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Help 18,19

 Apr 25, 2018
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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

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