Question

I did these questions but I'm not sure if I did them right especially C:
Forest type 1 has an average diameter of 35 cm and a standard deviation of 6 cm. Forest type 2 has an average diameter of 25 cm and a standard deviation of 5 cm. The distribution of the diameters is normal in both types.

Answer 1

A What is the probability that a sample mean will be greater than 36 cm from type 1 if 20 trees are measured?
B What is the probability that a sample variance is less than 19 cm2 from type 1 if 20 trees are measured?
C Find the probability that the sample mean computed from 20 measurements from type 1 will exceed the sample mean computed from 25 measurements from type 2 by at least 11 cm.
D Find the probability that the ratio of sample variances (S12/S22) computed from 10 measurements from type 1 (S12) and 8 measurements from type 2 (S22) will exceed 5.4.

My answers:
FT1 μ=35cm σ=6cm
FT2 μ=25cm σ=5cm

(A)
σx=σ/sqrt(n)
=6/SQRT(20)

Z= (x-Ux)/σx
=(36-35)/6/SQRT(20)
= 1/6/SQRT(20)
= 0.745355992

P(Z>0.75) = P(Z<-0.75)
=0.2266

(B)
P(S²<19cm²) = P(X²20-1 <[(20-1)19]/62 = P(X²19 <[(19)19]/36 ≈ 10.03
P(X²19<10.03) ≈ 0.05

(C)
σx̄1- x̄2 = sqrt( 62 /20 +52 /25) =sqrt(1.8 +1) ≈ 1.673

P(X1-X2)<11)= P(Z<[( x̄1- x̄2)-(μ1 -μ2)]/ σx̄1- x̄2
P(Z< [(11) - (35-25)]/1.673)
P(Z< 0.60) ≈ 0.2743

(note: even though the sample size is <30 we can simply use a Z transformation because the problem stated that the population distribution is normal in both cases)

(D)
P(S21/S22>5.4) = P(F(9,7)> 5.4 (25/36)) = P(F(9,7)> 3.75) ≈ 0.05