Question

Poverty and Dietary Calcium. Calcium is the most abundant mineral in the human body and has several important functions. Recommendations for calcium are provided in Dietary Reference Intakes, developed by the Institute of Medicine of the National Academy of Sciences, The recommended adequate intake (RAI) of calcium for adults (ages $19-50$) is $1000$milligrams (mg) per day. If adults with incomes below the poverty level have a mean calcium intake equal to the RAI. what percentage of all samples of $18$ such adults have mean calcium intakes of at most $947,4\mathrm{mg}$? Assume that $\sigma =188\mathrm{mg}$State any assumptions that you are making in solving this problem.

Short answer

$11.7\%$ of all samples of 18 such persons have mean calcium intakes of at least $947.4\mathrm{mg}$

Step-by-step solution

Given information

$1000$milligrams (mg) of calcium per day is the recommended acceptable intake (RAI) for adults (ages $19-50$). Assume that people with incomes below the poverty line consume the same amount of calcium as the RAl. Assume that $\sigma =188\mathrm{mg}$ is correct.

Calculation

We wish to know what percentage of all $18$adult samples had a mean calcium consumption of at least $947.4\mathrm{mg}$

We assume that calcium intake is roughly normally distributed among adults with incomes below the poverty line.

We have$\mu =1000,\sigma =188$ and $n=18$ based on the above data.

Calculation

Adults' mean calcium intakes are likely to be at least $947.4\mathrm{mg}$

$$\begin{gathered} P(\overline{X}\le 947.7)=P\left(\frac{\overline{X}-\mu }{\sigma /\sqrt{n}}\le \frac{947.4-1000}{188/\sqrt{18}}\right) \\ =P\left(Z\le \frac{-52.6}{44.31202}\right) \\ =P(Z\le -1.19) \\ =0.1170 \end{gathered}$$

As a result, $11.7\%$ of all samples of $18$ such persons have mean calcium intakes of at least $947.4\mathrm{mg}$