Question

A scientist claims that 9% of viruses are airborne. If the scientist is accurate, what is the probability that the proportion of airborne viruses in a sample of 533 viruses would differ from the population proportion by less than 3%

Answer 1

The probability that the sample proportion of airborne viruses in differ from the population proportion by greater than 3% is 0.006.

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

μρ=ρ

The standard deviation of this sampling distribution of sample proportion is:

σρ=$\sqrt \frac{p(1-p)}{n}$

The information provided is:

n = 533

p = 0.09

As the sample size is quite large, i.e. n = 533 > 30, the central limit theorem can be used to approximate the sampling distribution of sample proportion by a Normal distribution.

The mean and standard deviation are:

μρ=ρ=0.09

σρ=$\sqrt \frac{p(1-p)}{n}$=$\sqrt\frac{0.09(1-0.09)}{533}=0.01239$

Compute the probability that the sample proportion of airborne viruses in differ from the population proportion by greater than 3% as follows:

$p(p^-p > 0.03)=P(p^ > 0.12) \\=P(z > \frac{0.12-0.9}{0.012}) \\= P(Z > 2.5) \\= 1-P(Z > 2.5) \\ = 1- 0.993 \\ = 0.00621$

Thus, the probability that the sample proportion of airborne viruses in differ from the population proportion by greater than 3% is 0.006.

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