3. Assume the yearly acorn yield for adult oak trees is normally distributed wit

ID: 3173521 • Letter: 3

Question

3. Assume the yearly acorn yield for adult oak trees is normally distributed with mean 1000 grams, with a standard deviation of 200 grams.

(a) What are the bounds containing the yearly yield for approximately 95% of adult oak trees?

(b) What is the probability that an oak tree yields less than 980 grams of acorns?

(d) Calculate P(More than 800 grams|less than 980 grams) for the yield of a single adult oak tree.

(e) Calculate P(More than 800 grams|less than 980 grams or greater than 1100) for the yield of a single adult oak tree.

(f) Repeat parts a)-e) using the average of 16 oak trees instead of a single tree.

a) for 95% CIm z=1.96

hence confidence interval =mean +/- z*std deviation =608.0072 ; 1391.9928

b) P(X<980)=P(Z<(980-1000)/200)=P(X<-0.1)=0.4602

c)P(X<980)+P(X>1100)=P(Z<-0.1)+P(Z>0.5)=0.4602+0.3085=0.7687

d)P(X>800|X<980)=P(800<X<980)/P(X<980)=P(-1<Z<-0.1)/P(Z<-0.1)=(0.4602-0.1587)/0.4602=0.6552

e) P(An(BUC)=P(AnB)UP(AnC)

P(X>800 and P(X<980 U X>1000) =P(800<X<980 U (X>800 n X>1000) =P(800<X<980 U (X>1000)

P(X>800|P(X<980UX>1100)=P(800<X<980UP(X>100)/P(X<980UX>1100) =(0.4602-0.1587+0.3085)/(0.7687)=0.7935

f) for 16 oak trees,std deviation=(200/(16)1/2 =50

a) confidence interval =1000+/-1.96*50=902.0018 ; 1097.9982

b) P(X<980)=P(Z<(980-1000)/50)=P(Z<-0.4)=0.3446

c)P(X<980)+P(X>1100)=P(Z<-0.4)+P(Z>2)=0.3446+0.02275=0.3673

d)P(X>800|X<980)=P(-4<Z<-0.4)/P(Z<-0.4)=(0.3446-0.00003)/0.3446=0.999908

e)=P(0.3446-0.00003+0.02275)/(0.3673)=0.999914

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