Question

Ten percent of the population is left-handed. A class of 100 students is selected. Convert the binomial probability PX 12) to a normal probability by using the correction for continuity. b.) Ten percent of the population is left-handed. A class of 5350 students is selected. Convert the binomial probability PX s 22) to a normal probability by using the correction for continuity.

Answer 1

To convert binomial probability to normal probability using the correction for continuity, follow these steps: calculate the continuity correction, calculate the z-score, and use the standard normal distribution table or calculator. For PX(12) with n=100 and p=0.1, the continuity correction would be (11.5, 12.5). For PX(22) with n=5350 and p=0.1, the continuity correction would be (21.5, 22.5).

Step-by-step explanation:

To convert binomial probability to normal probability using the correction for continuity, follow these steps:

1. Calculate the continuity correction by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of the desired binomial probability.
2. Calculate the z-score using the continuity corrected values, mean (np), and standard deviation (sqrt(npq)).
3. Use the standard normal distribution table or calculator to find the area to the left of the z-score.

For part (a), PX(12) with n=100 and p=0.1, the continuity correction would be (11.5, 12.5). The mean (np) is 100 * 0.1 = 10, and the standard deviation (sqrt(npq)) is sqrt(100 * 0.1 * 0.9) = 3. For part (b), PX(22) with n=5350 and p=0.1, the continuity correction would be (21.5, 22.5), the mean (np) is 5385 * 0.1 = 538.5, and the standard deviation (sqrt(npq)) is sqrt(5385 * 0.1 * 0.9) = 24.63.

Answer 2

(A) P(x > 11.5) ≈ P(z > (11.5 - 10) / 3) = P(z > 0.50)

(B) P(x ≤ 22.5) ≈ P(z ≤ (22.5 - 535) / 23.09) = P(z ≤ -22.05)

a. For a class of 100 students:

Identify the parameters:

Sample size (n) = 100

Probability of success (left-handed) (p) = 0.10

Binomial probability to convert: P(x > 12)

Apply continuity correction:

Since we're dealing with "greater than," we subtract 0.5 from x: P(x > 12) ≈ P(x > 11.5)

Approximate with a normal distribution:

Mean (μ) = np = 100 * 0.10 = 10

Standard deviation (σ) = √(np(1-p)) = √(100 * 0.10 * 0.90) ≈ 3

Convert to a normal probability:

P(x > 11.5) ≈ P(z > (11.5 - 10) / 3) = P(z > 0.50)

b. For a class of 5350 students:

Identify the parameters:

n = 5350

p = 0.10

Binomial probability to convert: P(x ≤ 22)

Apply continuity correction:

Since we're dealing with "less than or equal to," we add 0.5 to x: P(x ≤ 22) ≈ P(x ≤ 22.5)

Approximate with a normal distribution:

μ = np = 5350 * 0.10 = 535

σ = √(np(1-p)) ≈ 23.09

Convert to a normal probability:

P(x ≤ 22.5) ≈ P(z ≤ (22.5 - 535) / 23.09) = P(z ≤ -22.05)