Independent solution

How to solve this Normal Distribution question

Setup

Setup

Let S be the total amount across the 24 independent, identically distributed fills. Add their means and variances.

E[S]=24(12)=288E[S]=24(12)=288
Var⁡(S)=24σ2\operatorname{Var}(S)=24\sigma^2

Model

Model

A sum of independent normal variables is normal, and the given upper-tail probability places the threshold at the 80th percentile.

S∼N(288,24σ2)S\sim N(288,24\sigma^2)
Pr⁡(S>290)=0.20\Pr(S>290)=0.20
290−28824σ=Φ−1(0.80)\frac{290-288}{\sqrt{24}\sigma}=\Phi^{-1}(0.80)

Compute

Compute

Insert the standard-normal quantile and solve for the individual standard deviation.

Φ−1(0.80)=0.8416212336\Phi^{-1}(0.80)=0.8416212336
σ=224(0.8416212336)=0.4850736581\sigma=\frac{2}{\sqrt{24}(0.8416212336)}=0.4850736581

Answer

Answer

The individual-fill standard deviation rounds to 0.49.

0.49(C)\boxed{0.49\quad\text{(C)}}