Q.Find the mean deviation about the median for the following data: 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

The mean deviation about the median for this distribution is 11.36 — found from the median (28) and the frequency-weighted absolute deviations of each class mark from it.

We are given a grouped frequency distribution:

Step 1: Find $N$ and the cumulative frequencies.

$N = 6+7+15+12+6+4 = 50$

Cumulative frequencies (class by class): $6,\ 13,\ 28,\ 40,\ 46,\ 50$.

Step 2: Locate the median class.

$\dfrac{N}{2} = 25$. Scanning the cumulative frequencies, the value $25$ is first reached (exceeded) in the class $20$-$30$, where the cumulative frequency is $28$. So the median class is $20$-$30$, with:

• $L = 20$ (lower boundary of median class)
• $cf = 13$ (cumulative frequency of the class just before the median class)
• $f = 15$ (frequency of the median class)
• $h = 10$ (class width)

Step 3: Compute the median.

$\text{Median} = L + \frac{\frac{N}{2} - cf}{f} \times h = 20 + \frac{25 - 13}{15} \times 10 = 20 + \frac{12}{15}\times 10 = 20 + 8 = 28$

…

The median works out to about 27.86; averaging $f_i|x_i - \text{median}|$ over all 50 girls gives a mean deviation of about 10.34.

Step 1 — Set up the table (class marks $x_i$ = midpoints):

$N = 50$, so $N/2 = 25$.

Step 2 — Find the median class. The cumulative frequency first reaches/exceeds 25 in the class 20-30 (cumulative freq 28). So the median class is $20$–$30$, with $L=20$, cumulative freq before it $cf=14$, class frequency $f=14$, class width $h=10$.

$\text{Median} = L + \dfrac{\frac{N}{2}-cf}{f}\times h = 20 + \dfrac{25-14}{14}\times10 = 20+\dfrac{110}{14} = 20+7.857 \approx 27.857$

Step 3 — Compute $f_i|x_i-\text{Median}|$:

| $x_i$ | $|x_i-27.857|$ | $f_i$ | $f_i|x_i-\text{Median}|$ | …