Question

Question 2 [12 marks Let X be the maximum temperature in the month of January in Melbourne and let Y be the maximum temperature in the same month in Sydney. Historical records show that the distribution of X and Y have probability density functions fx(x), fy(y), respectively, shown in Figure 1 (a) By eye-balling the plot, explain why the population mean and variances of the maximum temperatures in Melbourne and Sydney are Hx 25°C, Hy28°C σ -1°C, and σ-2.25°C (b) Assume now that you have 10 independent measurements of maximum temperature X1o, and 5 independent measurements of maximum temp in Melbourne Λι, er ature in Sydney Yi,... . Ys (i) What are the distributions of TO 25) andY -28)? (ii) What is the probability that the sample average of the measured temperatures in Melbourne, denoted as X, will lie between 24.37°C and 25.32°C? (iii) What is the distribution of Y-X, where Υ is the sample average of the measured temperatures in Sydney? (iv) What critical temperature T, will yield PO-x T) 0.05? (X-25) and s30-28), where 陨and S루 and Y, respectively? (v) What are the distributions of the sample variances of X

Answer 1

a)
i) the mean for x is 25 because peak of f(x) occurs at x = 25
similary mean for y is 28

Now 95% lies between 2 sd of mean
for X
we see that 95% lies between 23 and 27
hence
2*sd = 27-25 = 2
sd = 1
similary
for Y
2 *sd = 31 - 28 = 3
sd = 1.5
hence
variance of Y = 1.5^2 = 2.25

b)
i)
Z = (Xbar - mu)/(sigma_x/sqrt(n)) follow standard normal distribution
hence
sqrt(10)(Xbar - 25) follow standard normal distribution

similarly
Z = (ybar - 28)/(1.5/sqrt(5))
= sqrt(20)/3 * (ybar - 28) follow standard normal distribution

ii)
P(24.37 < Xbar < 25.32)
= P ( −1.99<Z<1.01 )
=0.8205

iii)
Ybar - Xbar = N(28-25, 1^2/10 + 1.5^2/5)
= N(3 , 0.55)

iv)
P(Ybar - Xbar > Tc) < 0.05
P(Z > z) <= 0.05
z = 1.645
hence
Tc = 3 + 1.645*sqrt(0.55)
= 4.21996

v)
these follow t-distribution with 9 and 4 degree of freedom respectively