The frequency table of the tright of 20 children inaclass
is given below find the mean of their height.
height(in cm). frequency
151. 7
153. 4
155. 6
157. 3
Answers
Answer:
Given data:
The frequency chart of heights of few children in a class is as follows -
\begin{gathered} \: \: \: \: \: \: \: \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}\end{gathered}
height
height
151cm
153cm
155cm
157cm
frequency
frequency
7
4
6
3
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}x
1
,x
2
,...x
n
are the \sf{n}n items given, then
\:\:\sf{A.M. = \dfrac{x_1 + x_2 + ... + x_n}{n} = \dfrac{\sum\limits_{i=1}^{n} x_i}{n}}A.M.=
n
x
1
+x
2
+...+x
n
=
n
i=1
∑
n
x
i
or briefly,
\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ \dfrac{\sum x}{n}}
n
∑x
A.M. is usually denoted by \sf{ \overline{x}}
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}}}}
A.M.=
Totalnumberofobservations
Sumofobservations
A.M.=
Totalnumberofobservations
Sumofobservations
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.
\begin{gathered}\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}}}}}\end{gathered}
⟼A.M.=
7+4+6+3
(151×7)+(153×4)+(155×6)+(157×3)
⟼A.M.=
7+4+6+3
1057+612+930+471
⟼A.M.=
11+9
1669+1401
⟼A.M.=
2
0
307
0
⟼A.M.=
2
307
⟼
A.M.=153.5
A.M.=153.5
\begin{gathered} \\ \therefore \underline{ \sf{ \pmb{The \: mean \: height \: of \: the \: 20 \: children \: is \: \orange{153.5 \: cm} .}}}\end{gathered}
∴
Themeanheightofthe20childrenis153.5cm.
Themeanheightofthe20childrenis153.5cm.
_________________________________________