Answer
Since $P(ρ ̂\geq0.11)\gt0.05$, it is not an unusual event.
Work Step by Step
$μ_{ρ ̂}=0.10$
$σ_{ρ ̂}=\sqrt {\frac{p(1-p)}{n}}=\sqrt {\frac{0.10(1-0.10)}{1100}}=0.00905$
The proportion in the sample: $ρ ̂=\frac{121}{1100}=0.11$
Let's find the z-score for 0.11:
$z=\frac{ρ ̂-μ_{ρ ̂}}{σ_{ρ ̂}}=\frac{0.11-0.10}{0.00905}=1.10$
According to Table V, the area of the standard normal curve to the left of z-score equal to 1.10 is 0.8643.
But, we want the area of the standard normal curve to the right of z-score equal to 1.10:
$P(ρ ̂\geq0.11)=1-0.8643=0.1357$
