Question

cans of soda vary slightly in weight. given below are the measured weights of seven ca considered an outlier? what are the mean and median weights if the outlier is excluded 0.8159 0.8195 0.8164 0.8173 0.7912 0.8143 0.8123 find the mean and median of the weights. the mean is (round to five decimal places as needed.)

Explanation:

Step1: Calculate the sum of all weights

$0.8159 + 0.8195+0.8164 + 0.8173+0.7912+0.8143+0.8123=5.6869$

Step2: Calculate the mean

The mean $\bar{x}=\frac{5.6869}{7}\approx0.81241$

Step3: Arrange the data in ascending - order

$0.7912, 0.8123, 0.8143, 0.8159, 0.8164, 0.8173, 0.8195$
The median is the middle - value. Since $n = 7$ (odd), the median is the 4th value, which is $0.8159$.

To find the outlier, we can use the inter - quartile range method. But a quick look shows that $0.7912$ is far from the other values.

Excluding the outlier $0.7912$:

Step4: Calculate the sum of non - outlier weights

$0.8159 + 0.8195+0.8164 + 0.8173+0.8143+0.8123 = 4.9957$

Step5: Calculate the new mean

The new mean $\bar{x}=\frac{4.9957}{6}\approx0.83262$

Step6: Arrange the non - outlier data in ascending - order

$0.8123, 0.8143, 0.8159, 0.8164, 0.8173, 0.8195$
Since $n = 6$ (even), the median is the average of the 3rd and 4th values. Median$=\frac{0.8159 + 0.8164}{2}=0.81615$

The original mean is approximately $0.81241$ and the original median is $0.8159$. After excluding the outlier ($0.7912$), the new mean is approximately $0.83262$ and the new median is $0.81615$.

Answer:

Original mean: $0.81241$, Original median: $0.8159$, New mean (excluding outlier): $0.83262$, New median (excluding outlier): $0.81615$