Answer:
C. n = 90; p = 0.8
Step-by-step explanation:
According to the Central Limit Theorem, the distribution of the sample means will be approximately normally distributed when the sample size, 'n', is equal to or larger than 30, and the shape of sample distribution of sample proportions with a population proportion, 'p' is normal IF n·p ≥ 10 and n·(1 - p) ≥ 10
Analyzing the given options, we have;
A. n = 45, p = 0.8
∴ n·p = 45 × 0.8 = 36 > 10
n·(1 - p) = 45 × (1 - 0.8) = 9 < 10
Given that for n = 45, p = 0.8, n·(1 - p) = 9 < 10, a normal distribution can not be used to approximate the sampling distribution
B. n = 90, p = 0.9
∴ n·p = 90 × 0.9 = 81 > 10
n·(1 - p) = 90 × (1 - 0.9) = 9 < 10
Given that for n = 90, p = 0.9, n·(1 - p) = 9 < 10, a normal distribution can not be used to approximate the sampling distribution
C. n = 90, p = 0.8
∴ n·p = 90 × 0.8 = 72 > 10
n·(1 - p) = 90 × (1 - 0.8) = 18 > 10
Given that for n = 90, p = 0.9, n·(1 - p) = 18 > 10, a normal distribution can be used to approximate the sampling distribution
D. n = 45, p = 0.9
∴ n·p = 45 × 0.9 = 40.5 > 10
n·(1 - p) = 45 × (1 - 0.9) = 4.5 < 10
Given that for n = 45, p = 0.9, n·(1 - p) = 4.5 < 10, a normal distribution can not be used to approximate the sampling distribution
A sampling distribution Normal Curve
45 × (1 - 0.8) = 9
90 × (1 - 0.9) = 9
90 × (1 - 0.8) = 18
45 × (1 - 0.9) = 4.5
Now we will investigate the shape of the sampling distribution of sample means. When we were discussing the sampling distribution of sample proportions, we said that this distribution is approximately normal if np ≥ 10 and n(1 – p) ≥ 10. In other words
Therefore;
A normal curve can be used to approximate the sampling distribution of only option C. n = 90; p = 0.8
