Consider a binomial experiment with 20 trials and a probability of 0.45 on a single trial.Use the normal distribution to find the probability of exactly 10 successes.

(20 C 10) (.45)^10 * (.55)^10

0.9907

To find the probability of exactly 10 successes in a binomial experiment using the normal distribution approximation, we can use the concept of the normal approximation to the binomial distribution. The normal approximation can be applied when both np (mean) and nq (variance) are greater than or equal to 5, where n is the number of trials and p is the probability of success.

For this problem, we have n = 20 and p = 0.45. So, np = 20 * 0.45 = 9 and nq = 20 * (1 - 0.45) = 11.

To apply the normal approximation, we convert the binomial distribution to a normal distribution using the mean μ = np and the standard deviation σ = sqrt(npq).

In this case, the mean μ = 9 and the standard deviation σ = sqrt(9 * 0.55) = 2.04.

Now, to find the probability of exactly 10 successes, we need to find the area under the normal curve between the values X = 9.5 and X = 10.5. We add and subtract 0.5 to account for the continuity correction.

To find this probability, we can use the z-score formula:

Z = (X - μ) / σ

For X = 9.5:
Z = (9.5 - 9) / 2.04 = 0.2451

For X = 10.5:
Z = (10.5 - 9) / 2.04 = 0.7353

Now, we look up the probabilities associated with these z-scores in the standard normal distribution table or use a calculator. The table gives us the probability to the left of the z-score. So, we need to find P(Z ≤ 0.2451) and P(Z ≤ 0.7353) and then subtract the two probabilities to find the area between them.

Using the standard normal distribution table or a calculator, we find:
P(Z ≤ 0.2451) = 0.5964
P(Z ≤ 0.7353) = 0.7689

Therefore, the probability of exactly 10 successes in the given binomial experiment is:
P(9.5 ≤ X ≤ 10.5) = P(Z ≤ 0.7353) - P(Z ≤ 0.2451) = 0.7689 - 0.5964 = 0.1725

So, the probability of exactly 10 successes is approximately 0.1725.