Question

2. It is known that the wing span of a certain species of hawk has mean and a standard deviation of 2.5 inches.

a. If an entomologist takes a sample of 40 hawks, what is the probability that the average wing length for the sample differs from µ by more than 0.5 inch.

b. How large a sample is needed to be 98% sure that the average wing length for the sample differs from µ by less than 0.5 inch

Answer 1

2)

here std error =std deviation/sqrt(n)=2.5/sqrt(40)

also as z score =(X-mean)/std error

hence  probability that the average wing length for the sample differs from µ by more than 0.5 inch

=P(Z>0.5/(2.5/sqrt(40)))+P(Xbr<-0.5/(2.5/sqrt(40)))=P(Z>1.26)+P(Z<1.26)=0.1038+0.1038 =0.2076

b)

for 98 % CI value of z=	2.326
standard deviation =	2.50
margin of error E =	0.5
required sample size n=(z/E)2 =	136.0