Consider the following frequency distribution : \begin{tabular}{|l|l|l|l|l|l|} \hline Class: & $10-20$ & $20-30$ & $30-40$ & $40-50$ & $50-60$ \\ \hline Frequency : & $\alpha$ & 110 & 54 & 30 & $\beta$ \\ \hline \end{tabular} If the sum of all frequencies is 584 and median is 45 , then $|\alpha-\beta|$ is equal to

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Statistics

Consider the following frequency distribution :

\begin{tabular}{|l|l|l|l|l|l|}
\hline Class: &

10 - 20

20 - 30

30 - 40

40 - 50

50 - 60

\\
\hline Frequency : &

α

& 110 & 54 & 30 &

β

\\
\hline
\end{tabular}

If the sum of all frequencies is 584 and median is 45 , then

∣ α - β ∣

is equal to

Answer: 164

The frequencies add to

584

α + 110 + 54 + 30 + β = 584 \Rightarrow α + β = 390

A median of

45

places the median class at

40

–

50

, so

L = 40

f = 30

h = 10

\frac{N}{2} = 292

, and the cumulative frequency before that class is

α + 110 + 54 = α + 164

. The median formula gives

45 = 40 + \frac{292 - ( α + 164 )}{30} \times 10 \Rightarrow 15 = 128 - α \Rightarrow α = 113

Then

β = 390 - 113 = 277

, so

∣ α - β ∣ = 164

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Integer answer