The weight of a male bald eagle has approximately a normal distribution with mean μ= 7.4 pounds and a standarddeviation of o=1.4 pounds. Suppose we take a random sample of n=81 male bald eagles. Let X be the randomvariable representing the mean weight in pounds of the 81 sampled male bald eagles. Let Xtot be the randomvariable representing sum of the weights of the 81 sampled bald eagles.a) About what proportion of male bald eagles have weights between 7.0 and 7.8 pounds?b) About what proportion of male bald eagles have weights greater than 7.8 pounds?c) About how many of the 81 sampled male bald eagles would you expect to have weight greater than 7.8 pounds?(nearest integer)?d) About how many of the 81 sampled male bald eagles would you expect to have weight between 7.0 and 7.8pounds? (nearest integer)?e) What is the standard deviation of the distribution of X (in pounds)?f) What is the standard deviation of the distribution of Xtot (in pounds)?g) What is the probability that 7.38 < X<7.42 pounds?h) What is the probability that Xtot >600 pounds?

Since you have posted a question with multiple sub parts, we will provide the solution only to the first three sub parts as per our Q&A guidelines. Please repost the remaining sub parts separately. From the provided information, Mean (µ) = 7.4 Standard deviation (σ) = 1.4 X~N (7.4, 1.4)a) The proportion of male bald eagles that will have weights between 7.0 and 7.8 pounds can be obtained as: P7.0<X<7.8=P7.0-7.41.4<X-μσ<7.8-7.41.4=P-0.2857<Z<0.2857=PZ<0.2857-PZ<-0.2857=PZ<0.2857-1-PZ<0.2857=2PZ<0.2857-1=20.612446-1 From excel =NORM.S.DIST(0.2857,TRUE)=0.224892≈0.2249 The excel output is as follow: Thus, the required proportion is 0.2249.b) The proportion of male bald eagles which have weights greater than 7.8 pounds can be obtained as: PX>7.8=PX-μσ>7.8-7.41.4=PZ>0.2857=1-PZ<0.2857=1-0.612446 From excel =NORM.S.DIST(0.2857,TRUE)=0.387554≈0.3876 The excel output is as follow: Thus, the required proportion is 0.3876. c) Sample size (n) = 81 Proportion of weight greater than 7.8 pounds is (p) = 0.3876 The expected number of male which have weight greater than 7.8 pounds = np = 81×0.3876 = 32 (approximately). Thus, 32 male bald eagles expected to have a weight greater than 7.8 pounds.