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katrin [286]
3 years ago
5

Suppose 42% of the population has myopia. If a random sample of size 442 is selected, what is the probability that the proportio

n of persons with myopia will differ from the population proportion by less than 3%
Mathematics
1 answer:
sineoko [7]3 years ago
7 0

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%.

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