Question

Suppose that the height of Australian men is normally distributed with a mean of 175cm and standard deviation of 5cm.

i. What is the probability that a Australian man's height will be between 180cm and 190cm?   

ii. What is the probability that a Australian man's height will be less than 190cm?

iii. Ten percent (10%) of Australian men were taller than what height?

Answer 1

Solution:

Given, X follows Normal distribution with,

   = 175

= 5

i. )

P(180 < x< 190)

= P(X < 190) - P(X < 180 )

=  P[(X - )/ <  (190 - 175)/5] -   P[(X - )/ <  (180 - 175)/5]

= P[Z < 3.00] - P[Z < 1.00]

= 0.9987 - 0.8413 ..Use z table

= 0.1574

Answer :  0.1574

ii. )

P[X < 190]

=  P[(X - )/ <  (190 - 175)/5]

= P[Z < 3.00]

= 0.9987

Answer :  0.9987

iii.)

For top 10% data , let x be the required cut-off.

P(X > x) = 10%

P(X > x) = 0.10

P(X < x) = 1 - 0.10

P(X < x) = 0.90

For the standard normal variable z , P(Z < z) = 0.90

Use z table , see where is 0.90  probability and then see the corresponding z value.

P(Z < 1.282) = 0.90

So z = 1.282

Now using z score formula ,

x = + (z * ) = 175 + (1.282 * 5) = 181.41

Answer : 181.41 cm