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:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Standard error due to sampling:
[tex]\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{27}{\sqrt{36}} = 4.5[/tex]
P(weights lies between 163 and 170)
[tex]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\%[/tex]
[tex]P(163 \leq x \leq 170) = 56.1\%[/tex]
0.561 is the probability that the sample mean of their weights lies between 163 and 170.
