Question

The following data were collected as part of a quality control study for the analysis of sodium in serum; results are concentrations of Na+ in mmol/L.

140, 143, 141, 137, 132, 157, 143, 149, 118, 145

Report the mean, the median, the range, the standard deviation, and the variance for this data.

Answer 1

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Answer 2

Given:

Data of concentrations of Na+ in $mmoll^{-1}$140, 143, 141, 137, 132, 157, 143, 149, 118, 145

To Find:

Mean,  median,  range, standard deviation, and  variance for this data.

Solution:

Mean-  All the data added together and divided by the no. of data.

$Mean=\dfrac{140+143+ 141+137,+132+157+143+149+118+145}{10} = \dfrac{1405}{10}$$= 140.5$ $mmoll^{-1}$

Median- Arranging the 10 data from the smaller value  to the larger value  of data

118, 132, 137, 140, 141, 143, 143, 145, 149, 157  

For a data set of 10 number median is av. of the $5^{th}$ and $6^{th}$ values

$Median = \dfrac{(141 + 143)}{2} = 142 mmoll^{-1}$

Range- Diff. of the larger and smaller value of data. $Range= 157- 118 = 39 mmoll^{-1}$

Standard Deviation- Calculating the diff. of each values and the mean value (140.5), squaring the diff. ,adding all together.  

Differences are    –0.5, 2.5, 0.5, –3.5, –8.5, 16.5, 2.5, 8.5 –22.5 4.5  

Squaring the diff. value are   0.25, 6.25, 0.25, 12.25, 72.25, 272.25, 6.25, 72.25, 506.25, 20.25  

Total sum of squares=  968.50

Standard deviation is dividing this total sum of square by no. of data -1 and then taking square root

$s= \sqrt{\dfrac{968.50}{10-1} }$

$= 10.4$

Variance-

Variance= the square of the s = 108.