A frequency distribution of 100 students' marks is given. Complete the frequency table and find y and z.

Here is the solution to the problem.

First, let's reconstruct the complete frequency distribution table based on the given information and the student's work for part (a).
The total number of students is 100.
The given frequencies are: 6, 9, 12, y, 20, 7, 4, z (where z is the frequency for $70 \le x < 80$).

*a) Determine the value of $y$.
The sum of all frequencies must be 100.
From the student's work, $100 - 86 = 14$, which implies $y=14$.
This means the sum of all other frequencies (excluding $y$) is 86.
Let's verify this by finding $z$.
$6 + 9 + 12 + 20 + 7 + 4 + z = 86$
$58 + z = 86$
$z = 86 - 58 = 28$.
So, the frequency for the class $70 \le x < 80$ is 28.
Now, let's sum all frequencies with $y=14$ and $z=28$:
$6 + 9 + 12 + 14 + 20 + 7 + 4 + 28 = 100$. This is correct.

The complete frequency distribution table is: | Marks ($x$) | No. of students ($f$) | Midpoint ($x_m$) | $fx_m$ |