Question

Suppose cattle in a large herd have a mean weight of 1169lbs and a variance of 17,161.

What is the probability that the mean weight of the sample of cows would differ from the population mean by greater than 12lbs if 145 cows are sampled at random from the herd? Round your answer to four decimal places.

Answer 1

Solution :

Given that ,

mean = = 1169

standard deviation = = 131

= / n = 131 / 145 = 10.8790

= 1 - P[(-12) /10.8790 < ( - ) / < (12) / 10.8790)]

= 1 - P(-1.10 < Z < 1.10)

= 1 - P(Z < 1.10) - P(Z < -1.10)

= 1 - P (0.8643 - 0.1357)  

= 1 - 0.7286

= 0.2714

Probability = 0.2714