Question

Suppose that 90% of all dialysis patients will survive for at least 5 years. In a simple random sample of 100 new dialysis patients, what is the probability that the proportion surviving for at least five years will exceed 80%, rounded to 5 decimal places

Answer 1

Answer:

The  probability   $P(\^ p > 0.80) = 0.99957$

Step-by-step explanation:

From the question we are told that

     The population proportion is  p  =  0.90

     The sample size is  n  =  30

  Generally mean of the sampling distribution is  $\mu_{\= x } = p = 0.90$

Generally the standard deviation is mathematically represented as

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

=>    $\sigma = \sqrt{\frac{0.90( 1 -0.90 )}{100} }$

=>    $\sigma = 0.03$

Generally the he probability that the proportion surviving for at least five years will exceed 80%, rounded to 5 decimal places is mathematically represented as

       $P(\^ p > 0.80) = P(\frac{ \^ p - p }{ \sigma } > \frac{0.80 - 0.90}{0.03} )$

Generally $\frac{\^p - p}{\sigma} = Z(The standardized \ value \ of \^ p)$

So  

    $P(\^ p > 0.80) = P(Z > -3.33 )$

From the z-table $P(Z > -3.33 ) = 0.99957$

So

    $P(\^ p > 0.80) = 0.99957$