The table below summarises the times taken by female participants to complete a 10 km race.

1. Find the class midpoints and form the total estimated time.

The class midpoints are
$35,45,55,65,75$.

Using $\text{midpoint} \times \text{frequency}$,
$\sum ft=20(35)+39(45)+16(55)+20(65)+75x$
$=700+1755+880+1300+75x=4635+75x.$

2. Use the estimated mean formula and solve for $x$.

Total frequency:
$20+39+16+20+x=95+x.$

Given estimated mean $=50.1$,
$\frac{4635+75x}{95+x}=50.1$
$4635+75x=50.1(95+x)$
$4635+75x=4759.5+50.1x$
$24.9x=124.5$
$x=5.$

1. Use the class midpoints and frequencies to calculate the estimated standard deviation.

Class midpoints: $35,45,55,65,75$

Frequencies: $20,39,16,20,5$, so $\sum f=100$.

Using the estimated mean from part (a), $\bar{x}=50.1$,

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

$\sum f x^2=20(35^2)+39(45^2)+16(55^2)+20(65^2)+5(75^2)$

$=24500+78975+48400+84500+28125=264500$

$\text{standard deviation}=\sqrt{\frac{264500}{100}-50.1^2}$

$=\sqrt{2645-2510.01}=\sqrt{134.99}=11.618\ldots$

1. Compare the average race times using the means.

Female mean $=50.1$ min, male mean $=45.3$ min.

Since $50.1>45.3$, female participants took longer on average than male participants.

2. Compare the consistency of race times using the standard deviations.

Female standard deviation $=13.8$ min, male standard deviation $=12.6$ min.

Since $13.8>12.6$, female participants' times were more spread out, so they were less consistent than male participants' times.