Question

note: please make sure to properly format your answers. all dollar figures in the answers need to include the dollar sign and any amount over 1.000 should include the comma ($2,354.67). all percentage values in the answers need to include a percentage sign (%). for all items without specific rounding instructions, round your answers to two decimal places, show both decimal places (5.06).
the following list of prices is for a used original radio for a 1955 thunderbird. the prices vary depending on the condition of the radio. $210, $210, $320, $200, $300, $10, $340, $300, $245, $325, $700, $250, $240, $200
a. find the mean of the radio prices. round to the nearest cent.
b. find the median of the radio prices.
c. find the mode of the radio prices. (place in ascending order.)
d. find the four quartiles.
q1
q2
q3
q4
e. find the interquartile range for this data set.
f. find the boundary for lower outliers. what is the lower outlier?
g. find the boundary for upper outliers. what is the upper outlier?

Explanation:

Step1: Sort the data

First, sort the data in ascending order: \$10, \$200, \$200, \$210, \$210, \$240, \$245, \$250, \$300, \$300, \$320, \$325, \$340, \$700.

Step2: Calculate the mean

The mean formula is \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\). Here \(n = 14\) and \(\sum_{i=1}^{14}x_{i}=10 + 200+200+210+210+240+245+250+300+300+320+325+340+700=4050\). So \(\bar{x}=\frac{4050}{14}\approx289.29\).

Step3: Calculate the median

Since \(n = 14\) (even), the median \(Q_{2}=\frac{x_{\frac{n}{2}}+x_{\frac{n}{2}+1}}{2}\). \(\frac{n}{2}=7\) and \(\frac{n}{2}+1 = 8\). \(x_{7}=245\), \(x_{8}=250\). So \(Q_{2}=\frac{245 + 250}{2}=247.5\).

Step4: Calculate the mode

The mode is the most - frequently occurring value. Here \(\$200\) and \(\$210\) and \(\$300\) each occur twice.

Step5: Calculate quartiles

For \(Q_{1}\): \(n = 14\), \(Q_{1}\) is the value at position \(\frac{n + 1}{4}=\frac{14+1}{4}=3.75\). \(x_{3}=200\), \(x_{4}=210\). \(Q_{1}=200+(210 - 200)\times0.75=207.5\).
\(Q_{2}=247.5\) (already found).
For \(Q_{3}\): position \(\frac{3(n + 1)}{4}=\frac{3\times(14 + 1)}{4}=11.25\). \(x_{11}=320\), \(x_{12}=325\). \(Q_{3}=320+(325 - 320)\times0.25=321.25\).
\(Q_{4}\) is the maximum value, \(Q_{4}=700\).

Step6: Calculate the inter - quartile range (IQR)

\(IQR=Q_{3}-Q_{1}\). \(IQR = 321.25-207.5 = 113.75\).

Step7: Calculate lower and upper outlier boundaries

Lower boundary \(=Q_{1}-1.5\times IQR\). \(207.5-1.5\times113.75=207.5 - 170.625 = 36.875\). The lower outlier is \(\$10\) (since \(10<36.875\)).
Upper boundary \(=Q_{3}+1.5\times IQR\). \(321.25+1.5\times113.75=321.25 + 170.625=491.875\). The upper outlier is \(\$700\) (since \(700>491.875\)).

Answer:

a. \(\$289.29\)
b. \(\$247.50\)
c. \(\$200\), \(\$210\), \(\$300\)
d. \(Q_{1}=\$207.50\), \(Q_{2}=\$247.50\), \(Q_{3}=\$321.25\), \(Q_{4}=\$700\)
e. \(\$113.75\)
f. Lower boundary: \(\$36.88\), Lower outlier: \(\$10\)
g. Upper boundary: \(\$491.88\), Upper outlier: \(\$700\)