$A$ total of $25$ patients admitted to a hospital are tested for blood sugar levels $(mg/dl)$ and the results obtained are as follows:
$\begin{array}{lllll} 87 & 71 & 83 & 67 & 85 \\ 77 & 69 & 76 & 65 & 85 \\ 85 & 54 & 70 & 68 & 80 \\ 73 & 78 & 68 & 85 & 73 \\ 81 & 78 & 81 & 77 & 75 \end{array}$
Find the mean,median,and mode $(mg/dl)$ of the above data.
$\begin{array}{lllll} 87 & 71 & 83 & 67 & 85 \\ 77 & 69 & 76 & 65 & 85 \\ 85 & 54 & 70 & 68 & 80 \\ 73 & 78 & 68 & 85 & 73 \\ 81 & 78 & 81 & 77 & 75 \end{array}$
Find the mean,median,and mode $(mg/dl)$ of the above data.
(N/A) Mean: Sum of all observations $= 87+71+83+67+85+77+69+76+65+85+85+54+70+68+80+73+78+68+85+73+81+78+81+77+75 = 1891$
Number of observations,$n = 25$
$\text{Mean } (\bar{x}) = \frac{\sum x_i}{n} = \frac{1891}{25} = 75.64$
Median: Arranging the observations in ascending order:
$54, 65, 67, 68, 68, 69, 70, 71, 73, 73, 75, 76, 77, 77, 78, 78, 80, 81, 81, 83, 85, 85, 85, 85, 87$
Since $n = 25$ (odd),the median is the $\left(\frac{n+1}{2}\right)^{th}$ term.
$\text{Median} = \left(\frac{25+1}{2}\right)^{th} = 13^{th}$ term.
The $13^{th}$ term in the ordered sequence is $77$.
Mode: The observation that occurs most frequently is the mode.
$85$ appears $4$ times,which is the highest frequency.
$\text{Mode} = 85$.
Final Answer: $\text{Mean} = 75.64, \text{Median} = 77, \text{Mode} = 85$.
Number of observations,$n = 25$
$\text{Mean } (\bar{x}) = \frac{\sum x_i}{n} = \frac{1891}{25} = 75.64$
Median: Arranging the observations in ascending order:
$54, 65, 67, 68, 68, 69, 70, 71, 73, 73, 75, 76, 77, 77, 78, 78, 80, 81, 81, 83, 85, 85, 85, 85, 87$
Since $n = 25$ (odd),the median is the $\left(\frac{n+1}{2}\right)^{th}$ term.
$\text{Median} = \left(\frac{25+1}{2}\right)^{th} = 13^{th}$ term.
The $13^{th}$ term in the ordered sequence is $77$.
Mode: The observation that occurs most frequently is the mode.
$85$ appears $4$ times,which is the highest frequency.
$\text{Mode} = 85$.
Final Answer: $\text{Mean} = 75.64, \text{Median} = 77, \text{Mode} = 85$.
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