Question

In the exercise, answer the question by rounding z-scores to the nearest hundredth and then finding the required A values using the table. The cholesterol levels of a group of young women at a university are normally distributed, with a mean of 189 and a standard deviation of 39. What percent of the young women have the following cholesterol levels? (For each answer, enter a number. Round your answers to one decimal place.)

(a) greater than 216 %

(b) between 194 and 222 %

Answer 1

Solution :

Given that,

mean = = 189

standard deviation = = 39

a ) P (x > 216 )

= 1 - P (x < 216 )

= 1 - P ( x -  / ) < ( 216 - 189 / 39)

= 1 - P ( z < 27 / 39 )

= 1 - P ( z < 0.69)

Using z table

= 1 - 0.7549

= 0.2451

Probability = 0.2451 = 24.5%

b ) P (194 < x < 222 )

P ( 194 - 189 / 39 ) < ( x -  / ) < ( 222 - 189 / 39)

P ( 5 / 39 < z < 33 / 39 )

P ( 0.13 < z < 0.85 )

P ( z < 0.85 ) - P ( z < 0.13)

Using z table

= 0.8023 - 0.5517

= 0.2506

Probability = 0.2506 = 25.1%