Question

find the mean median mode range and standard deviation of the data set of Tain after multiplying each value by four

Answer 1

first we need the values multiplying by 4

so we order the values ​​from lowest to highest and multiply

`\begin{gathered} 4\longrightarrow16 \\ 4\longrightarrow16 \\ 6\longrightarrow24 \\ 7\longrightarrow28 \\ 8\longrightarrow32 \\ 9\longrightarrow36 \\ 11\longrightarrow44 \\ 12\longrightarrow48 \\ 13\longrightarrow52 \\ 15\longrightarrow60 \end{gathered}`

the mean is the average, so sum all the number and divide by the quantity

the total sum is 356 and must divide by 10(the number of values)

`\begin{gathered} (356)/(10) \\ =35.6 \end{gathered}`

the mean is 35.6

the median is the average between the 2 centrak numbers on this case 32 and 36

so

`\begin{gathered} (32+36)/(2) \\ =34 \end{gathered}`

the median is 34

the mode is the value that is repeated the most within all

we can notice that 16 is twice and the other values ​​only once

so the mode is 16

The range is the substrac from the larger number and the smaller number

so

`\begin{gathered} 60-16 \\ =44 \end{gathered}`

then the range is 44

The standard deviation must calcuate with this equation

`\sqrt[](\sum ^(\square)_(\square)`

the symbol that looks like a 3 on the contrary means "the sum of", X is a value, u is the mean and N the total of numbers

so, first we need to calculate the sum

`\sum ^(\square)_(\square)|x-u|^2`

replacing x for each number and finding the solution

to 16

`\begin{gathered} |16-35.6|^2 \\ =384.16 \end{gathered}`

to 24

`\begin{gathered} |24-35.6|^2 \\ =134.56 \end{gathered}`

to 28

`\begin{gathered} |28-35.6|^2 \\ =57.56 \end{gathered}`

to 32

`\begin{gathered} |32-35.6|^2 \\ =12.96 \end{gathered}`

to 36

`\begin{gathered} |36-35.6|^2 \\ =0.16 \end{gathered}`

to 44

`\begin{gathered} |44-35.6|^2 \\ =70.56 \end{gathered}`

to 48

`\begin{gathered} |48-35.6|^2 \\ =153.76 \end{gathered}`

to 52

`\begin{gathered} |52-35.6|^2 \\ =268.96 \end{gathered}`

to 60

`\begin{gathered} |60-35.6|^2 \\ =595.36 \end{gathered}`

now, we add everything

and the result is

`\sum ^(\square)_(\square)|x-u|^2=1678.13`

and replace all on the equation

`\begin{gathered} \sqrt[]{(1678.13)/(10)} \\ \\ =12.95 \end{gathered}`

the standard deviation is 12.95