Question

The class intervals in a frequency distribution table is 44 - 48, 49 - 53, 54 - 58 with corresponding frequencies as 4, 5 and 7 respectively. The width of the second class interval is ______.​

Answer 1

The width of the second class interval is 5

Given :

The class intervals in a frequency distribution table is 44 - 48, 49 - 53, 54 - 58 with corresponding frequencies as 4, 5 and 7 respectively

To find :

The width of the second class interval

Solution :

Step 1 of 2 :

Form the frequency table

Here it is given that the class intervals in a frequency distribution table is 44 - 48, 49 - 53, 54 - 58 with corresponding frequencies as 4, 5 and 7 respectively

The given frequency table is in inclusive form.

Now we convert it into exclusive form

$\begin{gathered} \begin{array}{|c|c| c| } \bf{Class \: Interval} & \bf{Class \: Boundary }& \bf{Frequency } \\ \\ 44 - 48 & 43.5 - 48.5 & 4 \\ \\49 - 53 & 48.5-53.5 & 5 \\ \\54 - 58 & 53.5 - 58.5& 7 \\ \end{array}\end{gathered}$

Step 2 of 2 :

Find the width of the second class interval

The second class interval (Class boundary) is 48.5 - 53.5

For the class 48.5 - 53.5

Upper Class boundary = 53.5

Lower class boundary = 48.5

We know that Class Length or Class Size or Class Width of the class interval of any class is defined to be the difference between the lower and upper class boundaries ( not class limits) of that particular class interval.

∴ Class width of the second class interval

= Upper Class boundary - Lower class boundary

= 53.5 - 48.5

= 5

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Answer 2

The required class width is 5

Given

• class intervals
• 44 - 48, 49 - 53, 54 - 58

To find

• width of the second class interval

Solution

we are provided with class intervals which are not continuous and asked to determine the class width of the second class.

First let us make the class endeavours continuous.

This could be done by adding 0.5 to the upper limit and reducing 0.5 from the lower limit of each class.

first class : 44 - 48

or, 44- 0.5 - 48 + 0.5

or, 43.5 - 48.5

After employing the procedure we get the class interval as

43.5 - 48.5

48.5 - 53.5

53.5 - 58.5

Now, the class width is defined as the difference between the upper limit and the lower limit.

therefore the class width of the second class would be,

53.5 - 48.5

or, 5

Therefore, the required class width is 5

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