The table shows the distribution of the weights of 40 students.

1. Form and solve the total frequency equation

The total number of students is $40$.

$6+8+a+11+3=40$

$28+a=40$

$a=40-28=12$

1. Find the estimated mean using class midpoints and frequencies

Class midpoints: $45,55,65,75,85$

Frequencies: $6,8,12,11,3$

$\text{Estimated mean}=\frac{45(6)+55(8)+65(12)+75(11)+85(3)}{40}$

$=\frac{270+440+780+825+255}{40}$

$=\frac{2570}{40}=64.25$

To 3 significant figures, the estimated mean is $64.3\text{ kg}$.

Alternatively, if using exact calculator display, $64.25\text{ kg}$.

1. Calculate the estimated standard deviation using grouped-data midpoints

Using midpoints $45,55,65,75,85$ and frequencies $6,8,12,11,3$,

$\sum fx^2=6(45^2)+8(55^2)+12(65^2)+11(75^2)+3(85^2)=170600$

With $n=40$ and the estimated mean from part (b)(i), $\bar{x}=64.5$,

$\text{standard deviation}=\sqrt{\frac{\sum fx^2}{n}-\bar{x}^2}$

$=\sqrt{\frac{170600}{40}-64.5^2}$

$=\sqrt{4265-4160.25}$

$=\sqrt{104.75}=10.234\ldots$

1. State the effect of correcting all recorded weights by subtracting 2 kg

Since every recorded weight was $2\text{ kg}$ more than the actual weight, rectifying the error means subtracting $2\text{ kg}$ from every value.

Therefore, the mean decreases by $2\text{ kg}$:
$64.5 - 2 = 62.5\text{ kg}.$

The standard deviation is unchanged because subtracting the same constant from every value does not change the spread of the data.