$100$ surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Number of letters	$1-4$	$4-7$	$7-10$	$10-13$	$13-16$	$16-19$
Number of surnames	$6$	$30$	$40$	$16$	$4$	$4$

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also,find the modal size of the surnames.

(N/A) The cumulative frequencies are as follows:

Number of letters	Frequency $(f)$	Cumulative frequency $(cf)$
$1-4$	$6$	$6$
$4-7$	$30$	$36$
$7-10$	$40$	$76$
$10-13$	$16$	$92$
$13-16$	$4$	$96$
$16-19$	$4$	$100$

$1$. Median: $n=100$,so $n/2 = 50$. The cumulative frequency just greater than $50$ is $76$,which corresponds to the class $7-10$.
Median $= l + [(\frac{n}{2} - cf) / f] \times h = 7 + [(50 - 36) / 40] \times 3 = 7 + (14/40) \times 3 = 7 + 1.05 = 8.05$.
$2$. Mean: Using assumed mean $a = 11.5$ and $h = 3$,we calculate $x_i$ (midpoints): $2.5, 5.5, 8.5, 11.5, 14.5, 17.5$.
$u_i = (x_i - 11.5)/3$: $-3, -2, -1, 0, 1, 2$.
$f_i u_i$: $-18, -60, -40, 0, 4, 8$. Sum $\sum f_i u_i = -106$.
Mean $= a + h \times (\sum f_i u_i / \sum f_i) = 11.5 + 3 \times (-106/100) = 11.5 - 3.18 = 8.32$.
$3$. Mode: The modal class is $7-10$ (highest frequency $40$).
Mode $= l + [(f_1 - f_0) / (2f_1 - f_0 - f_2)] \times h = 7 + [(40 - 30) / (80 - 30 - 16)] \times 3 = 7 + (10/34) \times 3 = 7 + 0.88 = 7.88$.