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use the display of data to find the mean, median, mode, and midrange. t…

Question

use the display of data to find the mean, median, mode, and midrange.
the mean of the data is □. (round to the nearest tenth as needed.)
the median of the data is □. (round to the nearest tenth as needed.)
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
oa. the mode of the data is □.
(use a comma to separate answers as needed.)
ob. there is no mode for the given data.
the midrange of the data is □. (round to the nearest tenth as needed.)

Explanation:

Step1: List out the data set

From the graph, we have the following data points (score - frequency):

  • Score \(9\): frequency \(3\)
  • Score \(10\): frequency \(3\)
  • Score \(11\): frequency \(4\)
  • Score \(12\): frequency \(5\)
  • Score \(13\): frequency \(6\)
  • Score \(14\): frequency \(3\)
  • Score \(15\): frequency \(2\)

The data set is \(9,9,9,10,10,10,11,11,11,11,12,12,12,12,12,13,13,13,13,13,13,14,14,14,15,15\)

Step2: Calculate the mean

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\)
\(\sum_{i=1}^{n}x_{i}=9\times3 + 10\times3+11\times4+12\times5+13\times6+14\times3+15\times2\)
\(=27+30 + 44+60+78+42+30\)
\(=311\)
\(n=3 + 3+4+5+6+3+2=26\)
\(\bar{x}=\frac{311}{26}\approx12.0\)

Step3: Calculate the median

Since \(n = 26\) (even), the median is the average of the \(\frac{n}{2}\)th and \((\frac{n}{2}+1)\)th values.
\(\frac{n}{2}=13\), \(\frac{n}{2}+1 = 14\)
The \(13\)th value is \(12\) and the \(14\)th value is \(12\)
Median \(=\frac{12 + 12}{2}=12.0\)

Step4: Calculate the mode

The mode is the value that appears most frequently. The score \(13\) has the highest frequency (\(6\))
Mode \(=13\)

Step5: Calculate the mid - range

The formula for the mid - range is \(\text{Mid - range}=\frac{\text{Minimum}+\text{Maximum}}{2}\)
Minimum \(=9\), Maximum \(=15\)
\(\text{Mid - range}=\frac{9 + 15}{2}=12.0\)

Answer:

The mean of the data is \(12.0\).
The median of the data is \(12.0\).
A. The mode of the data is \(13\).
The midrange of the data is \(12.0\).