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hoa [83]
3 years ago
7

A scientist claims that 7% 7 % of viruses are airborne. If the scientist is accurate, what is the probability that the proportio

n 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.
Mathematics
1 answer:
Reptile [31]3 years ago
7 0

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

     

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