Answer:
The sample mean would not be considered unusual because the probability is greater than or equal to 0.50 of the sample mean being within the range.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 12750
Standard Deviation, σ = 1.7
Sample size, n = 36
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Standard error due to sampling:
[tex]\dfrac{\sigma}{\sqrt{n}} = \dfrac{1.7}{\sqrt{36}} = 0.283[/tex]
P(sample mean being less than 12,750 or greater than 12,753)
[tex]1 - P( 12750 < x < 12753) = 1-P(\displaystyle\frac{12750 - 12750}{0.283} \leq z \leq \displaystyle\frac{12753-12750}{0.283})\\\\ = 1-(P(0 \leq z \leq 10.06))\\\\= 1-P(z \leq 10.06) +P(z < 0)\\= 1 - 1 + 0.500 = 0.500[/tex]
Thus, the sample mean would not be considered unusual because the probability is greater than or equal to 0.50 of the sample mean being within the range.
