The distribution of the heights of 50 children are given. If the mean height for the distribution is 117.8 cm, then complete the following table:
$\begin{matrix} H e i g h t & 110 & 115 & x_{1} & 120 & 121 & 125 N u m b e r o f s t u d e n t s & 6 & 8 & 14 & f_{1} & 4 & 3 \end{matrix}$

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Solution

$\begin{matrix} x_{i} & f_{i} & f_{i} x_{i} 110 & 6 & 660 115 & 8 & 920 x_{1} & 14 & 14 x_{1} 120 & f_{1} & 120 f_{1} 121 & 4 & 484 125 & 3 & 375 T o t a l & \sum f_{i} = 35 + f_{1} & \sum f_{1} x_{1} = 2439 + 14 x_{1} + 120 f_{1} \end{matrix}$

$Given, \sum f_{i} = 50$
$\Rightarrow 35 + f_{1} = 50 \Rightarrow f_{1} = 15$
$We know that, ¯ x = \frac{\sum x_{1} f_{i}}{\sum f_{i}}$
$\Rightarrow 117.8 = \frac{2439 + 14 x_{1} + 120 f_{1}}{50} \Rightarrow 5890 = 2439 + 14 x_{1} + 120 \times 15 [∵ f_{1} = 15] \Rightarrow 5890 - 4239 = 14 x_{1} \Rightarrow 1651 = 14 x_{1} ∴ x_{1} = 117.93 \approx 118$

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Q. The distribution of the heights of 50 children are given. If the mean height for the distribution is 117.8 cm, then find

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The distribution of the heights of 50 children are given. If the mean height for the distribution is 117.8 cm, then complete the following table: Height110115x1120121125Number of students6814f143

xififixi11066601158920x11414x1120f1120f112144841253375Total∑fi=35+f1∑f1x1=2439+14x1+120f1 Given, ∑fi=50 ⇒35+f1=50⇒f1=15 We know that, ¯x=∑x1fi∑fi ⇒117.8=2439+14x1+120f150⇒5890=2439+14x1+120×15[∵f1=15]⇒5890−4239=14x1⇒1651=14x1∴x1=117.93≈118

The distribution of the heights of 50 children are given. If the mean height for the distribution is 117.8 cm, then complete the following table:
$\begin{matrix} H e i g h t & 110 & 115 & x_{1} & 120 & 121 & 125 N u m b e r o f s t u d e n t s & 6 & 8 & 14 & f_{1} & 4 & 3 \end{matrix}$