Question

consider the following sample data, which represent weights walnuts in grams: { 13.2, 13.2, 14.1, 14.1, 14.3, 14.4, 14.4, 14.4, 14.9, 15.8, 16, 16, 16.1, 16.2, 17.4, 17.5, 18.4, 18.6, 19.2 , 20.9 }. first, give the mean of the data set. 15.955 part 2 of 6 next, give the median of the data set. 15.9 part 3 of 6 now give the mode of the data set. if there is more than one, write them in order, separated by commas. 14.4 part 4 of 6 finally, give the midrange of the data set. 17.05 part 5 of 6 given the relationship between the mean and median above, what shape is the distribution likely to be? the distribution will probably be skewed to the left. the distribution will be roughly symmetric. the distribution will probably be skewed to the right. part 6 of 6 suppose the last value in the data set is mistakenly recorded as 209. how would this affect the mean? the mean would get smaller. the mean would get larger. the mean would not change. how would this affect the median? the median would get smaller. the median would not change. the median would get larger.

Explanation:

Step1: Recall mean formula

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $x_{i}$ are data - points and $n$ is the number of data - points. Here $n = 20$, and $\sum_{i=1}^{20}x_{i}=13.2 + 13.2+14.1+14.1+14.3+14.4+14.4+14.4+14.9+15.8+16+16+16.1+16.2+17.4+17.5+18.4+18.6+19.2+20.9 = 319.1$, so $\bar{x}=\frac{319.1}{20}=15.955$.

Step2: Recall median formula

For $n = 20$ (an even - numbered data set), the median is the average of the $\frac{n}{2}$th and $(\frac{n}{2}+1)$th ordered data - points. First, order the data. The 10th and 11th ordered data - points are 15.8 and 16, so the median $M=\frac{15.8 + 16}{2}=15.9$.

Step3: Recall mode formula

The mode is the data - point that appears most frequently. In the data set, 14.4 appears 3 times, more frequently than any other value, so the mode is 14.4.

Step4: Recall mid - range formula

The mid - range is $\frac{\text{min}+\text{max}}{2}$. Here, $\text{min}=13.2$ and $\text{max}=20.9$, so the mid - range $=\frac{13.2 + 20.9}{2}=17.05$.

Step5: Analyze distribution shape

When the mean ($15.955$) is greater than the median ($15.9$), the distribution is probably skewed to the right because the mean is pulled in the direction of the tail.

Step6: Analyze effect on mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. If the last value changes from 20.9 to 209, the sum $\sum_{i = 1}^{n}x_{i}$ increases. Since $n$ is constant, the mean will get larger.

Step7: Analyze effect on median

For $n = 20$ (an even - numbered data set), the median is the average of the 10th and 11th ordered data - points. Changing the largest value (20.9 to 209) does not change the 10th and 11th ordered data - points in the ordered data set. So the median will not change.

Answer:

Part 5 of 6: The distribution will probably be skewed to the right.
Part 6 of 6:
The mean would get larger.
The median would not change.