Independent solution

How to solve this Normal Distribution question

Setup

Setup

Convert the stated variance to a standard deviation before standardizing.

μ=10,σ=4=2\mu=10,\qquad \sigma=\sqrt4=2

Model

Model

Let q be the desired percentile and match its standardized value to the lower-tail normal quantile.

Pr⁡(X≤q)=0.12\Pr(X\le q)=0.12
q−102=Φ−1(0.12)\frac{q-10}{2}=\Phi^{-1}(0.12)

Compute

Compute

Evaluate the standard-normal quantile and transform it to the original units.

Φ−1(0.12)=−1.1749868\Phi^{-1}(0.12)=-1.1749868
q=10+2(−1.1749868)=7.6500264q=10+2(-1.1749868)=7.6500264

Answer

Answer

The 12th percentile is approximately 7.65 years.

7.65(B)\boxed{7.65\quad\text{(B)}}