Given a random sample size of nequals900 from a binomial probability distribution with Pequals0.20, complete parts a. through e. below.
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Part 1
a. Find the probability that the number of successes is greater than 220.
P(Xgreater than220)equals
enter your response here (Round to four decimal places as needed.)
To find the probability that the number of successes is greater than 220, we need to use the normal approximation to the binomial distribution.
First, calculate the mean (μ) and standard deviation (σ) of the binomial distribution:
μ = n * p = 900 * 0.20 = 180
σ = √(n * p * (1 - p)) = √(900 * 0.20 * (1 - 0.20)) = √(900 * 0.20 * 0.80) = √144 = 12
Next, convert the number of successes (220) to a z-score:
z = (x - μ) / σ
z = (220 - 180) / 12
z = 40 / 12
z = 3.3333
Now, use the standard normal distribution table to find the probability:
P(X > 220) = P(Z > 3.3333)
Looking up the closest z-score in the table, we find that P(Z > 3.33) is approximately 0.0004.
Therefore, the probability that the number of successes is greater than 220 is approximately 0.0004.