The serum cholesterol levels in milligrams/deciliter (mg/dL) in a certain Medite

Question

The serum cholesterol levels in milligrams/deciliter (mg/dL) in a certain Mediterranean population are found to be normally distributed with a mean of 154 and a standard deviation of 28. Answer the following: (a) Determine the z-score for a person from this population that has a cholesterol level of 135. Then find the z-score for someone whose cholesterol level is 217. (b) If x represents a possible cholesterol level from this population, find P(x > 135). (c) Find P(135 < x < 217) and give an interpretation of this value. (d) The top 11% of all people in this group have cholesterol levels that make them "at-risk" for heart problems. Determine the raw-score cholesterol level which separates the at-risk people from the rest of the group. The serum cholesterol levels in milligrams/deciliter (mg/dL) in a certain Mediterranean population are found to be normally distributed with a mean of 154 and a standard deviation of 28. Answer the following: (a) Determine the z-score for a person from this population that has a cholesterol level of 135. Then find the z-score for someone whose cholesterol level is 217. (b) If x represents a possible cholesterol level from this population, find P(x > 135). (c) Find P(135 < x < 217) and give an interpretation of this value. (d) The top 11% of all people in this group have cholesterol levels that make them "at-risk" for heart problems. Determine the raw-score cholesterol level which separates the at-risk people from the rest of the group.

Explanation / Answer

mean = 154

standard deviation = 28

now the distribution is normal

therefore the formula to be used =

z = (x-mean)/standard deviation

A) z score for 135 = (135 - 154)/28 = -19/28 = -0.67

the z score for 217 = (217 - 154)/28 = 2.25

b)p(x>135) = p(z> -0.67)

= 1 - p(z> -0.67 = 1 - 0.2514 = 0.7486

c)p(135<x<217)

as z score for 135 = -0.67

z score for 217 = 2.25

therefore the p(135<x<217) = p(-0.67<z<2.15) = p(2.25) - p(-0.67)

= 0.9878 - 0.2514 = 0.7364

d) we need to find the x for the top11% meand for 89%

formula to be used =

x = mean + z*standard deviation

z score for 89% = 1.23

x = 154 + 1.23*28

therefore

x = 188.44

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.