Calculate the mean, variance, and standard deviation for the given grouped data.

The class intervals are 8-12, 13-17, 18-22, 23-27, 28-32, 33-37.
The frequencies ($f$) are 5, 3, 6, 7, 5, 4.

Midpoints ($x$):

• For 8-12: $x = \frac{8+12}{2} = 10$
• For 13-17: $x = \frac{13+17}{2} = 15$
• For 18-22: $x = \frac{18+22}{2} = 20$
• For 23-27: $x = \frac{23+27}{2} = 25$
• For 28-32: $x = \frac{28+32}{2} = 30$
• For 33-37: $x = \frac{33+37}{2} = 35$

Total frequency:

• $5 \times 10 = 50$
• $3 \times 15 = 45$
• $6 \times 20 = 120$
• $7 \times 25 = 175$
• $5 \times 30 = 150$
• $4 \times 35 = 140$

Sum of $f \cdot x$:

The formula for the mean of grouped data is $\bar{x} = \frac{\sum f \cdot x}{\sum f}$.

Rounding to two decimal places:

The mean time is $\boxed{\text{22.67 minutes}}$.

• For $x=10$: $x^2 = 10^2 = 100 \implies f \cdot x^2 = 5 \times 100 = 500$
• For $x=15$: $x^2 = 15^2 = 225 \implies f \cdot x^2 = 3 \times 225 = 675$
• For $x=20$: $x^2 = 20^2 = 400 \implies f \cdot x^2 = 6 \times 400 = 2400$
• For $x=25$: $x^2 = 25^2 = 625 \implies f \cdot x^2 = 7 \times 625 = 4375$
• For $x=30$: $x^2 = 30^2 = 900 \implies f \cdot x^2 = 5 \times 900 = 4500$
• For $x=35$: $x^2 = 35^2 = 1225 \implies f \cdot x^2 = 4 \times 1225 = 4900$

The formula for variance of grouped data is $\text{Variance} = \frac{\sum f \cdot x^2}{\sum f} - \left(\frac{\sum f \cdot x}{\sum f}\right)^2$.
We know $\sum f = 30$, $\sum f \cdot x = 680$, and $\sum f \cdot x^2 = 17350$.

Rounding to two decimal places:

The variance is $\boxed{\text{64.56 minutes}^2}$.

The standard deviation is the square root of the variance.

Rounding to two decimal places:

The standard deviation is $\boxed{\text{8.03 minutes}}$.

Question Two

Distance ($D$) = 240 km
New Time ($T$) = $1\frac{1}{2}$ hours = 1.5 hours

The formula for speed is $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$.

The speed the aeroplane must use is $\boxed{\text{160 km/h}}$.