Answer
$P(X\geq3.6)=0.0071$
Since $P(X\geq3.6)\gt0.05$, this result is unusual.
Unusual is not impossible. This sample contains many ten-gram portions of peanut butter with a high number of insect fragments.
Work Step by Step
$μ_{X ̅}=3$ and $σ_{X ̅}=0.245$
Let's find the z-score for 3.6:
$z=\frac{3.6-3}{0.245}=2.45$
According to Table V, the area of the standard normal curve to the left of z-score equal to 2.45 is 0.9929.
But, we want the area of the standard normal curve to the right of z-score equal to 2.45:
$1-0.9929=0.0071$
$P(X\geq3.6)=0.0071\lt0.05$. It is an unusual event.
