Question

Construct a frequency distribution and a relative frequency histogram for the accompanying data set using five classes. Which class has the greatest relative frequency and which has the least relative​ frequency?Complete the table below. Use the minimum data entry as the lower limit of the first class.Class Frequency, f Relative frequencyx-x x xx-x x xx-x x xx-x x xx-x x x sumf= X?​(Type integers or decimals. Round to the nearest thousandth as​ needed.)DATA:Triglyceride levels of 26 patients​ (in milligrams per deciliter of​ blood)138 199 240 143 294 175 240 216 223180 138 266 161 175 402 172 459 147391 152 199 294 188 320 421 161

Answer 1

`\begin{array}{cc}{Class}& {Frequency} & 138 - 202 & 14 & 203 - 267 & 5 & 268 - 332 & 3 & 333 - 397 & 1 & 398 - 462 & 3 \ \end{array}`

The class with the greatest is 138- 202 and the class with the least relative frequency is 333 - 397

Explanation:

Solving (a): The frequency distribution

Given that:

`Lowest = 138` --- i.e. the lowest class value

`Class = 5` --- Number of classes

From the given dataset is:

`Highest = 459`

So, the range is:

`Range = Highest - Lowest`

`Range = 459 - 138`

`Range = 321`

Divide by the number of class (5) to get the class width

`Width = 321 / 5`

`Width = 64.2`

Approximate

`Width = 64`

So, we have a class width of 64 in each class;

The frequency table is as follows:

`\begin{array}{cc}{Class}& {Frequency} & 138 - 202 & 14 & 203 - 267 & 5 & 268 - 332 & 3 & 333 - 397 & 1 & 398 - 462 & 3 \ \end{array}`

Solving (b) The relative frequency histogram

First, we calculate the relative frequency by dividing the frequency of each class by the total frequency

So, we have:

`\begin{array}{ccc}{Class}& {Frequency} & {Relative\ Frequency} & 138 - 202 & 14 & 0.53 & 203 - 267 & 5 & 0.19 & 268 - 332 & 3 & 0.12 & 333 - 397 & 1 & 0.04 & 398 - 462 & 3 & 0.12 \ \end{array}`

See attachment for histogram

The class with the greatest is 138- 202 and the class with the least relative frequency is 333 - 397