Question

Suppose the distribution of board lengths in a stack of lumber is normal, with a mean of 20 ft and a standard deviation of 0.1 ft.

(a) Find the probability that an individual board from the stack is longer than 19.7 ft.

(b) Find the 80th percentile for the distribution of board lengths.

(c) The middle 40% of board lengths fall between what two lengths?

Answer 1

Solution :

Given that ,

a) P(x > 19.7) = 1 - p( x< 19.7)

=1- p P[(x - ) / < (19.7 - 20) / 0.1]

=1- P(z < -3.00)

= 1 - 0.0013

= 0.9987

b) Using standard normal table,

P(Z < z) = 80%

= P(Z < 0.842 ) = 0.80  

z = 0.842

Using z-score formula,

x = z * +

x = 0.842 * 0.1 + 20

x = 20.08

80 th percentile = 20.08 feet.

c) Using standard normal table,

P( -z < Z < z) = 40%

= P(Z < z) - P(Z <-z ) = 0.40

= 2P(Z < z) - 1 = 0.40

= 2P(Z < z) = 1 + 0.40

= P(Z < z) = 1.40 / 2

= P(Z < z) = 0.70

= P(Z < 0.524) = 0.70

= z  ± 0.524

Using z-score formula,

x = z * +

x = -0.524 * 0.1 + 20

x = 19.95 feet

Using z-score formula,

x = z * +

x = 0.524 * 0.1 + 20

x = 20.05 feet

The middle 40% of board lengths fall between 19.95 feet and 20.05 feet.