91Ó°ÊÓ

Daily allowance,$x=1200mgor1800mg$

Mean,$\mathrm{μ}=2000mg$

Standard deviation,$\mathrm{σ}=500mg$

The formula used:$z=\frac{x−\mathrm{μ}}{\mathrm{σ}}$

Calculate the Z − score,

$z=\frac{x−\mathrm{μ}}{\mathrm{σ}}=\frac{1200mg−2000mg}{500mg}=−1.60$

or

$z=\frac{x−\mathrm{μ}}{\mathrm{σ}}=\frac{1800mg−2000mg}{500mg}=−0.40$

To find the equivalent probability, use the normal probability table in the appendix.

In the typical normal probability table for$P(z<−1.60)$, look at the row that starts with -1.6 and the column that starts with.00.

Or

The usual normal probability table for$P(z<−0.40)$has a row that starts with -0.4 and a column that starts with.00.

$$\begin{gathered} P(1200<x<1800)=P(−1.60<z<−0.40) \\ =P(z<−0.40)−P(z<−1.60) \\ =0.3446−0.0548 \\ =0.2898 \\ =28.98\% \end{gathered}$$

Therefore,

Around $28.98\%$ of the meals ordered contained between $1200$ mg and $1800$ mg of sodium.