Question

Suppose that the height of Australian males is a normally distributed random variable with a mean of 176.8cm and a standard deviation of 9.5cm.

a. If the random variable X is the height of an Australian male, identify the distribution of X and state the value/s of its parameter/s.

b. Calculate (using the appropriate statistical tables) the probability that a randomly selected Australian man is more than two metres tall.

c. To become a jockey, as well as a passion for the sport, you need to be relatively small, generally between 147cm and 168cm tall. Calculate (using the appropriate statistical tables) the proportion of Australian males who fit this height range.

d. Some of the smaller regional planes have small cabins, consequently the ceilings can be quite low. Calculate (using the appropriate statistical tables) the ceiling height of a plane such that at most 2% of the Australian men walking down the aisle will have to duck their heads.

e. Verify your answers to parts b., c. and d. using the appropriate Excel statistical function and demonstrate you have done this by including the Excel formula used.

f. A random sample of forty Australian males is selected. State the type of distribution and the value/s of the parameter/s for the mean of this sample.

g. Calculate (using the appropriate statistical tables) the probability that the average height of this sample is less than 170cm.

Answer 1

a.

The distribution of X is normal distribution with a mean of 176.8cm and a standard deviation of 9.5cm.

X ~ Normal( = 176.8cm, = 9.5cm)

b.

Probability that a randomly selected Australian man is more than two metres tall = P(X > 200cm)

= P[z > (200 - 176.8) / 9.5]

= P[z > 2.44]

= 0.0073

c.

Proportion of Australian males who fit the height range = P[147cm < X <  168cm]

= P[X < 168cm] - P[X < 147cm]

= P[z < (168 - 176.8) / 9.5] - P[z < (147 - 176.8) / 9.5]

= P[z < -0.9263] - P[z < -3.1368]

= 0.1771 - 0.0009

= 0.1762

d.

Let x be the ceiling height of a plane such that at most 2% of the Australian men walking down the aisle will have to duck their heads. Then, P(X > x) = 0.02

=> P[z > (x - 176.8) / 9.5] = 0.02

=> (x - 176.8) / 9.5 = 2.0537

=> x = 176.8 + 2.0537 * 9.5 = 196.3102 cm

e.

Using excel function, =1 - NORM.DIST(200,176.8,9.5,TRUE)

P(X > 200cm) = 0.007301

Using excel function, = NORM.DIST(168,176.8,9.5,TRUE) - NORM.DIST(147,176.8,9.5,TRUE)

P[147cm < X <  168cm] = 0.176287

Using excel function, =NORM.INV(1-0.02,176.8,9.5) we get the value of x = 196.3106 cm such that

P(X > x) = 0.02

f.

By Central limit theorem, the distribution of sample mean is normal distribution with a mean of 176.8cm and a standard deviation of 9.5/ = 1.5cm.

~ Normal( = 176.8cm, = 1.5cm)

g.

Probability that the average height of this sample is less than 170cm = P( < 170)

= P[z < (170 - 176.8) / 1.5]

= P[z < -4.53]

= 0.0000