Suppose that the serum cholesterol levels of 17yearolds foll

Suppose that the serum cholesterol levels of 17-year-olds follow a normal distribution with a mean of 152 mg/dLi and a standard deviation of 29.0 mg/dLi. For each of the problems below, find the appropriate probability.
Note: For problems A, B, C, and D round your answers to four decimal places.

A) P(L > 117.49)   

B) P(134.60 L < 194.05)   



C) The government is offering a free program to help 17-year-olds lower their cholesterol. Those in the 97^th percentile of all 17-year-olds are eligible for the program. What is the minimum cholesterol level one would need to qualify for the program?  

D) The serum cholesterol levels that enclose the central 98% of the distribution are:
    Note: Round your answers to two decimal places for this problem.
    Low:  to High:

Solution

Mean ( u ) =152
Standard Deviation ( sd )=29
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X > 117.49) = (117.49-152)/29
= -34.51/29 = -1.19
= P ( Z >-1.19) From Standard Normal Table
= 0.883                  
b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 134.6) = (134.6-152)/29
= -17.4/29 = -0.6
= P ( Z <-0.6) From Standard Normal Table
= 0.27425
P(X < 194.05) = (194.05-152)/29
= 42.05/29 = 1.45
= P ( Z <1.45) From Standard Normal Table
= 0.92647
P(134.6 < X < 194.05) = 0.92647-0.27425 = 0.6522                  
c)
P ( Z > x ) = 0.03
Value of z to the cumulative probability of 0.03 from normal table is 1.88
P( x-u/ (s.d) > x - 152/29) = 0.03
That is, ( x - 152/29) = 1.88
--> x = 1.88 * 29+152 = 206.549                  
the minimum cholesterol level one would need to qualify for the program is ~ 207