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Solve the Problem =(= ( Lower Limit of Median Class )+() + (

Question 216

Multiple Choice

Solve the problem.
-When data are summarized in a frequency distribution, the median can be found by first identifying the median class (the class that contains the median) . We then assume that the values in that class are evenly
Distributed and we can interpolate. This process can be described by median =(= ( lower limit of median class ) +() + ( class width ) (n+12(m+1)  frequency of median class ) ) \left( \frac { \frac { \mathrm { n } + 1 } { 2 } - ( \mathrm { m } + 1 ) } { \text { frequency of median class } } \right)
where n\mathrm { n } is the sum of all class frequencies and m\mathrm { m } is the sum of the class frequencies that precede the median class. Use this procedure to find the median of the frequency distribution below:
 Score  Frequency 505921606924707922808916909917\begin{array}{c|c}\text { Score } & \text { Frequency } \\\hline 50-59 & 21 \\60-69 & 24 \\70-79 & 22 \\80-89 & 16 \\90-99 & 17\end{array}


A) 72.572.5
B) 72.072.0
C) 71.871.8
D) 74.574.5

Correct Answer:

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