Question

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.)

Answer 1

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.)

Answer:

P(17 ≤ x ≤ 23) = 0.5467

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.

Hope this Helps!!!

Answer 2

d)

We first get the z score for the two values. As z = (x - u) / s, then as

x1 = lower bound = 16.5

x2 = upper bound = 23.5

u = mean = 20

s = standard deviation = 4

Thus, the two z scores are

z1 = lower z score = (x1 - u)/s = -0.875

z2 = upper z score = (x2 - u) / s = 0.875

Using table/technology, the left tailed areas between these z scores is

P(z < z1) = 0.190786953

P(z < z2) = 0.809213047

Thus, the area between them, by subtracting these areas, is

P(z1 < z < z2) = 0.618426094 [ANSWER]