the owner of a busy coffee shop wanted to see if it was worth keeping tea on the menu. she logged the number…

the owner of a busy coffee shop wanted to see if it was worth keeping tea on the menu. she logged the number of cups of tea she sold each day for seven days. 6 12 5 7 7 3 9 calculate the mean, median, range, and midrange of the number of cups of tea sold for the week. mean, 7.5; median, 7; range, 7; midrange, 9 mean, 9; median, 7.5; range, 7; midrange, 7 mean, 7; median, 9; range, 7.5; midrange, 7 mean, 7; median, 7; range, 9; midrange, 7.5

the owner of a busy coffee shop wanted to see if it was worth keeping tea on the menu. she logged the number of cups of tea she sold each day for seven days. 6 12 5 7 7 3 9 calculate the mean, median, range, and midrange of the number of cups of tea sold for the week. mean, 7.5; median, 7; range, 7; midrange, 9 mean, 9; median, 7.5; range, 7; midrange, 7 mean, 7; median, 9; range, 7.5; midrange, 7 mean, 7; median, 7; range, 9; midrange, 7.5

Answer

Explanation:

Step1: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. Here, $n = 7$, and the data - set is $6,12,5,7,7,3,9$. So, $\sum_{i=1}^{7}x_{i}=6 + 12+5 + 7+7+3+9=49$, and $\bar{x}=\frac{49}{7}=7$.

Step2: Calculate the median

First, arrange the data in ascending order: $3,5,6,7,7,9,12$. Since $n = 7$ (odd), the median is the $\left(\frac{n + 1}{2}\right)$-th value. $\frac{7+1}{2}=4$-th value, which is $7$.

Step3: Calculate the range

The range is the difference between the maximum and minimum values. The maximum value is $12$ and the minimum value is $3$. So, the range $=12 - 3=9$.

Step4: Calculate the mid - range

The mid - range is $\frac{\text{Maximum}+\text{Minimum}}{2}$. So, the mid - range $=\frac{12 + 3}{2}=7.5$.

Answer:

Mean, 7; Median, 7; Range, 9; Midrange, 7.5