Answer :
The correct answer is A)For this group of women, weight has greater variation than height
1. Understanding Z-Scores:
- A z-score represents the number of standard deviations a given data point is from the mean.
- Given the height and weight distributions, we can calculate Frieda's z-scores for height and weight:
- Z-Score for Height:
[tex]\( z = \frac{x - \mu}{\sigma} \) \( z = \frac{67 - 65}{2.5} = \frac{2}{2.5} = 0.8 \)[/tex]
- Z-Score for Weight:
[tex]\( z = \frac{135 - 125}{10} = \frac{10}{10} = 1.0 \)[/tex]
2. Analyzing Coefficients of Variation (CV):
- The CV is calculated as the ratio of the standard deviation to the mean, often expressed as a percentage. It indicates relative variation.
- CV for Height:
[tex]\( CV = \frac{2.5}{65} \approx 0.0385 \approx 3.85\% \)[/tex]
- CV for Weight:
[tex]\( CV = \frac{10}{125} = 0.08 = 8\% \)[/tex]
The CV for weight is higher than for height, but both are below 10%.
3. Considering Each Statement:
- A) For this group of women, weight has greater variation than height:
Given the coefficients of variation calculated above, this statement is true. Weight has a higher CV than height, indicating relatively greater variation.
- B) The variation coefficient exceeds 10 percent for both height and weight:
As seen from the coefficients of variation calculated above, neither height nor weight has a CV exceeding 10 percent. This statement is incorrect.
- C) Frieda is taller and thinner than women in her age group:
Frieda's height is above average (z-score of 0.8), but her weight is also above average (z-score of 1.0). Saying she's taller and thinner might not be accurate; in fact, she's heavier than the mean. This statement is incorrect.
- D) Frieda's height is more unusual than her weight:
The z-score for height (0.8) is lower than the z-score for weight (1.0), indicating that Frieda's weight is more unusual relative to the group. This statement is incorrect.
Given these analyses, the correct answer is A.
