What is the probability of 17 to 23 successes? Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.)

The complete question is

A binomial probability distribution has p = 0.20 and n = 100.

What is the probability of 17 to 23 successes? Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.)

P(17 ≤ x ≤ 23) = 0.5467

Step-by-step explanation:

Using the normal approximation to this binomial distribution problem,

The mean = μ = (sample size) × (proportion) = np = 100 × 0.2 = 20.

The standard deviation = σ = √[np(1-p)] = √(100×0.2×0.8) = 4.

To now find the probability that there will be between 17 to 23 successes inclusive.

P(17 ≤ x ≤ 23)

We first normalize/standardize/obtain the z-scores of 17 and 23.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 17,

z = (x - μ)/σ = (17 - 20)/4 = - 0.75

For 23,

z = (x - μ)/σ = (23 - 20)/4 = 0.75

To determine the probability that there will be between 17 to 23 successes inclusive.

P(17 ≤ x ≤ 23) = P(-0.75 ≤ z ≤ 0.75)

We'll use data from the normal probability table for these probabilities

P(17 ≤ x ≤ 23) = P(-0.75 ≤ z ≤ 0.75)

= P(z ≤ 0.75) - P(z ≤ -0.75)

= 0.77337 - 0.22663 = 0.54674 = 0.5467 to 4 d.p.

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