Question

in the table below, heart beats of 30 women are recorded. if mean of the data is 76, find the missing frequency x and y
No. of heart No. of women
65-68 x
68-71 4
71-74 3
74-77 7
77-80 8
80-83 4
83-86 y ​

Answer 1

Answer:

x = 2,y=2

Step-by-step explanation:

Answer 2

Concept Introduction:-

In a collection of numeral, the mean are the average or most common value.

Given Information:-

We have been given that a table.

To Find:-

We have to find that the missing frequency.

Solution:-

According to the problem

$\sum \mathrm{f}_{\mathrm{i}}=30 \quad$ [Given ]

$\sum \mathrm{f}_{\mathrm{i}}=\mathrm{x}+\mathrm{y}+26$

$30=\mathrm{x}+\mathrm{y}+26\\30-26=\mathrm{x}+\mathrm{y}\\\mathrm{x}+\mathrm{y}=4$

$\therefore \mathrm{x}+\mathrm{y}=4$

$\Rightarrow \mathrm{x}=4-\mathrm{y}.....(i)$

$\Rightarrow \mathrm{Mean}(\overline{\mathrm{x}})=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}$

$\Rightarrow 76=\frac{1978+66.5 \mathrm{x}+84.5 \mathrm{y}}{30}\\\Rightarrow 2280=1978+66.5 \mathrm{x}+84.5 \mathrm{y}\\\Rightarrow 302=66.5 \mathrm{x}+84.5 \mathrm{y}\\\Rightarrow 3020=665 \mathrm{x}+845\mathrm{y}\\\Rightarrow 604=133(\mathrm{4-y})+169\mathrm{y}\\\Rightarrow 604=532-133\mathrm{y}+169\mathrm{y}\\\Rightarrow 604=532+36\mathrm{y}\\\Rightarrow 604-532=36\mathrm{y}\\\Rightarrow 72=36\mathrm{y}\\\Rightarrow \mathrm{y}=\frac{72}{36}\\\Rightarrow \mathrm{y}=2$  

Substituting value of $y$ in $(i)$

$x=4-y\\x=4-2\\x=2$

Final Answer:-

The value of $x$ and $y$ are $2$ and $2$ respectively.

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