Question

7. Let X be the random variable denoting the height of a randomly chosen adult individ- ual. If the individual is male, then X has a normal distribution with mean of = 70 inches with standard deviation of σ| 3.5 inches: while if the individual is female. then X has a normal distribution with mean μ0-66 inches and standard deviation of ơ0 3 inches. |Note: For computing probabilities and quantiles for the normal distribution, use the R functions pnorm, dnorm, and qnorm (a) Draw the two normal density functions in the same plot frame, that is, overlay (b) If (or conditional that) the individual is male, what is the probability that the (c) If the individual is female, what is the probability that the height is at least 68 the two density functions. height is at least 68 inches? inches (d) Assume that there is a 55% chance that the individual is male, so a 45% chance that she is female. Without knowing the gender of the individual, what is the probability that the height of the individual is at least 68 inches? Hint: Theorem (e) If you observe that the height is at least 68 inches, what is your posterior proba- (f) Suppose that you want a rule to classify an individual into male or female de- Classification Rule: Classify individual into male if the height of the person is What should be c so that the probability that you will classify a male person as (g) For the classification rule in (f), what is the probability that a female person will of Total Probability. bility that the individual is male? Hint: Bayes Theorem.] pending on the person's height. Consider the classification rule given by: at least c inches female (an error in decision, called Type I is 0.05? be classified as male (also an error of decision called Type II)? (h) If you want to decrease the probability of a Type II error to be equal to 0.05, what should c be and what will then happen to the probability of a Type I error?

Answer 1

a. Using R language

x <- rnorm(1000, mean = 70,sd = 3.5)
y <- rnorm(1000, mean = 66,sd = 3)
hx1 <- density(x)
hx2 <- density(y)
plot(hx1,col="red", ylim = c(0,0.15), main = "")
lines(hx2, col="green")

55 60 65 70 75 80 85 N= 1000 Bandwidth= 0.8095

b.

P(x<K) = pnorm(k, mean, sd)

P(x>=68) = 1 - pnorm(68, 70, 3.5)

                 = 0.7161454

c.

P(x<K) = pnorm(k, mean, sd)

P(x>=68) = 1 - pnorm(68, 66, 3)

       =  0.2524925

55 60 65 70758085 N -1000 Bandwidth 0.7963