Question

Obtain the original frequency distribution.

Mid values	$25$	$105$	$230$	$400$	$650$	$900$	Total
Frequency	$10$	$30$	$40$	$60$	$80$	$30$	$250$
Class length	$50$	$110$	$140$	$200$	$300$	$200$

Answer 1

Hint: Here we have to find the original class interval in the given data.
Using the first set of mid values and class length, then we can find the first class interval.
Then we can find the second class interval we have to use only the class length data. Similarly we can find all of it.
Finally we get the required frequency distribution.

Complete step-by-step solution:
To obtain an original frequency distribution, we need the class intervals. With the given details,

Mid values	$25$	$105$	$230$	$400$	$650$	$900$	Total
Frequency	$10$	$30$	$40$	$60$	$80$	$30$	$250$
Class length	$50$	$110$	$140$	$200$	$300$	$200$

Now we have to First class interval:
Let us consider the upper limit and lower limit be x and y,
Here we have a data for Mid-value= $25$
So we can write it as $\dfrac{{{\text{x + y}}}}{{\text{2}}}{\text{ = 25}}$
Taking cross multiplication we get
$ \Rightarrow {\text{x + y = 25}} \times {\text{2 }}$
Let us multiply the terms and we get
$ \Rightarrow {\text{x + y = 50 }}...{\text{(1)}}$
Now we use the data Class length = $50$
So we can write it as, ${\text{x - y = 50 }}...{\text{(2)}}$
Now adding ${\text{(1)}}$ and ${\text{(2)}}$ we get,
$ \Rightarrow {\text{x + y + x - y = 50 + 50}}$
Cancel the same term and adding the RHS we get,
$ \Rightarrow 2{\text{x = 100}}$
Let us divide \[2\] on both sides we get
$ \Rightarrow {\text{x = 50}}$
Putting the value in ${\text{(2)}}$ we get
$ \Rightarrow {\text{y = 0}}$
Hence we can conclude the first class interval is $0 - 50$
Second class interval:
Let us consider the upper limit be ${\text{x}}$ and lower limit is ${\text{50}}$ (we found in above case)
Class interval = $110$
So we can write it as,
$ \Rightarrow {\text{x - 50 = 110 }}$
On adding \[50\] on both sides we get
$ \Rightarrow {\text{x = 110 + 50}}$
Let us adding we get,
$ \Rightarrow {\text{x = 160}}$
Hence we conclude the second class interval is ${\text{160}}$.
Third class interval:
Let the upper limit be ${\text{x}}$and lower limit is $160$(we found in above case)
Class length = $140$
So we can write it as,
$ \Rightarrow {\text{x - 160 = 140}}$
 On adding \[160\] on both sides we get
$ \Rightarrow {\text{x = 140 + 160}}$
Let us add the RHS we get
$ \Rightarrow {\text{x = 300}}$
So the third class interval is $300$.
Fourth class interval:
Let the upper limit be ${\text{x}}$ and lower limit is $300$ (we found in above case)
Class length = $200$
$ \Rightarrow {\text{x - 300 = 200}}$
On adding \[300\] on both sides we get
$ \Rightarrow {\text{x = 200 + 300}}$
Let us add the RHS we get
$ \Rightarrow {\text{x = 500}}$
So the fourth class interval is $500$.
Fifth class interval:
Let the upper limit be ${\text{x}}$ and lower limit is $500$ (we found in above case)
Class length = $300$
$ \Rightarrow {\text{x - 500 = 300}}$
On adding \[500\] on both sides we get
$ \Rightarrow {\text{x = 300 + 500}}$
Let us add the RHS we get
$ \Rightarrow {\text{x = 800}}$
So the fifth class interval is $800$.
Sixth class interval:
Let the upper limit be ${\text{x}}$ and lower limit is $800$ (we found in above case)
Class length = $200$
$ \Rightarrow {\text{x - 800 = 200}}$
On adding \[800\] on both sides we get
$ \Rightarrow {\text{x = 200 + 800}}$
Let us add the RHS we get
$ \Rightarrow {\text{x = 1000}}$
So the sixth class interval is $1000$.
Now, the original frequency distribution,

Class interval	$0 - 50$	$50 - 160$	$160 - 300$	$300 - 500$	$500 - 800$	$800 - 1000$	Total
Frequency	$10$	$30$	$40$	$60$	$80$	$30$	$250$

Note: In this question we have an alternative method as follows:
After finding the first class interval as did in the above solution, using mid-value we will find the other class intervals.
Let the upper limit be ${\text{x}}$ and lower limit is ${\text{50}}$ (we already found)
Mid-value= $105$
So we can write it as,
 $ \Rightarrow \dfrac{{{\text{x + 50}}}}{{\text{2}}}{\text{ = 105}}$
Taking cross multiplication we get
$ \Rightarrow {\text{x + 50 = 105}} \times {\text{2 }}$
On multiplying the RHS and subtract \[50\] on both sides we get
$ \Rightarrow {\text{x = 210 - 50}}$
Let us subtract the RHS we get
$ \Rightarrow {\text{x = 160}}$
So the second class interval is ${\text{160}}$.
Similarly we can find the entire class interval