Hint: To find the median class of the given data, we will find the cumulative frequency by adding the frequency in the current step with the cumulative frequency found in the previous step. Next find the sum of frequencies denoted by . Next, we have to find and check the cumulative frequency which is nearest to or greater than this value. The corresponding class will be the median class.Complete step-by-step answer:We need to find the median class of the given data. We can do this by finding the cumulative frequency. This is shown below:Number of carsFrequencyCumulative frequency (cf)0-107710-20920-301330-402140-501250-601560-70470-8012We can find cumulative frequency by adding the frequency in the current step with the cumulative frequency found in the previous step.Now, let us find or the sum of frequencies. We can find this by adding the frequencies or this is same as the cumulative frequency in the last column.Hence, \\[\\sum{f}=93\\] .Now, we have to find . That is,Now, we have to check the cumulative frequency which is nearest to or greater than 46.5. From the table. We can see the cumulative frequency=50. Hence, the median class will be 30-40.So, the correct answer is “Option B”.Note: When finding cumulative frequency, you may add the frequency of the previous class and the current class frequency. This will lead to wrong results.We have to add the current class frequency with previous cumulative frequency. When checking the corresponding median class of value, do not check in the frequency column. You have to check in the cumulative frequency column.

Hint: We solve this problem by first recognizing that given data about the number of persons is cumulative frequency of greater than typical. Then we find the frequency of every class by subtracting the cumulative frequency of the next class from the cumulative frequency of the considered class. Then we find the cumulative frequency of less than type from the frequencies of the classes found. Then we consider the formula for the median, $Median=l+\\dfrac{\\dfrac{N}{2}-C}{f}\\times h$, and then find the median class by finding the class which has the cumulative frequency just greater than $\\dfrac{N}{2}$. Then we substitute the values accordingly and find the required value of the median.Complete step-by-step solution:First, let us acknowledge that given data about the number of persons is the cumulative frequency of greater than typical.Cumulative frequency of greater than type is the sum of frequencies of all the classes succeeding the class considered.As we need to find the median, let us find the cumulative frequency of each class.From the table given in the question, we can say that the total number of persons, that is total frequency is 230, as it is a cumulative frequency of greater than typeSo, let us find the frequencies and cumulative frequencies of each class and record them in a table.Class IntervalCumulative Frequency (greater than)FrequencyCumulative Frequency (less than)0-10230230 – 218 =121210-20218218 – 200 = 183020-30200200 – 165 = 356530-40165165 – 123 = 4210740-50123123 – 73 = 5015750-607373 – 28 = 4520260-702828 – 8 = 2022270+88230Now let us consider the formula for the median of the above frequency distribution.$Median=l+\\dfrac{\\dfrac{N}{2}-C}{f}\\times h$where $l=$ Lower limit of the median class$h=$ width of the class interval$f=$ frequency of the median class$N=$ Sum of all frequencies$C=$ Cumulative frequency of the class preceding median classThe median class is the class that has the cumulative frequency just greater than or equal to $\\dfrac{N}{2}$.Here $N=230$Here $\\dfrac{N}{2}=\\dfrac{230}{2}=115$The class having the frequency just greater than $\\dfrac{N}{2}$ is 40-50.So, the median class is 40-50.So, we get the values of $l,h,f,C$ as,$\\begin{align}  & \\Rightarrow l=40 \\\\  & \\Rightarrow h=10 \\\\  & \\Rightarrow f=50 \\\\  & \\Rightarrow C=107 \\\\ \\end{align}$So, substituting these values in the above formula of the median, we get,\\[\\begin{align}  & \\Rightarrow Median=40+\\left( \\dfrac{115-107}{50} \\right)\\times 10 \\\\  & \\Rightarrow Median=40+\\left( \\dfrac{8}{50} \\right)\\times 10 \\\\ \\end{align}\\]Calculating the above value, we get,\\[\\begin{align}  & \\Rightarrow Median=40+\\dfrac{80}{50} \\\\  & \\Rightarrow Median=40+\\dfrac{8}{5} \\\\  & \\Rightarrow Median=40+1.6 \\\\  & \\Rightarrow Median=41.6 \\\\ \\end{align}\\]So, we get the value of median as 41.6. Hence the answer is Option C.Note: The common mistake one makes while solving this question is one might not recognize that the given data is a cumulative frequency of greater than type and just solve it by taking it as normal frequency and applying the formula.