Question

Suppose cattle in a large herd have a mean weight of 1228lbs and a variance of 8100. What is the probability that the mean weight of the sample of cows would be greater than 1240lbs if 115 cows are sampled at random from the herd? Round your answer to four decimal places.

Answer 1

µ = 1228

sd = sqrt(8100) = 90

n = 115

$P(\bar{X}>1240)=P(\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}>\frac{1240-\mu}{\sigma/\sqrt{n}})$

                              $=P(Z>\frac{1240-1228}{90/\sqrt{115}})$

                              = P(Z > 1.43)

                              = 1 - P(Z < 1.43)

                              = 1 - 0.9236

                              = 0.0764