Answer:
The probability is [tex]P(|p-\^{p}| > 0.03) = 0.0040[/tex]
Step-by-step explanation:
From the question we are told that
The population proportion is [tex]p = 0.07[/tex]
The mean of the sampling distribution is [tex]\mu_p = 0.07[/tex]
The sample size is n = 600
Generally the standard deviation is mathematically represented as
[tex]\sigma_p = \sqrt{\frac{p (1 -p)}{n} }[/tex]
=> [tex]\sigma_p = \sqrt{\frac{0.07(1 -0.07)}{600} }[/tex]
=> [tex]\sigma_p = 0.010416 [/tex]
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
[tex]P(|p-\^{p}| > 0.03) = 1 - P(|p -\^{p}| \le 0.03)[/tex]
=> [tex]P(|p-\^{p}| > 0.03) = 1 - P(-0.03 \le p -\^{p} \le 0.03 )[/tex]
Now add p to both side of the inequality
=> [tex]P(|p-\^{p}| > 0.03) = 1 - P( 0.07-0.03 \le \^{p} \le 0.03+ 0.07 )[/tex]
=> [tex]P(|p-\^{p}| > 0.03) = 1 - P(0.04 \le \^{p} \le 0.10 )[/tex]
Now converting the probabilities to their respective standardized score
=> [tex]P(|p-\^{p}| > 0.03) = 1 - P(\frac{0.04 - 0.07}{0.010416} \le Z \le \frac{0.10 -0.07}{0.010416} )[/tex]
=> [tex]P(|p-\^{p}| > 0.03) = 1 - P(-2.88 \le Z \le 2.88 )[/tex]
=> [tex]P(|p-\^{p}| > 0.03) = 1 - [P(Z \le 2.88) - P(Z \le -2.88)][/tex]
From the z-table
[tex]P(Z \le 2.88) = 0.9980[/tex]
and
[tex]P(Z \le -2.88) = 0.0020[/tex]
So
[tex]P(|p-\^{p}| > 0.03) = 1 - [0.9980 - 0.0020][/tex]
=> [tex]P(|p-\^{p}| > 0.03) = 0.0040[/tex]
