Question

Lucas recorded his lunch expenditure each day for one week in the table below.SEE IMAGEFind the mean, standard deviation, and variance of Lucas’s lunch expenditures. Round to the nearest thousandth.

Answer 1

Mean)

In general, the formula is

$\operatorname{mean}=\frac{\text{ sum of elements in the dataset}}{\text{ number of elements in the dataset}}$

Thus, in our case,

$\begin{gathered} \Rightarrow X=\operatorname{mean}=\frac{4.85+5.10+5.50+4.75+4.5+5+6}{7} \\ \Rightarrow X=\operatorname{mean}=5.1 \end{gathered}$

Thus, the mean is 5.1

Variance)

In general, the formula is

$\text{Variance}=S^2=\sum ^{}_{}\frac{(x_i-X)^2}{N-1}$

Where x_i is each of the elements in the data set, X is the mean, and N is the number of elements in the data set.

Therefore, in our case,

$\begin{gathered} N=7,X=5.1 \\ \Rightarrow S^2=\frac{1}{7-1}((4.85-5.1)^2+(5.1-5.1)^2+(5.5-5.1)^2+(4.75-5.1)^2+(4.5-5.1)^2+(5-5.1)^2+(6-5.1)^2) \\ \Rightarrow S^2=0.2541666\ldots \\ \Rightarrow S^2\approx0.254 \end{gathered}$

The variance is 0.254

Standard deviation)

If S^2 is the variance; then,

$\text{ standard deviation}=\sigma=\sqrt[]{S^2}$

Hence, in our case,

$\begin{gathered} \Rightarrow\sigma=\sqrt[]{0.2541666\ldots}\approx0.504149\ldots \\ \Rightarrow\sigma\approx0.504 \end{gathered}$

The standard deviation is 0.504