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(D) We know that the mid-point $(m)$ of a class interval is calculated as:$m = \frac{	ext{Lower limit} + 	ext{Upper limit}}{2}$Given that the upper

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$2 m - l$

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(D) We know that the mid-point $(m)$ of a class interval is calculated as:$m = \frac{	ext{Lower limit} + 	ext{Upper limit}}{2}$Given that the upper class limit is $l$,we substitute it into the formula:$m = \frac{	ext{Lower limit} + l}{2}$Multiplying both sides by $2$,we get:$2m = 	ext{Lower limit} + l$Rearranging the equation to solve for the lower class limit:$	ext{Lower limit} = 2m - l$Therefore,the lower class limit of the class is $2m - l$.

Ages (in years)	Number of teachers
$20-24$	$10$
$25-29$	$28$
$30-34$	$32$
$35-39$	$48$
$40-44$	$50$
$45-49$	$35$
$50-54$	$12$

Let $m$ be the mid-point and $l$ be the upper class limit of a class in a continuous frequency distribution. The lower class limit of the class is:

Similar Questions

If the sum of $25$ observations is $900$, find the mean of those observations.

Draw a histogram to represent the following grouped frequency distribution:

Ages (in years) Number of teachers

$20-24$ $10$

$25-29$ $28$

$30-34$ $32$

$35-39$ $48$

$40-44$ $50$

$45-49$ $35$

$50-54$ $12$

Find the median of the observation: $62, 67, 82, 71, 75, 77, 63, 91$.

If the mean of $18, 2x, 27, 12, 3x, 5x$ and $21$ is $24,$ then find $x$.

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