For a binomial distribution with n=20 trials and probability of success p=.05, we let r= number of successes out of 20 trials. Use the normal approximation to estimate P(6<=r<=12). How does this value compare with the corresponding probability obtained using the binomial table?

I already found the mean and standard deviation as well as the equivalent in X but I don't know what to do next because my table answers don't match the answer key. Help?

To estimate the probability using the normal approximation, you'll need to use the mean (µ) and standard deviation (σ) that you calculated. The formula for the mean of a binomial distribution is µ = np, and the formula for the standard deviation is σ = √(np(1-p)).

In this case, you mentioned that n = 20 and p = 0.05. Therefore, the mean µ = 20 * 0.05 = 1, and the standard deviation σ = √(20 * 0.05 * (1-0.05)) = √(0.95).

To estimate the probability of r successes using the normal approximation, you need to convert the interval [6,12] to the corresponding interval in terms of X. If X denotes the number of successes in the binomial distribution, you can use the formula Z = (X - µ) / σ to standardize the values.

For X = 6: Z = (6 - 1) / √(0.95) ≈ 5.26
For X = 12: Z = (12 - 1) / √(0.95) ≈ 11.84

Now, you can use a standard normal distribution table or a calculator to look up the probabilities associated with these Z-scores. Find the probability associated with Z = 5.26 and subtract the probability associated with Z = 11.84 from it.

The corresponding probability obtained using the binomial table represents the exact probability calculation based on the binomial distribution, while the normal approximation is an estimation. It's possible that the two values might not match precisely due to the nature of the approximation.

If your answers still don't match the answer key, double-check your calculations for mean and standard deviation, and ensure that you accurately converted the interval to Z-scores.