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Help 18,19 ​

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Help 18,19

Apr 25, 2018

#1
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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.

• Find the mean
• For every data point, subtract the mean from it. Then, square it.
• Sum the values from step 2
• Divide by the number of data points minus 1
• Take the square root of the result

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.

Apr 25, 2018