Answer :
Answer:
0.500 is the required probability.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 12,749
Standard Deviation, σ = 1.5
Let X be the distribution of sample means.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Standard error due to sampling =
[tex]=\dfrac{\sigma}{\sqrt{n}} = \dfrac{1.5}{\sqrt{37}} = 0.2465[/tex]
We have to evaluate:
P(x less than 12,749 or greater than 12,752)
[tex]= 1 - P(12749\leq x \leq 12752)[/tex]
Now,
[tex]P(12749\leq x \leq 12752)\\\\ = P(\displaystyle\frac{12749 - 12749}{0.2465} \leq z \leq \displaystyle\frac{12752-12749}{0.2465}) \\\\= P(0 \leq z \leq 12.17)\\\\= P(z \leq 12.17) - P(z < 0)\\= 1.000 - 0.500 = 0.500[/tex]
P(x less than 12,749 or greater than 12,752)
[tex]= 1 - P(12749\leq x \leq 12752)\\=1-0.500\\=0.500[/tex]
Thus, 0.500 is the required probability.