Q. 3.136 - Elementary Statistics

Question

Body mass index (BMI) is a measure of body fat based on height and weight. According to the document Dietary Guidelines for Americans published by the U.S. Department of Agriculture and the U.S. Department of Health and Human Services, for adults, a BMI of greater than $25$ indicates an above healthy weight (i.e., overweight or obese). The BMIs of $75$ randomly selected U.S. adults have a mean of $26.0$ and a standard deviation of $5.0$

a. Construct a graph
b. Apply Chebyshev's rule with $k=2$ to make pertinent statements about the observations in the sample.
c. repeat part (b) with $k=3$

Step-by-Step Solution

Answer

Part(a) Required figure is given below.

Part(b) At least $56$ persons are above healthy weight
Part(c) At least $67$ persons are above healthy weight

Part(a) Step 1 : Given information

We are given that the BMIs of $75$ U.S. adults is selected.

Mean $\overline{x}=26.0$

Standard deviation $s=5.0$

Part(a) Step 2 : Simplify

According to Chebysev's Rule 

$$\begin{gathered} \overline{x}=26 \\ \overline{x}-s=21 \\ \overline{x}-2s=16 \\ \overline{x}-3s=11 \\ \overline{x}+s=31 \\ \overline{x}+2s=36 \\ \overline{x}+3s=41 \end{gathered}$$

Required graph is

Part(b) Step 1 : Given information

We are given that the BMIs of $75$ U.S. adults is selected.

Mean $\overline{x}=26.0$

Standard deviation $s=5.0$

Part(b) Step 2 : Simplify

According to Chebyshev's rule ,if

$k=2$, then at least $75\%$ of the adult in the sample within two standard deviations to either side of the mean.

 Now,

$75\times 0.75\approx 56$

 Now, two standard deviations to either side of the mean is from 

$16$ to  $36.0$

Interpretation- At least $56$ out of $75$ adult  in the sample are above healthy weight between $16.0$and $36.0$

Part(c) Step 1 : Given information

We are given that the BMIs of $75$ U.S. adults is selected.

Mean $\overline{x}=26.0$

Standard deviation $s=5.0$

Part(c) Step 2 : Simplify

According to Chebyshev's rule ,if

$k=3$, then at least $89\%$ of the adult females in the sample within two standard deviations to either side of the mean.

 Now,

$75\times 0.89\approx 67$

 Now, two standard deviations to either side of the mean is from $15.5$ to $36.5$

Interpretation- At least $67$ out of $75$ adult in the sample are above healthy weight in between $15.5$ and $36.5$