+2440 In order to find the mean, find the sum of all the data points and divide by the number of data points. Since this is a "lunch expenditure for a week," I would assume that this is only a sample, so I will x bar as the mean.
\(\overline{x}=\frac{4.85+5.10+5.50+4.75+4.50+5.00+6.00}{7}=$5.10\)
Yet again, I am making the assumption that this data is a sample because that seems like a reasonable assumption. In order to find the standard deviation, this formula will provide that answer:
\( \text{SD}_{\text{sample}}=\sqrt{\frac{\Sigma_{i=1}^{n} (x_{i}-\overline{x})^2}{n-1}}\)
Now, this is a loaded formula! Let's first define a few variables:
\(x_i\) is the individual data point
\(\overline{x}\) is the average of the data set
\(n\) is the number of data points
Let's break this formula down. There are 5 steps, one of which we have already done.
I will show the work with a table. This takes care of steps 2 and 3 simultaneously.
| Day | \(x_i\) | \(x_i-\overline{x}\) | \((x_i-\overline{x})^2\) | ||
| Monday | 4.85 | -0.25 | 0.0625 | ||
| Tuesday | 5.10 | 0.00 | 0.0000 | ||
| Wednesday | 5.50 | 0.40 | 0.1600 | ||
| Thursday | 4.75 | -0.35 | 0.1225 | ||
| Friday | 4.50 | -0.60 | 0.3600 | ||
| Saturday | 5.00 | -0.10 | 0.0100 | ||
| Sunday | 6.00 | 0.90 | 0.8100 | ||
| Total | 1.5250 | ||||
Now, I will do the remaining steps.
\(\text{SD}_{\text{sample}}\sqrt{\frac{1.525}{7-1}}\approx0.504\)
The variance is just the square of the standard deviation.
\(\text{S}^2=\left(\sqrt{\frac{1.525}{6}}\right)^2=\frac{1.525}{6}\approx0.2542\)
19) Picking a random number, say 2932, is independent to a coin flip; one has no effect on the other.
