If a typical competitor has a B/C ratio with mean=1.05 and standard deviation=0.09, what is the markup associated with a 30% probability of beating four typical competitor based on  (1) Friedman’s method (1.5 point) and  (2) Gate’s method (1.5 point)?

Number of members chosen by Helena in her homeroom = 5

Mean number of tickets = 16

So, total candidates who had purchased ticket = 16 × 5= 80

Now, number of candidates chosen randomly = 10

So, total candidates who had purchased ticket = 22.5 × 10=225

No, one sample can't be a better representation by choosing randomly a group of members.

In Sample 1 , we are considering only 5 drama numbers.

And In Sample 2, we are considering 10 random samples among 75 members.

The mean in two cases varies by (22.5-16=6.5) which is large because we are considering number of tickets sold by members of club.

It would have been better if she has taken more than 3 random samples of 15  students ,among 75 students, then she would have got appropriate mean or value near the mean.