The above table depicts the age group of people living in an apartment. Since the group width is unequal find the length of rectangle with respect to age group width 5 per interval.

$\begin{matrix} a g e g r o u p & n u m b e r o f p e o p l e 10 - 19 & 12 20 - 24 & 13 25 - 29 & 36 30 - 34 & 15 \end{matrix}$
[3 MARKS]

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Solution

Concept: 1 Mark
Application: 2 Marks

$\begin{matrix} A g e g r o u p & N u m b e r o f p e o p l e & W i d t h & L e n g t h o f r e c t a n g l e 10 - 19 & 12 & 10 & (\frac{12}{10}) \times 5 = 6 20 - 24 & 13 & 5 & (\frac{13}{5}) \times 5 = 13 25 - 29 & 36 & 5 & (\frac{36}{5}) \times 5 = 36 30 - 44 & 15 & 15 & (\frac{15}{15}) \times 5 = 5 \end{matrix}$

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The above table depicts the age group of people living in an apartment. Since the group width is unequal find the length of rectangle with respect to age group width 5 per interval.

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Question 4
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The above table depicts the age group of people living in an apartment. Since the group width is unequal find the length of rectangle with respect to age group width 5 per interval. age groupnumber of people10−191220−241325−293630−3415 [3 MARKS]

Concept: 1 Mark Application: 2 Marks Age groupNumber of peopleWidthLength of rectangle10−191210(1210)×5=620−24135(135)×5=1325−29365(365)×5=3630−441515(1515)×5=5