Question

label and answer all parts

Serum cholesterol is an important risk factor for coronary disease. Suppose that serum cholesterol is approximately normally distributed with mean 212.9 mg/dL and standard deviation 51.78 mg/dL. a. Cholesterol levels in the top 20% are classified as high enough to warrant risk for heart disease. What cholesterol level corresponds to this cut-off? -mg/dL cholesterol- b. Suppose cholesterol levels in the bottom 40% are classified as normal. What cholesterol level corresponds to a normal reading? o cholesterol- mg/dL C. Between what two values do the middle 48% of cholesterol readings fall? o between mg/dL and mg/dL Note: z-scores should include 2 decimal places; all other answers should be rounded to 3 decimal places

Answer 1

We have,

         $\mu=212.9$           $\sigma=51.78$

a) Consider,

              $P(X>x)=0.20$

             $P(\frac{X-\mu}{\sigma}>\frac{x-212.9}{51.78})=0.20$

              $P(Z>\frac{x-212.9}{51.78})=0.20$        ..............(1)

       From Z-table(standard normal table), P(Z>0.84)=0.20045=0.20     ..........(2)

            Compare (1) and (2), we get

                         $\frac{x-212.9}{51.78}=0.84$

                         $x=212.9+0.84(51.78)$

                        $x=256.395$

   Hence,

                 Z = 0.84

              Cholesterol = 256.395 mg/dL

b) Consider,

             $P(X<x)=0.40$

             $P(\frac{X-\mu}{\sigma}<\frac{x-212.9}{51.78})=0.4$

              $P(Z<\frac{x-212.9}{51.78})=0.4$

                $P(Z>-(\frac{x-212.9}{51.78}))=0.4$             symmetry of Z ..............(3)

        From Z-table, P(Z>0.25) =0.40129=0.4           .................(4)

                Compare (3) and (4) , we get

                   $-(\frac{x-212.9}{51.78})=0.25$

                    $x=212.9-0.25(51.78)$

                    $x=199.955$

     Hence,

               Z = 0.25

             Cholesterol = 199.955 mg/dL

c) Consider,

                $P(-x<X<x)=0.48$

                 $P(\frac{-x-212.9}{51.78}<\frac{X-\mu}{\sigma}<\frac{x-212.9}{51.78})=0.48$

                 $P(\frac{-x-212.9}{51.78}<Z<\frac{x-212.9}{51.78})=0.48$

                 $P(Z>\frac{-x-212.9}{51.78})-P(Z>\frac{x-212.9}{51.78})=0.48$

                 $1-P(Z>\frac{x-212.9}{51.78})-P(Z>\frac{x-212.9}{51.78})=0.48$

                 $1-2P(Z>\frac{x-212.9}{51.78})=0.48$

                   $P(Z>\frac{x-212.9}{51.78})=\frac{1-0.48}{2}$

                   $P(Z>\frac{x-212.9}{51.78})=0.26$    .......................(5)

     From Z-table, P(Z>0.64)=0.26109 = 0.26....................(6)

                  Compare (5) and (6) , we get

                         $\frac{x-212.9}{51.78}=0.64$

                       $x=212.9&plus;0.64(51.78)$

                      $x=246.039$       ..................(*)

        Due to symmetry of normal,

                       $x=212.9-0.64(51.78)$

                       $x=179.761$    .................(**)

Hence,

                    $Z=\pm 0.64$

               Between : 179.761 mg/dL and 246.039 mg/dL