Answer
P( $\bar{x}$ < 12750 or $\bar{x}$ > 12753) $\approx$ 0.5
Work Step by Step
n = 36
$\sigma$ = 1.7
$\mu$ = 12750
Want to find P($\bar{x}$ < 12750 or $\bar{x}$ > 12753)
i) Find the z score corresponding to 12750
z = $\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt n}}$
z = $\frac{12750 - 12750}{\frac{1.7}{\sqrt 36}}$
z = 0
ii) Find the z score corresponding to 12753
z = $\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt n}}$
z = $\frac{12753 - 12750}{\frac{1.7}{\sqrt 36}}$
z = 10.59
iii) P( z < 0 or z > 10.59) = $P(z < 0)$ + $P(z > 10.59)$
P( z < 0 or z > 10.59) $\approx$ 0.5 + 0 (Values were found using technology)
P( z < 0 or z > 10.59) $\approx$ 0.5
Therefore: P( $\bar{x}$ < 12750 or $\bar{x}$ > 12753) $\approx$ 0.5
