Question

a random sample of 20 employees of a grocery delivery service was asked about the total number of items they delivered on their most recent trip. the responses of the employees are shown. 11 20 9 23 16 12 25 36 19 14 41 26 49 18 16 29 12 32 20 27 calculator part a summarize the data set by entering the values in the table minimum lower quartile median upper quartile maximum

Explanation:

Step1: Sort the data

First, sort the data set in ascending order: \(9, 11, 12, 12, 14, 16, 16, 18, 19, 20, 20, 23, 25, 26, 27, 29, 32, 36, 41, 49\)

Step2: Find the minimum and maximum

The minimum value is the first - value in the sorted list. The maximum value is the last - value in the sorted list.
Minimum: \(9\)
Maximum: \(49\)

Step3: Find the median

Since \(n = 20\) (even), the median is the average of the \(\frac{n}{2}\)th and \((\frac{n}{2}+1)\)th values. \(\frac{n}{2}=10\), \(\frac{n}{2}+1 = 11\). The \(10\)th value is \(20\) and the \(11\)th value is \(20\). Median: \(\frac{20 + 20}{2}=20\)

Step4: Find the lower quartile

The lower half of the data is \(9, 11, 12, 12, 14, 16, 16, 18, 19, 20\). Since \(n_1=10\) (even), the lower quartile \(Q_1\) is the average of the \(5\)th and \(6\)th values. The \(5\)th value is \(14\) and the \(6\)th value is \(16\). \(Q_1=\frac{14 + 16}{2}=15\)

Step5: Find the upper quartile

The upper half of the data is \(20, 23, 25, 26, 27, 29, 32, 36, 41, 49\). Since \(n_2 = 10\) (even), the upper quartile \(Q_3\) is the average of the \(5\)th and \(6\)th values. The \(5\)th value is \(27\) and the \(6\)th value is \(29\). \(Q_3=\frac{27+29}{2}=28\)

Answer:

Minimum: \(9\)
Lower Quartile: \(15\)
Median: \(20\)
Upper Quartile: \(28\)
Maximum: \(49\)