Question

The cholesterol levels, among young adults between the age of 18 and 24, follows a normal distribution with an average of 178mg/dL and a standard deviation of 41mg/dL.

a) Find the probability that a randomly selected young adult has a cholesterol level between 155 and 200.

b) Suppose a doctor wants to prescribe a certain medication only to young adults with a cholesterol level in the top 12% of the population. What is the minimum cholesterol level for which the medication will be prescribed?

c) Construct a symmetric interval of cholesterol levels around the mean that contain 96% of the young adult population.

d) Suppose a random sample of 25 young adults is selected. Find the probability that the average cholesterol level in this sample is above 190.

e)Again, suppose a random sample of 25 young adults is selected. What is the probability that exactly 10 of them have a cholesterol level above 190?

Answer 1

a) P(155 < X < 200)

= P((155 - )/ < (X - )/ < (200 - )/)

= P((155 - 178)/41 < Z < (200 - 178)/41)

= P(-0.56 < Z < 0.54)

= P(Z < 0.54) - P(Z < -0.56)

= 0.7054 - 0.2877

= 0.4177

b) P(X > x) = 0.12

or, P((X - )/ < (x - )/) = 0.12

or, P(Z > (x - 178)/41) = 0.12

or, P(Z < (x - 178)/41) = 0.88

or, (x - 178)/41 = 1.175

or, x = 1.175 * 41 + 178

or, x = 226.175

c) P(X < x) = 0.02

or, P((X - )/ < (x - )/) = 0.02

or, P(Z < (x - 178)/41) = 0.02

or, (x - 178)/41 = -2.05

or, x = -2.05 * 41 + 178

or, x = 93.95

P(X > x) = 0.02

or, P((X - )/ < (x - )/) = 0.02

or, P(Z > (x - 178)/41) = 0.02

or, P(Z < (x - 178)/41) = 0.98

or, (x - 178)/41 = 2.05

or, x = 2.05 * 41 + 178

or, x = 262.05

d) P( > 190)

= P(( - )/() > (190 - )/())

= P(Z > (190 - 178)/(41/))

= P(Z > 0.65)

= 1 - P(Z < 0.65)

= 1 - 0.7422

= 0.2578

d) P(X > 190)

= P((X - )/ > (190 - )/)

= P(Z > (190 - 178)/41))

= P(Z > 0.29)

= 1 - P(Z < 0.29)

= 1 - 0.6141

= 0.3859

n = 25

p = 0.3859

It is a binomial distribution

P(X = x) = nCx * p^x * (1 - p)^{n - x}

P(X = 10) = 25C10 * (0.3859)^10 * (1 - 0.3859)^15 = 0.1595