Question

After an article appeared online criticizing Starbucks for the differences in pay for their employees, you remembered that you worked with similar data on #4 of this quest. Below is the sample shown previously: Barista: $12.50, Lead Barista: $13.75, Trainee: $11.25, Shift supervisor: $15, Barista: $12.50, Store Manager: $33 a) What is the standard deviation of this data? What is the range of this data? Now calculate the standard deviation and the range of the above data WITHOUT using the store manager’s wage rate. How different is it from your calculations for a and b? Does this support the argument that the manager wages are significantly higher than others’ wages? Be sure to reference your calculations in your explanation.

Answer 1

Given:

Barista: $12.50,

Lead Barista: $13.75,

Trainee: $11.25,

Shift supervisor: $15,

Barista: $12.50,

Store Manager: $33

Required:

To find the standard deviation and range of the given data.

Step-by-step explanation:

Standard deviation :

`SD=\sqrt{\sum_^((x-x1)^2)/(n-1)}`

Here x1 is the mean.

Now,

`\begin{gathered} x1=(12.50+13.75+11.25+12.50+33)/(4) \\ \\ =(83)/(5) \\ \\ =16.6 \end{gathered}`

So,

`\begin{gathered} =(12.5-16.6)^2+(13.75-16.6)^2+(11.25-16.6)^2+(12.5-16.6)^2+(33-16.6)^2 \\ \\ =8.1225+28.6225+268.96 \\ \\ =305.705 \end{gathered}`

Now

`\begin{gathered} SD=(305.705)/(5-1) \\ \\ =(305.705)/(4) \\ \\ =76.426 \end{gathered}`

Range:

To find the range, order all values in your data set from low to high. Subtract the lowest value from the highest value.

Therefore,

`11.25,12.50,12.50,13.75,33`

The range is

`\begin{gathered} R=33-11.25 \\ \\ =21.75 \end{gathered}`

Final Answer:

Standard deviation = 76.426

Range = 21.75