Question

Suppose the horses in a large stable have a mean weight of 975lbs, and a standard deviation of 52lbs.What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbsif 31horses are sampled at random from the stable

Answer 1

Answer: The required probability = 0.8926

Explanation:

Given:
`\mu=975\text{ lbs}`,
`\sigma= 52\text{ lbs}`

Let x = weight of horse stable.

Sample size : n= 31

Then, the probability that the mean weight of the sample of horses would differ from the population mean by less than 15 will be:

`P(-15<P(\overline{x}-\mu)<15)=P((-15)/((52)/(√(31)))<\frac{\overline{x}-\mu}{(\sigma)/(√(n))}<(15)/((52)/(√(31))))\\\\=P(-1.61<z<1.61)\ \ \ [z=\frac{\overline{x}-\mu}{(\sigma)/(√(n))}]\\\\=2(z<1.61)-1\ \ \ [P(-z<Z<z)=2(Z<z)-1]\\\\=2( 0.9463)-1\\\\=0.8926`

Hence, the required probability = 0.8926