Question

The weights of a population of workers have mean 167 and standard deviation 27. If a sample of 36 workers is chosen, approximate the probability that the sample mean of their weights lies between 163 and 170?

Answer 1

0.561 is the probability that the sample mean of their weights lies between 163 and 170.

Explanation:

We are given the following information in the question:

Mean, μ = 167

Standard Deviation, σ = 27

Since the sample size is large, by central limit theorem the distribution of means is a normal distribution.

We are given that the distribution of weights of a population of workers is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

Standard error due to sampling:

`\displaystyle(\sigma)/(√(n)) = (27)/(√(36)) = 4.5`

P(weights lies between 163 and 170)

`P(163 \leq x \leq 170) = P(\displaystyle(163 - 167)/(4.5) \leq z \leq \displaystyle(170-167)/(4.5)) = P(-0.889 \leq z \leq 0.667)\\\\= P(z \leq 0.667) - P(z < -0.889)\\= 0.748 - 0.187 = 0.561 = 56.1\%`

`P(163 \leq x \leq 170) = 56.1\%`

0.561 is the probability that the sample mean of their weights lies between 163 and 170.