0.5910 = 0.4090
\]
### Step 5: Round to the nearest thousandth
Finally, rounding this probability to the nearest thousandth gives:
\[
\boxed{0.409}
\]
This means the probability that 6 or more patrons will not show up is approximately 0.409.
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