Fifteen children of low number-talk parents scored 2,1,2,3,4,3,3,2,2,1,5,3,4,1,2 and fifteen children of high number-talk parents scored 3,4,5,4,5,4,2,3,5,4,5,3,4,5,4. Complete a frequency distribution table for scores 1 to 5 in each group.

Step-by-step solution

1. 1

   Set up the tally in a fixed order. List the possible scores $1$ through $5$ and pass through the data once, marking each value. Working score-by-score instead (scanning the list five times) is what usually causes a value to be counted twice or missed.

2. 2

   Tally the low group. Going through $2,1,2,3,4,3,3,2,2,1,5,3,4,1,2$:

   $X=1$ appears at positions $2$, $10$, $14$ → $3$ times

   $X=2$ appears at positions $1,3,8,9,15$ → $5$ times

   $X=3$ appears at positions $4,6,7,12$ → $4$ times

   $X=4$ appears at positions $5,13$ → $2$ times

   $X=5$ appears at position $11$ → $1$ time

3. 3

   Tally the high group. Going through $3,4,5,4,5,4,2,3,5,4,5,3,4,5,4$:

   $X=1$ → $0$ times; $X=2$ → $1$ time; $X=3$ → $3$ times; $X=4$ → $6$ times; $X=5$ → $5$ times.

4. 4

   Assemble the table.

   $X$	$f_{\text{low}}$	$f_{\text{high}}$
   5	1	5
   4	2	6
   3	4	3
   2	5	1
   1	3	0
   Total	15	15

5. 5

   Check both columns against n. $1+2+4+5+3=15$ ✓ and $5+6+3+1+0=15$ ✓. This total is the essential check: a frequently circulated version of this table lists the low column as $1,2,3,5,4$, which also sums to $15$ but swaps the counts of $X=1$ and $X=3$ — only re-tallying, not the total, catches that.

   $\boxed{f_{\text{low}}:\ 3,5,4,2,1\ \ (X=1\ldots5);\qquad f_{\text{high}}:\ 0,1,3,6,5}$

6. 6

   Read the comparison the table is for. The low group piles up at scores $1$–$2$ (eight of fifteen children) while the high group piles up at $4$–$5$ (eleven of fifteen), so the distributions are shifted rather than merely differently spread — which is the point the original study was making.