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

Answer:

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

Step-by-step 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:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling:

$\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{27}{\sqrt{36}} = 4.5$

P(weights lies between 163 and 170)

$P(163 \leq x \leq 170) = P(\displaystyle\frac{163 - 167}{4.5} \leq z \leq \displaystyle\frac{170-167}{4.5}) = P(-0.889 \leq z \leq 0.667)\\\\= P(z \leq 0.667) - P(z < -0.889)\\= 0.748 - 0.187 = 0.561 = 56.1\%$

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

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