Question

suppose cattle in a large herd have a mean weight of 914 lbs and a standard deviation of 149lbs what is the probability that the mean weight of the sample of cows would differ from the population means by greater than 7lbs if 57 cows are samples at random from the herd?

Answer 1

Here' the answer to the question. Please let me know in case you've doubts.

Normal dist parameters are given below:

Mean =914 lbs

Stdev =149 lbs.

n = sample size = 57

P(|xbar-Mu| > 7 lbs) = ?

first lets standardize the distribution using the formula:

Z = (xbar-Mu)/(stdev/sqrt(n))

So,

P(|xbar-Mu| > 7 lbs)

= P( |Z| > (7)/(149/sqrt(57)) )

= P( |z| > .3547) ( Lets look up the Z-tables to convert Z value to cumualtive probability

= 1- P( - 0.3547 < Z < + 0.3547)

= 1- (0.638589 - 0.361411)

= 1-0.2772

= .7228

Answer: 0.7228