Question

Every time you have your cholesterol measured, the measurement may be slightly different due to random fluctuations and measurement error. Suppose that for you, the population of possible cholesterol measurements if you are healthy has a mean of 190 and a standard deviation of 10. Further, suppose you know you should get concerned if your measurement ever gets up to the 97th percentile. What level of cholesterol does that represent?

Answer 1

Complete Question

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Answer:

a i

$P(X < 185 ) = 0.3085$

a ii

$P(X > 195 ) = 0.3085$

a iii

$P(185 < X < 195 ) = 0.3829$

b

$x = 208.8$

Step-by-step explanation:

From the question we are told that  

   The mean is  $\mu = 190$

   The standard deviation is  $\sigma = 10$

   

Generally the probability is  less than 185  is mathematically represented as

      $P(X < 185 ) = P(\frac{X - \mu }{\sigma } < \frac{185 - 190 }{10 } )$

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

=>   $P(X < 185 ) = P(Z< -0.5)$

From the z-table the p value  of  (Z< -0.5) is

    $P(Z< -0.5) = 0.3085$

So

    $P(X < 185 ) = 0.3085$

Generally the probability is  less than 185  is mathematically represented as

      $P(X > 195 ) = P(\frac{X - \mu }{\sigma } > \frac{195 - 190 }{10 } )$

=>   $P(X > 195 ) = P(Z > 0.5)$

From the z-table the p value  of  (Z > 0.5) is

    $P(Z > 0.5) = 0.3085$

So

    $P(X > 195 ) = 0.3085$

Generally the probability is  less than 185  is mathematically represented as

      $P(185 < X < 195 ) = P( \frac{185 - 190 }{10 } < \frac{X - \mu }{\sigma } < \frac{195 - 190 }{10 } )$

=>   $P(185 < X < 195 ) = P(-0.5 < Z > 0.5)$

=>   $P(185 < X < 195 ) = P(Z < 0.5 ) - P(Z < -0.5)$

From the z-table the p value  (Z <  0.5) and  (Z <  -0.5) is  

    $P(Z < 0.5) = 0.6915$

and  

   $P(Z < - 0.5) = 0.3085$

So

=>   $P(185 < X < 195 ) = 0.6915 - 0.3085$

=>   $P(185 < X < 195 ) = 0.3829$

Generally the level of cholesterol the 97th percentile represents  is mathematically evaluated as

    $P(X < x ) = 0.97$

=> $P(X < x ) = P(\frac{X - \mu}{\sigma } < \frac{x - 190}{10 } ) = 0.97$

=>   $P(X < x ) = P(Z < \frac{x - 190}{10 } ) = 0.97$

From the z-table  the z-score for  0.97  is  

      $z-score = 1.88$

=>

  $\frac{x - 190}{10 } = 1.88$

=>$x = 208.8$