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143SE

Expert-verifiedFound in: Page 142

Book edition
13th

Author(s)
James T. McClave, P. George Benson, Terry Sincich

Pages
888 pages

ISBN
9780134506593

**The amount spent on textbooks for the fall term was recorded for a sample of five university students. The following data were observed: $400, $350, $600, $525, and $450. Calculate each of the following values for this data:**

**a. ****The sample mean**

**b. ****The sample median**

**c. ****The sample range**

**d. ****The standard deviation**

a. Mean = $465

b. Median = $450

c. Range = $250

d. Standard Deviation = $99.37

*x = *400, 350, 600, 525, 450

$\begin{array}{c}\text{Mean =}\frac{\sum \text{\chi}}{\text{n}}\\ =\text{}\frac{400+350+600+525+450}{5}\\ =\text{}\frac{2325}{5}\\ =\text{}\$465\end{array}$

**Therefore, the mean value of books is $465.**

Arranging the data in ascending order,

350, 400, 450, 525, 600

As the sample size is odd, the central value is 450. **Therefore, $450 is the median.**

Maximum value = $600

Minimum value = $350

$\begin{array}{c}\text{Range}=\text{Max}\u2013\text{Min}\\ =\text{}600\text{}\u2013\text{}350\\ =\text{}\$250\end{array}$

**Therefore, the range is $250.**

$\begin{array}{c}\text{Variance =}\frac{\sum {\left(\text{\chi}-\right)}^{2}}{\text{n}-\text{1}}\\ \text{=}\frac{{\left(\text{400}-\text{465}\right)}^{\text{2}}\text{+}{\left(\text{350}-\text{465}\right)}^{\text{2}}\text{+}{\left(\text{600}-\text{465}\right)}^{\text{2}}\text{+}{\left(\text{525}-\text{465}\right)}^{\text{2}}\text{+}{\left(\text{450}-\text{465}\right)}^{\text{2}}}{\text{5}-\text{1}}\\ \text{=}\frac{\text{4225+13225+18225+3600+225}}{\text{4}}\\ \text{=}\frac{\text{39500}}{\text{4}}\end{array}$

$\text{= 9875}$

$\begin{array}{c}\text{Standard Deviation =}\sqrt{\text{Variance}}\\ =\sqrt{9875}\\ =99.37\end{array}$

**Therefore, the standard deviation is $99.37.**

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