Question

A forester measured 27 of the trees in a large forest on property that is for sale. He found a mean diameter of 10.4 inches and a standard deviation of 4.7 inches. Suppose that these trees provide an accurate description of the whole forest and that a Normal model applies.Determine:a) the percent of trees that has a diameter between 10.4 and 16.0 inches.b) the percent of trees that has a diameter greater than 16.0 inches. c) the percent of trees that has a diameter between 8.0 and 13.25 inches.d) the percent of trees that has a diameter between 8.0 and 10.0 inches. e) Any tree which has a diameter less than 4.00 inches is considered to be a “Small Tree.” Determine the percent of trees that receive this designation.f) The largest 5% of trees is designated as a “Big Tree”. Determine the diameter that a tree would need to have in order to receive this designation.g) The smallest 5% of trees are given extra nutrients in order to grow. Determine the diameter that a tree would need to have in order to be given the extra nutrients.

Answer 1

µ = 10.4, σ = 4.7

a) P(10.4 < X < 16) =

= P( (10.4-10.4)/4.7 < (X-µ)/σ < (16-10.4)/4.7 )

= P(0 < z < 1.1915)

= P(z < 1.1915) - P(z < 0)

Using excel function:

= NORM.S.DIST(1.1915, 1) - NORM.S.DIST(0, 1)

= 0.3833 = 38.33%

b)

P(X > 16) =

= P( (X-µ)/σ > (16-10.4)/4.7)

= P(z > 1.1915)

= 1 - P(z < 1.1915)

Using excel function:

= 1 - NORM.S.DIST(1.1915, 1)

= 0.1167 = 11.67%

c)

P(8 < X < 13.25) =

= P( (8-10.4)/4.7 < (X-µ)/σ < (13.25-10.4)/4.7 )

= P(-0.5106 < z < 0.6064)

= P(z < 0.6064) - P(z < -0.5106)

Using excel function:

= NORM.S.DIST(0.6064, 1) - NORM.S.DIST(-0.5106, 1)

= 0.4231 = 42.31%

d)

P(8 < X < 10) =

= P( (8-10.4)/4.7 < (X-µ)/σ < (10-10.4)/4.7 )

= P(-0.5106 < z < -0.0851)

= P(z < -0.0851) - P(z < -0.5106)

Using excel function:

= NORM.S.DIST(-0.0851, 1) - NORM.S.DIST(-0.5106, 1)

= 0.1613 = 16.13%

e)

P(X < 4) =

= P( (X-µ)/σ < (4-10.4)/4.7 )

= P(z < -1.3617)

Using excel function:

= NORM.S.DIST(-1.3617, 1)

= 0.0866 = 8.66%

f)

P(x > a) = 0.05

= 1 - P(x < a) = 0.05

= P(x < a) = 0.95

Z score at p = 0.95 using excel = NORM.S.INV(0.95) =1.6449

Value of X = µ + z*σ = 10.4 + (1.6449)*4.7 = 18.13

g)

P(x < a) = 0.05

Z score at p = 0.05 using excel = NORM.S.INV(0.05) =-1.6449

Value of X = µ + z*σ = 10.4 + (-1.6449)*4.7 = 2.67