Final answer:
To estimate the probability of more than 8 successes out of 30 trials with a success probability of 0.2 using the normal approximation to the binomial distribution, one should calculate the mean (μ = np), the standard deviation (σ = √npq), apply a continuity correction, and determine the z-score to use in a normal distribution cumulative probability function.
Step-by-step explanation:
Normal Approximation to the Binomial Distribution
When estimating probabilities for binomial distributions with a large number of trials (n), and a reasonable probability of success (p), we can use normal approximation for a more straightforward calculation. In the given problem, the student needs to find the probability of getting more than 8 successes out of 30 with a success probability of 0.2. Applying the normal approximation involves ensuring that both np and nq are greater than five, which allows us to approximate the binomial distribution by a normal distribution with mean μ = np and standard deviation σ = √npq.
In this case, we have:
- np = 30 × 0.2 = 6
- nq = 30 × 0.8 = 24
- μ = np = 6
- σ = √(30 × 0.2 × 0.8) = √(4.8) ≈ 2.19
To find the probability of more than 8 successes, we need to use the standard normal distribution. We'll convert the number of successes into a z-score and use normalcdf for calculations. First, we adjust the discrete success count by adding 0.5 to get a continuity correction, leading to 8.5. Thus, the z-score is given by:
Z = (8.5 - μ) / σ = (8.5 - 6) / 2.19 ≈ 1.14
We then use the normalcdf function with the z-score to estimate the desired probability:
Probability (X > 8) ≈ normalcdf(1.14, ∞) = 1 - normalcdf(-∞, 1.14)
Using a normal distribution probability table or calculator, we would get the probability for the z-score and hence the estimate for the binomial probability of having more than 8 successes.