Q) A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mean and median of the following data:

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(i) Let’s re-arrange the data with midpoint of each class, frequency, and multiply midpoint with frequency:

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We know that, mean of grouped data is given by:

Mean of grouped data = \frac{\sum fx}{\sum f} =  \frac{4070}{100} = 407

Therefore, Mean value of the given data is 407 

(ii) To find the median, we need to identify middle value of the data. Let’s rearrange the data.

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Step 1: we need to find the cumulative frequency in the frequency table to find the median. Its shown in last column.

Step 2: Next, total number of sub-divisions or Sum of the frequencies = 100. It shown at the bottom of middle column.

Step 3: Next, we need to identify Median Class.

Since the Median class is the class where the cumulative frequency crosses 50% of the half the total number of sub-divisions. Here, 50% of total frequency is 50 and cumulative frequency is crossing is crossing at 66 i.e. at class “40-50”. Hence, our Median class = 40-50

Next, To find the median, we use the formula:

Median = L + [(\frac{n}{2} - F) x \frac{h}{f} ]


L = Lower boundary of the median class = 40

n = Total number of sub-divisions = 100

F = Cumulative frequency of the class before the median class = 46

h = Class width = 50 – 40 = 10

f = Frequency of the median class = 20

hence, the Median = 40 + [(\frac{100}{2} - 46) x \frac{10}{20} ]

= 40 + [(50 – 46) x \frac{1}{2} ]

= 40 + 2

= 42

Therefore, Median of the given data is 42

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