Question

c. number of grocery store visits per month (data from 26 shoppers)
5 - # - summary:
min: __ q1: med: q3: max: __
the mean is __ which is __ than the median.
the shape of this data set is ____
grocery_store_visits_per_month_n_26

Explanation:

Step1: Find the minimum value

By looking at the dot - plot, the smallest value is 2.

Step2: Find the first quartile (Q1)

Since \(n = 26\), the position of Q1 is \(\frac{n + 1}{4}=\frac{26+1}{4}=6.75\). The 6th and 7th ordered data values are considered. Counting the dots from the left, Q1 = 6.

Step3: Find the median (Med)

The position of the median for \(n = 26\) is \(\frac{n+1}{2}=\frac{26 + 1}{2}=13.5\). The 13th and 14th ordered data values are considered. Counting the dots, the median is 8.

Step4: Find the third quartile (Q3)

The position of Q3 is \(3\times\frac{n + 1}{4}=3\times\frac{26+1}{4}=20.25\). The 20th and 21st ordered data values are considered. Counting the dots, Q3 = 9.

Step5: Find the maximum value

By looking at the dot - plot, the largest value is 10.

Step6: Calculate the mean

Let \(x_i\) be the number of visits and \(f_i\) be the frequency. We have:

\(x_i\)	\(f_i\)	\(x_if_i\)
3	1	3
4	1	4
5	1	5
6	2	12
7	2	14
8	4	32
9	6	54
10	5	50
11	3	33

The sum of \(f_i=26\) and the sum of \(x_if_i = 2 + 3+4 + 5+12+14+32+54+50+33=209\). The mean \(\bar{x}=\frac{\sum_{i = 1}^{k}x_if_i}{\sum_{i=1}^{k}f_i}=\frac{209}{26}\approx8.04\).
Since the mean (\(8.04\)) is greater than the median (8), and the tail of the distribution extends to the right (higher values), the shape of the data - set is right - skewed.

Answer:

Min: 2
Q1: 6
Med: 8
Q3: 9
Max: 10
The mean is 8.04 which is greater than the median.
The shape of this data set is right - skewed.