Question

Suppose cattle in a large herd have a mean weight of 1158lbs and a standard deviation of 92lbs . What is the probability that the mean weight of the sample of cows would differ from the population mean by less than 12lbs if 55 cows are sampled at random from the herd? Round your answer to four decimal places.

Answer 1

Solution :

Given that,

mean = = 1158

standard deviation = = 92

n = 55

= 1158

= / n = 92 / 55

P(1146 < < 1170) = P((1146 - 1158) / 92 / 55 <( - ) / < (1170 - 1158) / 92 / 55))

= P( -0.97 < Z < 0.97)

= P(Z < 0.97) - P(Z < -0.97) Using z table,

= 0.834 - 0.166

= 0.6680

Probability = 0.6680