Basic Statistics: Tales of Distributions 10th Edition

Published by Cengage Learning
ISBN 10: 0-49580-891-1
ISBN 13: 978-0-49580-891-6

Chapter 5 - Other Descriptive Statistics - Problems - Page 76: 5.5

Answer

to answer Hamlet: **“Tobe, indeed.”**

Work Step by Step

Let’s calculate the **z-scores** for Tobe’s apple and Zeke’s orange and determine who grew the more **outstanding** fruit (relative to the norms of their own fruit type). --- ## Tobe’s Apple * **Tobe’s apple weight**: $X = 9$ oz * **Mean weight of apples**: $\mu = 5$ oz * **Standard deviation of apples**: $\sigma = 1.0$ oz ### Z-score formula: $$ z = \frac{X - \mu}{\sigma} $$ ### Z-score for Tobe: $$ z_{\text{apple}} = \frac{9 - 5}{1.0} = \frac{4}{1} = \mathbf{4.00} $$ --- ## Zeke’s Orange * **Zeke’s orange weight**: $X = 10$ oz * **Mean weight of oranges**: $\mu = 6$ oz * **Standard deviation of oranges**: $\sigma = 1.2$ oz ### Z-score for Zeke: $$ z_{\text{orange}} = \frac{10 - 6}{1.2} = \frac{4}{1.2} \approx \mathbf{3.33} $$ --- ### Conclusion | Grower | Fruit | Z-Score | Relative Performance | | ------ | ------ | -------- | -------------------- | | Tobe | Apple | **4.00** | Higher | | Zeke | Orange | **3.33** | Lower | **Tobe wins** — his apple was a full **4 standard deviations** above the average apple weight, compared to Zeke’s orange, which was **3.33 standard deviations** above the average orange.
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