Question

One timber company claims that the amount of usable lumber in the trees being harvested in ClearCut is approximately normally distributed, with a mean of 256 cubic feet per tree and a standard deviation of 42 cubic feet.

a) What is the probability that the amount of usable lumber in a randomly selected tree is less than 250 cubic feet?

b) If a decision is made not to harvest the smallest 20% of the trees, they will be harvested when they are bigger. What is the smallest size of tree allowed to be cut?

c) If 49 trees are randomly selected from ClearCut, what is the probability that the average amount of usable lumber in the 49 selected trees is less than 250 cubic feet?

Answer 1

Solution :

Given that ,

mean = = 256

standard deviation = = 42

a) P(x < 250)

= P[(x - ) / < (250 - 256) / 42]

= P(z < -0.14)

Using z table,

= 0.4443

b) Using standard normal table,

P(Z < z) = 20%

= P(Z < z ) = 0.20

= P(Z < -0.84 ) = 0.20  

z = -0.84

Using z-score formula,

x = z * +

x = -0.84 * 42 + 256

x = 220.72

c) n = 49

=   = 256

= / n = 42 / 49 = 6

P( < 250) = P(( - ) / < (250 - 256) / 6 )

= P(z < -1.00)

Using z table

= 0.1587