Question

Assume the cholesterol levels of adult American women can be described by a Normal model with a mean of 188 mg/dL and a standard deviation of 24.

B. What percent of adult women do you expect to have cholesterol levels below 185 mg/dL?

C. Below what value is the lowest 5% of women’s cholesterol levels?

D. Estimate the interquartile range of the cholesterol levels.

Answer 1

Answer:

Given,

Mean = 188

Standard deviation = 24

b)

P(X < 185) = P((x-u)/s < (185 - 188)/24)

= P(z < - 0.125)

=  0.4502618 [since from z table]

= 0.4503

c)

P(X < x) = 095

P(Z < z) = 0.95

since from standard z table

z = 1.64

consider,

z = (x - u)/s

substitute values

1.64 = (x - 188)/24

x = 227.36

d)

Consider,

z = (x - u)/s

At 25th percentile, z = - 0.68

- 0.68 = (x - 188)/24

x = 171.68

At 75th percentile, z = 0.67

so,

0.67 = (x - 188)/24

x = 204.08

IQR = 75th - 25th percentile

= 204.08 - 171.68

= 32.4