Assume a binomial probability distribution has p = .60 and n = 200.

a)What are the mean and standard deviation (to 2 decimals)?

b)Why can the normal probability distribution be used to approximate this binomial distribution?


c)What is the probability of 100 to 110 successes (to 4 decimals)?


d)What is the probability of 130 or more successes (to 4 decimals)?


e)What is the advantage of using the normal probability distribution to approximate the binomial probabilities?

a. mean = np = 200(.6) =120

Standard deviation = sqrt((.6*.4*200))= sqrt(48)

a) To find the mean and standard deviation of a binomial probability distribution with p = 0.60 and n = 200, you can use the following formulas:

Mean (μ) = n * p
Standard Deviation (σ) = √(n * p * (1 - p))

Mean (μ) = 200 * 0.60 = 120
Standard Deviation (σ) = √(200 * 0.60 * (1 - 0.60)) = √(48) ≈ 6.93

Therefore, the mean is 120 and the standard deviation is approximately 6.93.

b) The normal probability distribution can be used to approximate a binomial distribution under certain conditions. One of the conditions is that the sample size (n) is large enough. In this case, n = 200, which is considered large enough.

Another condition is that both p (the probability of success) and q (the probability of failure, which is 1 - p) are not too close to 0 or 1. In this case, p = 0.60, which is not close to 0 or 1. Therefore, the normal probability distribution can be used to approximate this binomial distribution.

c) To find the probability of 100 to 110 successes, you can use the normal approximation to the binomial distribution. The formula for this approximation is:

P(100 ≤ X ≤ 110) ≈ P((100 - 0.5 - μ) / σ ≤ Z ≤ (110 + 0.5 - μ) / σ)

Here, Z is the standard normal random variable. Using the mean (μ) and standard deviation (σ) from part a):

P(100 ≤ X ≤ 110) ≈ P((100 - 0.5 - 120) / 6.93 ≤ Z ≤ (110 + 0.5 - 120) / 6.93)

P(-1.72 ≤ Z ≤ - .72) ≈ P(Z ≤ -0.72) - P(Z ≤ -1.72)

Using a standard normal table or a calculator, you can find the corresponding probabilities. Let's assume P(Z ≤ -0.72) = 0.2339 and P(Z ≤ -1.72) = 0.0427.

P(100 ≤ X ≤ 110) ≈ 0.2339 - 0.0427 = 0.1912 (approximately)

Therefore, the probability of 100 to 110 successes is approximately 0.1912.

d) To find the probability of 130 or more successes, you can again use the normal approximation to the binomial distribution. Using the same formula as in part c), we can find:

P(X ≥ 130) ≈ P(Z ≥ (130 + 0.5 - 120) / 6.93)

P(Z ≥ 1.44) ≈ 1 - P(Z ≤ 1.44)

Using a standard normal table or a calculator, you can find the corresponding probability. Let's assume P(Z ≤ 1.44) = 0.9251.

P(X ≥ 130) ≈ 1 - 0.9251 = 0.0749 (approximately)

Therefore, the probability of 130 or more successes is approximately 0.0749.

e) The advantage of using the normal probability distribution to approximate the binomial probabilities is that it simplifies the calculations. The normal distribution is a continuous distribution, while the binomial distribution is a discrete distribution. The normal distribution has a known and easily calculable mean and standard deviation. By approximating the binomial distribution with the normal distribution, we can use standard normal probabilities from tables or calculators to find the probabilities. This saves time and effort compared to calculating binomial probabilities directly using the binomial formula.