Chapter 2.2, Problem 57E

To determine

Percentage of the meals ordered contained between 1200 mg and 1800 mg of sodium.

Answer to Problem 57E

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

Explanation of Solution

Given information:

Daily allowance, $x=1200\text{mg}\text{or 1800}\text{mg}$

Mean, $\mu =2000\text{mg}$

Standard deviation, $\sigma =500\text{mg}$

Calculations:

Calculate the Z − score,

  $z=\frac{x-\mu }{\sigma }=\frac{1200\text{mg}-2000\text{mg}}{500\text{mg}}=-1.60$

Or

  $z=\frac{x-\mu }{\sigma }=\frac{1800\text{mg}-2000\text{mg}}{500\text{mg}}=-0.40$

Use normal probability table in the appendix, to find the corresponding probability.

See the row that starts with -1.6 and the column that starts with .00 of the standard normal probability table for $P\left(z<-1.60\right)$ .

Or

See the row that starts with -0.4 and the column that starts with .00 of the standard normal probability table for $P\left(z<-0.40\right)$ .

  $\begin{array}{l}P\left(1200<x<1800\right)=P\left(-1.60<z<-0.40\right)\\ =P\left(z<-0.40\right)-P\left(z<-1.60\right)\\ =0.3446-0.0548\\ =0.2898\\ =28.98\%\end{array}$

Therefore,

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