QUESTION IMAGE
Question
the graph below represents results of a survey in which students stated the number of minutes theyd spent watching tv the previous day. tv watch time which interval does the median number of minutes watched fall within?
Step1: Find the total number of students
Let the frequencies of the intervals \(0 - 29\), \(30 - 59\), \(60 - 89\), \(90 - 119\), \(120 - 149\), \(150 - 179\) be \(f_1\), \(f_2\), \(f_3\), \(f_4\), \(f_5\), \(f_6\) respectively. Assume \(f_1 = 2\), \(f_2 = 3\), \(f_3 = 1\), \(f_4 = 3\), \(f_5 = 7\), \(f_6 = 4\). The total number of students \(n=f_1 + f_2+f_3+f_4+f_5+f_6=2 + 3+1+3+7+4=20\) (Since \(n = 20\) (even), the median is the average of the \(\frac{n}{2}=10^{th}\) and \((\frac{n}{2}+ 1)=11^{th}\) ordered - data values.
Step2: Calculate the cumulative frequencies
The cumulative frequency of \(0 - 29\) is \(CF_1=f_1 = 2\). The cumulative frequency of \(30 - 59\) is \(CF_2=f_1 + f_2=2 + 3=5\). The cumulative frequency of \(60 - 89\) is \(CF_3=f_1 + f_2+f_3=2+3 + 1=6\). The cumulative frequency of \(90 - 119\) is \(CF_4=f_1 + f_2+f_3+f_4=2+3 + 1+3=9\). The cumulative frequency of \(120 - 149\) is \(CF_5=f_1 + f_2+f_3+f_4+f_5=2+3 + 1+3+7=16\).
Step3: Locate the median interval
Since \(CF_4=9\lt10\) and \(CF_5 = 16\geq10\) (also \(CF_5\geq11\)), the \(10^{th}\) and \(11^{th}\) values fall within the \(120 - 149\) interval.
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\(120 - 149\)