TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Bura)/Übungen 2019W/3.7

Drug company

Manufacturing and selling drugs that claim to reduce an individual's cholesterol level is big business. A company would like to market their drug to women if their cholesterol is in the top 15%. 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. Use R to answer the following questions.

(a) Draw and label the Normal model.

(b) What percent of adult women do you expect to have cholesterol levels over 200 mg/dL?

(c) What percent of adult women do you expect to have cholesterol levels between 150 and 170 mg/dL?

(d) Calculate the interquartile range of the cholesterol levels.

(e) Above what value are the highest 15% of women's cholesterol levels?

Hint: R commands pnorm(), qnorm() and dnorm() are useful.

> mean <- 188
> sd <- 24

(a)

> z_scores <- seq(100, 275, by = .1)
> dvalues <- dnorm(z_scores, mean, sd)
> plot(dvalues, xaxt = "n", type = "l", xlab= "Z-score") 
> axis(1, at=which(dvalues == dnorm(mean, mean, sd)), labels=c(0))
> axis(1, at=which(dvalues == dnorm(mean - sd, mean, sd)), labels=c(-1, 1))
> axis(1, at=which(dvalues == dnorm(mean - 2*sd, mean, sd)), labels=c(-2, 2))
> axis(1, at=which(dvalues == dnorm(mean - 3*sd, mean, sd)), labels=c(-3, 3))

Simpler Alternative (centered around the mean, not zero):

> x <- seq(mean - (4 * sd), mean + (4 * sd), by = .1)
> plot(x, dnorm(x, mean, sd))

(b)

> 1 - pnorm(200, mean, sd)
[1] 0.3085375

(c)

> pnorm(170, mean, sd) - pnorm(150, mean, sd)
[1] 0.1699546

(d)

> qnorm(0.75, mean, sd) - qnorm(0.25, mean, sd)
[1] 32.37551

(e)

> qnorm(0.85, mean, sd)
[1] 212.8744