TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2023W/HW04.2

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 desribed by a Normal model with a mean of 188 mg/dL and a standard deviation of 24. Use R to answer the following questions.

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

(a) Draw and label the cumulative distribution function of this model.

(b) Approximate the probability that cholesterol levels of adult American women are over 200 mg/dL?

(c) Approximate the probability that cholesterol levels of adult American women are between 155 and 170 mg/dL?

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

(e) Above what value are the highest 10% of the cholesterol levels of adult American women?

Dieses Beispiel ist als solved markiert. Ist dies falsch oder ungenau? Aktualisiere den Lösungsstatus (Details: Vorlage:Beispiel)

Hilfreiche Tipps

Ähnliche Beispiele:

• Übung 2019W - HW 3.7
• Übung 2020W - HW 4.6

Lösungsvorschlag von Simplex

(a)

mean <- 188
standardDeviation <- 24
mg <- (mean - 3 * standardDeviation):(mean + 3 * standardDeviation)
p <- pnorm(mg,mean,standardDeviation)
#X11() #output in extra window
plot(mg,p,type='l', main="mg/dL")

(b)

b <- 1 - pnorm(200, mean, standardDeviation)

$$P(X>200)\approx 30.85\%$$

(c)

c <- pnorm(170, mean, standardDeviation) - pnorm(155, mean, standardDeviation)

$$P(155<X\leq 170)\approx 14.21\%$$

(d)

d <- qnorm(0.75, mean=mean, sd=standardDeviation) - qnorm(0.25, mean=mean, sd=standardDeviation)

$${\text{IQR}}=x_{0.75}-x_{0.25}=32.376$$

(e)

e <- qnorm(0.9, mean=mean, sd=standardDeviation)

$$x_{0.9}=218.76$$

--Simplex 20:24, 2. Feb. 2023 (CET)