Question

A scientist claims that 7% 7 % of viruses are airborne. If the scientist is accurate, what is the probability that the proportion of airborne viruses in a sample of 600 600 viruses would differ from the population proportion by greater than 3% 3 % ? Round your answer to four decimal places.

Answer 1

Answer:

The probability is $P(|p-\^{p}| > 0.03) = 0.0040$

Step-by-step explanation:

From the question we are told that

   The population proportion is $p = 0.07$

   The mean of the sampling distribution is   $\mu_p = 0.07$

   The sample size is n = 600

Generally the standard deviation is mathematically represented as

     $\sigma_p = \sqrt{\frac{p (1 -p)}{n} }$

=>    $\sigma_p = \sqrt{\frac{0.07(1 -0.07)}{600} }$    

=>    $\sigma_p = 0.010416$    

Generally the probability that the proportion of airborne viruses in a sample of 600 viruses would differ from the population proportion by greater than 3% is mathematically represented as

      $P(|p-\^{p}| > 0.03) = 1 - P(|p -\^{p}| \le 0.03)$

=>   $P(|p-\^{p}| > 0.03) = 1 - P(-0.03 \le p -\^{p} \le 0.03 )$

Now  add p  to  both side of the inequality

=>   $P(|p-\^{p}| > 0.03) = 1 - P( 0.07-0.03 \le \^{p} \le 0.03+ 0.07 )$

=>   $P(|p-\^{p}| > 0.03) = 1 - P(0.04 \le \^{p} \le 0.10 )$

Now  converting the probabilities to their respective standardized score

=> $P(|p-\^{p}| > 0.03) = 1 - P(\frac{0.04 - 0.07}{0.010416} \le Z \le \frac{0.10 -0.07}{0.010416} )$

=> $P(|p-\^{p}| > 0.03) = 1 - P(-2.88 \le Z \le 2.88 )$

=>   $P(|p-\^{p}| > 0.03) = 1 - [P(Z \le 2.88) - P(Z \le -2.88)]$

From the z-table  

       $P(Z \le 2.88) = 0.9980$

and

       $P(Z \le -2.88) = 0.0020$

So

     $P(|p-\^{p}| > 0.03) = 1 - [0.9980 - 0.0020]$

=>   $P(|p-\^{p}| > 0.03) = 0.0040$