To analyze a situation involving the genetic mutation in a population of 600 people where 6% have the mutation, we can utilize the binomial distribution.
Step 1: Define Variables
Let \( n = 600 \) (the number of trials, or the number of people selected). Let \( p = 0.06 \) (the probability that a selected individual has the genetic mutation).
Step 2: Calculate Expected Values
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Expected number of people with the mutation (mean): \[ \text{Mean} = n \cdot p = 600 \cdot 0.06 = 36 \]
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Variance: \[ \text{Variance} = n \cdot p \cdot (1 - p) = 600 \cdot 0.06 \cdot 0.94 = 33.84 \]
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Standard Deviation: \[ \text{Standard Deviation} = \sqrt{Variance} = \sqrt{33.84} \approx 5.81 \]
Step 3: Probability Calculations
While you could calculate exact probabilities using the binomial formula, for large sample sizes, the binomial distribution can be approximated using the normal distribution.
Given that \( n \) is large and \( p \) is not too close to 0 or 1, you can use the normal approximation \( N(\mu, \sigma^2) \) where \( \mu = 36 \) and \( \sigma \approx 5.81 \).
Example Question with Normal Approximation
If you wanted to find the probability that exactly 40 people have the mutation, you would:
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Use the continuity correction to adjust for the normal approximation (i.e., look for \( P(39.5 < X < 40.5) \)).
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Convert to a z-score: \[ z = \frac{x - \mu}{\sigma} = \frac{39.5 - 36}{5.81} \approx 0.86 \] \[ z = \frac{40.5 - 36}{5.81} \approx 0.77 \]
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Look up these z-scores in the standard normal distribution table or use a calculator to find the respective probabilities and then calculate the difference to find the probability for the desired range.
Let me know if you have any specific questions or if you'd like additional calculations!