Question

Find the mean height of plants from following frequency distribution by short-cut method. \[\]

Height(in cm)	57	69	73	74	77
Number of plants	8	18	41	22	11

Answer 1

Hint: We take the height of the plants as data values ${{x}_{i}}$’s and the number of plants as the frequencies ${{f}_{i}}$’s. We take one of the mean say ${{x}_{3}}=73$ as the assumed mean $A=73$. We find the distances of $A$ from the rest of the data values as ${d_i} = {x_i} - A$. We find the mean in short cut method using the formula for mean as $\overline{x}=A+\dfrac{\sum\limits_{i=1}^{n}{{{f}_{i}}{{d}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}}$. \[\]

Complete step by step answer:
We know that mean or sample mean is the centre or expectation of the data sample set. It is denoted by $\overline{x}$. If there are $n$ number of data values with equal weights in the samples say ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ then the mean is calculated by first finding the sum of data values and then by dividing the sum by $n.$ So the sample mean is given by
\[\overline{x}=\dfrac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n}=\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}}}{n}\]
The number of times any data value occur in the data sample is called frequency and is denoted by $f.$ We denote the corresponding frequencies of data values ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ as ${{f}_{1}},{{f}_{2}},...,{{f}_{n}}.$ We can directly find the mean of frequented data as
\[\overline{x}=\dfrac{\sum\limits_{i=1}^{n}{{{f}_{i}}{{x}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}}\]
The short cut method or otherwise known as assumed mean method is a method to find the mean of large when the data values are large numbers. Let us take one of the data value ${{x}_{i}}$ as the assumed mean $A.$ We calculate the distance ${{d}_{i}}$ of the assumed mean $A$ from all of the ${{x}_{i}}$ where $i=1,2,...,n$. We have
\[{{d}_{i}}={{x}_{i}}-A\]
The mean in the short cut method is given by
\[\overline{x}=A+\dfrac{\sum\limits_{i=1}^{n}{{{f}_{i}}{{d}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}}\]
Let us observe the given table in the question. Here the height of the plants is the data value${{x}_{i}}$and the number of plants that have a particular height is the frequency${{f}_{i}}$. There are a total 5 different heights, so we have $n=5$. Let us assume then mean as $A=73$. Let us calculate the distances${{d}_{i}}$ ,${{f}_{i}}{{d}_{i}}$ , $\sum\limits_{i=1}^{5}{{{f}_{i}}{{d}_{i}}}$, $\sum\limits_{i=1}^{5}{{{f}_{i}}}$ and fill the frequency table.

Height(in cm) $\left( {{x}_{i}} \right)$	57	69	73	74	77
Number of plants $\left( {{f}_{i}} \right)$	8	18	41	22	11	$\sum\limits_{i=1}^{5}{{{f}_{i}}=100}$
${{d}_{i}}={{x}_{i}}-73$	-16	-4	0	1	4
${{f}_{i}}{{d}_{i}}$	-128	-72	0	22	44	$\sum\limits_{i=1}^{5}{{{f}_{i}}{{d}_{i}}=-134}$

So the mean by short cut method is
\[\overline{x}=A+\dfrac{\sum\limits_{i=1}^{5}{{{f}_{i}}{{d}_{i}}}}{\sum\limits_{i=1}^{5}{{{f}_{i}}}}=73+\left( \dfrac{-134}{100} \right)=73-1.34=71.66\]

Note: We note that mean is not unique for a sample having differently frequented data values as we can take different assumed mean. We take the median of data values as the assumed mean for easier calculations of ${{f}_{i}}{{d}_{i}}$. If the data is given classes then we take ${{x}_{i}}$ as midpoints of classes. We can alternatively solve using step deviation method as $\overline{x}=A+h\left( \dfrac{\sum\limits_{i=1}^{n}{{{f}_{i}}{{u}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}} \right)$ where ${{u}_{i}}=\dfrac{{{x}_{i}}-A}{h}$ and $h$ is the class width.