(D) First,calculate the total number of families:$Total = 150 + 400 + 40 + 250 + 160 = 1000$.The central angle for a category is calculated as:$\text{

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(D) First,calculate the total number of families:$Total = 150 + 400 + 40 + 250 + 160 = 1000$.The central angle for a category is calculated as:$\text{Central Angle} = \left( \frac{\text{Value of category}}{\text{Total value}} \right) \times 360^\circ$.For food and clothing,the value is $400$:$\text{Central Angle} = \left( \frac{400}{1000} \right) \times 360^\circ = 0.4 \times 360^\circ = 144^\circ$.Thus,the correct option is $D$.

Class 11 Mathematics Statistics Word problem -Statistics

Difficult

$A$ pie chart is to be drawn for representing the following data. The value of the central angle for food and clothing would be...$^o$

Items of expenditure Number of families

$Education$ $150$

$Food and clothing$ $400$

$House rent$ $40$

$Electricity$ $250$

$Miscellaneous$ $160$

A
$90$
B
$2.8$
C
$150$
D
$144$

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Word problem -Statistics

In a frequency distribution,if $d_i$ is the deviation of observations from $a$,and the mean is given by $\text{Mean} = a + \frac{\Sigma f_i d_i}{\Sigma f_i}$,then what is $a$?

Medium

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If the mean of the data:
Class $5 - 10$ $10 - 15$ $15 - 20$ $20 - 25$ $25 - 30$ $30 - 35$
Frequency $2$ $k$ $28$ $54$ $k + 1$ $5$

is $21$, then $k$ is one of the roots of the equation:

DifficultJEE MAIN 2026

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$A$ $100$ mark examination was administered to a class of $50$ students. Despite only integer marks being given,the average score of the class was $47.5$. The maximum number of students who could have received marks greater than the class average is:

MediumKVPY 2018

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The mean and the standard deviation of $10$ observations are $20$ and $2$ respectively. Each of these $10$ observations is multiplied by $p$ and then reduced by $q$,where $p \neq 0$ and $q \neq 0$. If the new mean and new standard deviation (s.d.) become half of the original values,then $q$ is equal to

DifficultMHT CET 2024

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The total expenditure incurred by an industry under different heads is best presented as a

Medium

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Items of expenditure	Number of families
$Education$	$150$
$Food and clothing$	$400$
$House rent$	$40$
$Electricity$	$250$
$Miscellaneous$	$160$

Class	$5 - 10$	$10 - 15$	$15 - 20$	$20 - 25$	$25 - 30$	$30 - 35$
Frequency	$2$	$k$	$28$	$54$	$k + 1$	$5$

$A$ pie chart is to be drawn for representing the following data. The value of the central angle for food and clothing would be...$^o$ Items of expenditure Number of families $Education$ $150$ $Food and clothing$ $400$ $House rent$ $40$ $Electricity$ $250$ $Miscellaneous$ $160$

Similar Questions

In a frequency distribution,if $d_i$ is the deviation of observations from $a$,and the mean is given by $\text{Mean} = a + \frac{\Sigma f_i d_i}{\Sigma f_i}$,then what is $a$?

If the mean of the data:Class$5 - 10$$10 - 15$$15 - 20$$20 - 25$$25 - 30$$30 - 35$Frequency$2$$k$$28$$54$$k + 1$$5$is $21$, then $k$ is one of the roots of the equation:

$A$ $100$ mark examination was administered to a class of $50$ students. Despite only integer marks being given,the average score of the class was $47.5$. The maximum number of students who could have received marks greater than the class average is:

The total expenditure incurred by an industry under different heads is best presented as a

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$A$ pie chart is to be drawn for representing the following data. The value of the central angle for food and clothing would be...$^o$

Items of expenditure Number of families

$Education$ $150$

$Food and clothing$ $400$

$House rent$ $40$

$Electricity$ $250$

$Miscellaneous$ $160$

If the mean of the data:
Class $5 - 10$ $10 - 15$ $15 - 20$ $20 - 25$ $25 - 30$ $30 - 35$
Frequency $2$ $k$ $28$ $54$ $k + 1$ $5$

is $21$, then $k$ is one of the roots of the equation: