Of two persons on a reducing diet, the first belongs to an age group for which the mean weight is 142 pounds with a standard deviation of 15 pounds, and the second belongs to an age group for which the mean weight is 138 pounds with a standard deviation of 13 pounds. If their respective weights are 161 pounds and 154 pounds, which of the two is more seriously overweight for his or her age group? (use z-scores and round to the hundredths, if necessary)
Solution
Let $X$ = weight of first group. $X\sim N(142,15^2)$.
For first group, the mean weight is $\mu = 142$ and standard deviation is $\sigma =15$.
The $Z$ score formula is $Z= \dfrac{X-\mu}{\sigma}$.
Weight of a person from first age group is $161$.
His/her z score for weight is
$$
\begin{aligned}
Z&= \frac{161-142}{15}\
&= 1.27.
\end{aligned}
$$
Let $Y$ = Weight of second group. Given that $Y\sim N(138, 13^2)$.
The Z-score formula is $Z= \dfrac{Y-\mu}{\sigma}$.
For second group, the mean weight is $\mu = 138$ and standard deviation is $\sigma =13$.
Weight of a person from second age group is $154$.
His/her z score for weight is
$$ \begin{aligned} Z&= \frac{154-138}{13}\\ &= 1.23 \end{aligned} $$
Because the z-score (1.27) of a person from first age group is larger than the z-score (1.23) of a person from second age group, a person from first age group is more seriously overweight for his or her age group.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
