Question

the frequency table of the heights of 20 children in a class is given below. find the mean of their height
height (1)151 (2) 153 (3)155 (4)157
frequency (1)7 (2)4 (3)6 (4)3
​

Answer 1

Given:

• the frequency table of the heights of 20 children in a class is given

To Find:

• their mean hieght

Frequency Table :

$\begin{gathered}\begin{gathered} \large\boxed{\begin{array}{ c |c} \rm{ \pmb{hieght}}& \rm{ \pmb{frequency}}\\ \\ \dfrac{\qquad\qquad}{ \sf 151cm}&\dfrac{\qquad\qquad}{ \sf 7 }& \\ \\ \dfrac{\qquad\qquad}{ \sf 153cm}& \dfrac{\qquad\qquad}{ \sf 4}& \\ \\\dfrac{\qquad\qquad}{ \sf 155cm}& \dfrac{\qquad\qquad}{ \sf 6}& \\ \\ \dfrac{\qquad\qquad}{ \sf 157cm}& \dfrac{\qquad\qquad}{ \sf 3}& \\ \\ \end{array}}\end{gathered}\end{gathered}$

Solution:

Here, we have given the hieghts of the students and the frequency of them, and said to find the mean of their hieghts

Let's use the below formula to find the mean of these observations:

$\longrightarrow \tt \: mean = \dfrac{sum \: of \: observations}{no.of \: observations}$

Now, let's substitute the given values and then find their mean

$\longrightarrow \sf \: mean = \frac{(151 \times 7) + (153 \times4) + (155 \times 6) + (157 \times 3) }{20} \\ \\ \\ \longrightarrow \sf \: mean = \frac{1057 + 612 + 930 + 471}{20} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \sf \: mean = \cancel\frac{3070}{20} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \: { \orange{ \underline{ \boxed{ \frak{mean = 153.5cm}}}} \bigstar} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

${\blue{\underline {\frak{Hence \: the\:Mean\:Of\:the\:observations = \pink{\pmb{153.5cm}}}}}}$

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Answer 2

ANALYSIS:-

Given data:

The frequency chart of heights of few children in a class is as follows -

$\: \: \: \: \: \: \: \begin{gathered} \footnotesize\boxed{\begin{array}{ c |c} \frak \red{ \pmb{height}}& \frak \red{ \pmb{frequency}}\\ \dfrac{\qquad\qquad}{ \sf{151 \: cm}}& \dfrac{\qquad\qquad}{ \sf{7}}& \\ \dfrac{\qquad\qquad}{ \sf{153 \: cm}}&\dfrac{\qquad\qquad}{ \sf{4}}& \\ \dfrac{\qquad\qquad}{ \sf{155 \: cm}}& \dfrac{\qquad\qquad}{ \sf{6}}& \\ \dfrac{\qquad\qquad}{ \sf{157 \: cm}}&\dfrac{\qquad\qquad}{ \sf{ 3}}& \\ \end{array}}\end{gathered}$

To find:

• The mean of the children's heights.

CONCEPT:-

‎ ‎ ‎ ‎ ‎ ‎Here, the concept used is one of the three most commonly used measure of central tendency, i.e, Arithmetic Mean or Mean (A.M.). The arithmetic mean of a statistical data is defined as the quotient obtained when the sum of all the observations is divided by the total number of items.

If $\sf{x_1, \: x_2, \: ... \: x_n}$ are the $\sf{n}$ items given, then

$\:\:\sf{A.M. = \dfrac{x_1 + x_2 + ... + x_n}{n} = \dfrac{\sum\limits_{i=1}^{n} x_i}{n}}$

or briefly,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ \dfrac{\sum x}{n}}$

A.M. is usually denoted by $\sf{ \overline{x}}$

SOLUTION:-

As we're asked to find the mean height of the children, we can find it by substituting the values in the formula:

$\small\boxed{ \sf{ \pmb{ A.M. = \dfrac{Sum \: of \: observations}{Total \: number \: of \: observations}}}}$

Here, the we've to sum up the measures of heights of each and every individual and then divide them with the total number of students, i.e, the sum of values given as the frequencies in the frequency chart.

$\longmapsto{ \sf{A.M. = \dfrac{(151 \times 7) + (153 \times 4) + (155 \times 6) +( 157 \times 3)}{7 + 4 + 6 + 3} }} \\ \\ \longmapsto{ \sf{ A.M. = \dfrac{1057 + 612 + 930 + 471}{7 + 4 + 6 + 3} }} \\ \\ \longmapsto{ \sf{ A.M. = \dfrac{1669 + 1401}{11 + 9} }} \\ \\ \longmapsto{ \sf{A.M. = \dfrac{307 \cancel{0}}{2 \cancel{0}}}} \\ \\ \longmapsto{ \sf{A.M. = \dfrac{ \cancel{307}}{ \cancel{2}} }}\\ \\ \longmapsto \underline{\boxed{ \tt{ \red{ \pmb{A.M. = 153.5}}}}}$

$\\ \therefore \underline{ \sf{ \pmb{The \: mean \: height \: of \: the \: 20 \: children \: is \: \orange{153.5 \: cm} .}}}$

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