Question

Serum cholesterol is an important risk factor for coronary disease. Suppose that serum cholesterol is approximately normally distributed with mean 221.4 mg/dL and standard deviation of 45.76 mg/dL.

1. Cholesterol levels in the top 16% are classified as high enough to warrant risk for heart disease. What cholesterol level corresponds to this cut-off?
   • z = _____
   • cholesterol = ________mg/dL
2. Suppose cholesterol levels in the bottom 32% are classified as normal. What cholesterol level corresponds to a normal reading?
   • z = ______
   • cholesterol = _________ mg/dL
3. Between what two values do the middle 22% of cholesterol readings fall?
   • z =_______
   • 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

Solution:-

Given that,

mean = = 221.4

standard deviation = = 45.76

Using standard normal table,

P(Z > z) = 16 %

= 1 - P(Z < z) = 0.16

= P(Z < z) = 1 - 0.16

= P(Z < z ) = 0. 84

= P(Z < 0.99 ) = 0. 84

z = 0.99

Using z-score formula,

x = z * +

x = 0.99 * 45.76 + 221.4

x = 266.702

P(Z < z) = 32 %

= P(Z < z) = 0.32  

= P(Z < 0.47 ) = 0.32

z = 0.47

Using z-score formula  

x = z * +

x = 0.47 *45.76 +221.4

x = 242.907

P( -z < Z < z) = 22%

= P(Z < z) - P(Z <-z ) = 0.22

= 2P(Z < z) - 1 = 0.22

= 2P(Z < z) = 1 + 0.22

= P(Z < z) = 1.22 / 2

= P(Z < z) = 0.61

= P(Z < 0.28 ) = 0. 61

= z  ± 0.28

Using z-score formula,

x = z * +

x = - 0.28 * 45.76 + 221.4

x = 208.587

Using z-score formula,

x = z * +

x = 0.28 * 45.76 + 221.4

x = 234.213

Between 208.587 and 234.213