jamie recorded the weights of 5 watermelons.\n- the heaviest watermelon weighed 24 pounds.\n- the mean of…
jamie recorded the weights of 5 watermelons.\n- the heaviest watermelon weighed 24 pounds.\n- the mean of the weights was 21 pounds.\n- the median of the weights was 20 pounds.\nshe realized she recorded the wrong weight for the heaviest watermelon, which actually weighed 28 pounds.\nhow did correcting the weight of the largest watermelon affect the mean and median?\n(a) the mean and the median increased.\n(b) the mean increased and the median decreased.\n(c) the median increased and the mean remained the same.\n(d) the mean increased and the median remained the same.
Answer
Explanation:
Step1: Recall the concept of mean
The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. Here $n = 5$. Initially, $\sum_{i=1}^{5}x_{i}=n\times\bar{x}=5\times21 = 105$ (using the wrong - recorded data). After correcting the weight of the heaviest watermelon (from 24 to 28 pounds), the new sum $\sum_{i = 1}^{5}x_{i}=105-24 + 28=109$. The new mean $\bar{x}_{new}=\frac{109}{5}=21.8$. Since the new sum is larger and $n$ is the same, the mean increases.
Step2: Recall the concept of median
The median is the middle - value when the data is arranged in ascending or descending order. Since we have 5 values (an odd number of values), the median is the 3rd - ranked value when the data is ordered. Changing only the largest value (the 5th - ranked value when ordered) does not affect the 3rd - ranked value. So the median remains the same.
Answer:
D. The mean increased and the median remained the same.