Question

e. number of dollars spent on field trip (data from 45 students) 5 - # - summary: min: __ q1: med: q3: max: the mean is which is than the median. the shape of this data set is __ dollars_spent_on_trip_n45

Explanation:

Step1: Find the minimum value

The minimum value (Min) is the smallest data - point. Looking at the dot - plot, Min = 0.

Step2: Calculate the position of Q1

There are \(n = 45\) data points. The position of the first quartile \(Q1\) is at \(\frac{n + 1}{4}=\frac{45+1}{4}=11.5\). The 11th and 12th ordered data points are considered. Counting the data points from the left, \(Q1 = 2\).

Step3: Calculate the position of the median

The position of the median (Med) for \(n = 45\) data points is at \(\frac{n + 1}{2}=\frac{45+1}{2}=23\). Counting the data points, Med = 8.

Step4: Calculate the position of Q3

The position of the third quartile \(Q3\) is at \(\frac{3(n + 1)}{4}=\frac{3\times(45 + 1)}{4}=34.5\). The 34th and 35th ordered data points are considered. Counting the data points, \(Q3 = 10\).

Step5: Find the maximum value

The maximum value (Max) is the largest data - point. Max = 12.

Step6: Calculate the mean

Let \(x_i\) be the value on the x - axis and \(f_i\) be the frequency (number of dots).
\[

$$\begin{align*} \sum_{i}f_ix_i&=0\times3+1\times7+2\times6+3\times3+4\times2+8\times2+9\times3+10\times7+11\times8+12\times4\\ &=0 + 7+12 + 9+8+16+27+70+88+48\\ &=285 \end{align*}$$

\]
The mean \(\bar{x}=\frac{\sum_{i}f_ix_i}{n}=\frac{285}{45}\approx6.33\).
Since \(6.33<8\), the mean is less than the median.

Step7: Determine the shape of the data - set

The left - hand side of the distribution (from 0 to 4) has more data points than the right - hand side (from 10 to 12). So the shape of the data set is left - skewed.

Answer:

Min: 0, Q1: 2, Med: 8, Q3: 10, Max: 12
The mean is 6.33 which is less than the median.
The shape of this data set is left - skewed.