Question

Can you please help me with this standard deviation related problem?

Answer 1

`\begin{gathered} \mu=13 \\ \sigma=1.1 \\ \end{gathered}`

So:

`\begin{gathered} P((X1-\mu)/(\sigma)\leq Z\leq(X2-\mu)/(\sigma)) \\ P((14.1-13)/(1.1)\leq Z\leq(15.2-13)/(1.1)) \\ P(1\leq Z\leq2)=0.1359 \end{gathered}`

Therefore:

`0.1352*3000\approx408`

Answer:

408 phones