Question

Suppose 42% of the population has myopia. If a random sample of size 442 is selected, what is the probability that the proportion of persons with myopia will differ from the population proportion by less than 3%

Answer 1

Answer:

0.7994 = 79.94% probability that the proportion of persons with myopia will differ from the population proportion by less than 3%.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Suppose 42% of the population has myopia.

This means that $p = 0.42$

Random sample of size 442 is selected

This means that $n = 442$

Mean and standard deviation:

$\mu = p = 0.42$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.42*0.58}{442}} = 0.0235$

What is the probability that the proportion of persons with myopia will differ from the population proportion by less than 3%?

Proportion between 0.42 + 0.03 = 0.45 and 0.42 - 0.03 = 0.39, which is the p-value of Z when X = 0.45 subtracted by the p-value of Z when X = 0.39.

X = 0.45

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.45 - 0.42}{0.0235}$

$Z = 1.28$

$Z = 1.28$ has a p-value of 0.8997

X = 0.39

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.39 - 0.42}{0.0235}$

$Z = -1.28$

$Z = -1.28$ has a p-value of 0.1003

0.8997 - 0.1003 = 0.7994

0.7994 = 79.94% probability that the proportion of persons with myopia will differ from the population proportion by less than 3%.