Question

temperatures in pittsburgh (°f) the temperatures in pittsburgh, pa for the first 12 days of july are shown in the stem - and - leaf plot. 7|2 6 9 9 8|3 5 5 5 6 9|1 2 5 key 7|6 = 76 find the mean, median, mode, range, and interquartile range (iqr) of these temperatures. press each hotspot.

Explanation:

Step1: List out the data

The data set is \(72,76,79,79,83,85,85,85,86,91,92,95\).

Step2: Calculate the mean

The sum of the data is \(72 + 76+79+79+83+85+85+85+86+91+92+95=1018\). There are \(n = 12\) data - points. The mean \(\bar{x}=\frac{1018}{12}\approx84.83\).

Step3: Calculate the median

Since \(n = 12\) (an even - numbered data set), the median is the average of the \(\frac{n}{2}=6\)th and \((\frac{n}{2}+1)=7\)th ordered data values. The 6th value is \(85\) and the 7th value is \(85\), so the median \(M=\frac{85 + 85}{2}=85\).

Step4: Calculate the mode

The mode is the number that appears most frequently. The number \(85\) appears \(3\) times, more frequently than any other number, so the mode is \(85\).

Step5: Calculate the range

The range \(R\) is the difference between the maximum and minimum values. The maximum value is \(95\) and the minimum value is \(72\), so \(R=95 - 72 = 23\).

Step6: Calculate the inter - quartile range (IQR)

First, find the lower half and upper half of the data. The lower half is \(72,76,79,79,83,85\) and the upper half is \(85,85,86,91,92,95\). The median of the lower half (\(Q_1\)) is the average of the 3rd and 4th values. The 3rd value is \(79\) and the 4th value is \(79\), so \(Q_1=\frac{79+79}{2}=79\). The median of the upper half (\(Q_3\)) is the average of the 3rd and 4th values of the upper - half. The 3rd value of the upper - half is \(86\) and the 4th value is \(91\), so \(Q_3=\frac{86 + 91}{2}=88.5\). The inter - quartile range \(IQR=Q_3 - Q_1=88.5 - 79 = 9.5\).

Answer:

Mean: \(84.83\), Median: \(85\), Mode: \(85\), Range: \(23\), IQR: \(9.5\)