A survey was carried out to find out the number of emails received in a week by each of a group of students. The table below represents the result of the survey.

Number of emails (n)	Frequency
$0 \le n < 10$	8
$10 \le n < 20$	13
$20 \le n < 30$	25
$30 \le n < 40$	30
$40 \le n < 50$	18
$50 \le n < 60$	6

1. Find the total number of students and the number who received at least 40 emails.

Total number of students
$8+13+25+30+18+6=100$
Number who received at least 40 emails
$18+6=24$

2. Calculate the probability that both selected students received at least 40 emails.

Since two students are chosen from the group, this is without replacement.
$P=\frac{24}{100}\times\frac{23}{99}$
$P=\frac{552}{9900}=\frac{46}{825}$

1. Find the median position and identify the class interval containing it.

Total frequency $=8+13+25+30+18+6=100$.

The median lies between the $50$th and $51$st values.

Cumulative frequencies:

$0\le n<10$: $8$

$10\le n<20$: $8+13=21$

$20\le n<30$: $21+25=46$

$30\le n<40$: $46+30=76$

Since the $50$th and $51$st values lie in $30\le n<40$, the median interval is $30\le n<40$.

1. Use class midpoints to estimate the mean from the grouped frequency table.

Class midpoints: $5, 15, 25, 35, 45, 55$

Total frequency:
$8+13+25+30+18+6=100$

$\text{Estimated mean}=\frac{5(8)+15(13)+25(25)+35(30)+45(18)+55(6)}{100}$

$=\frac{40+195+625+1050+810+330}{100}$

$=\frac{3050}{100}=30.5$

1. Use class midpoints to find $\sum fx^2$ for the grouped data.

Class midpoints: $5,15,25,35,45,55$

$\sum fx^2=8(5^2)+13(15^2)+25(25^2)+30(35^2)+18(45^2)+6(55^2)$

$=110100$

Also, $\sum f=100$ and from part (c), $\bar{x}=29.8$.

2. Substitute into the standard deviation formula and evaluate.

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

$=\sqrt{\frac{110100}{100}-29.8^2}$

$=\sqrt{1101-888.04}$

$=\sqrt{212.96}=14.593\ldots$