Question

Q-2 (7.6) An irruption is an irregular migration of birds to a region where they are not usually found. In North America, approximately every four years, there is a winter eruption that causes snowy owls to fly south in larger numbers than usual. Many Ontarians were fortunate to sight a snowy owl in 2014. Suppose that the size of the body of a snowy owl is normally distributed with a mean 61.5 cm and a standard deviation of 4.75 cm We select n 10 snowy owls from the population and compute the average body size (in cm) We consider this average as a point estimate of the population mean of the body size. (a) Compute the standard error of the mean. (b) What is the probability that the average body size of these 10 snowy owls will be less than 57cm? (c) Suppose that we randomly select 5 samples (each of size n = 10) from this population What is the probability that at most one of the 5 samples will have an average body size that is less than 57cm?

Answer 1

Let X follows N(u = 61.5,0 =4.75) a) The standard error of the mean is, 0x =%Tom b) =1.5027 P(X<57)=P 4 7) - DX-u 57-61.5 (0 1 .5021 = P(z<-3.00) (excelfunction(= NORMSDIST(-))) = 0.00135 0.0013 b) Here, X follows binomial distribution with parameters n= 5, p = 0.0013 P(X S1)=P(X = 0)+P( X =1) =(5)(0.0013)º(1–0.0013) - +((0.0013) (1–0.001394 = 0.993517 +0.006466 =1.0000