1. Judge A gives marks to 5 competitors of 10, 8, 7, 6, 3. Judge B gives marks to the corresponding competitors of 22, 17, 8, 19, 10. The value of Spearman’s Rank Correlation Coefficient is 0.A. Find A.
2. The Standard Deviation of 32, 38, 41, 26 and 45 is 6.71B when written to 3 decimal places. Find B.
3. A 3 sided dice has faces labeled 1, 2 and 3. If P(1)=0.3, P(2)=f, P(3)=g and the expectation of the score is 2.2, the variance of the score is 0.C6. Find C.
4. A geocacher counts the number of geocaches he finds in a month and discovers that it is normally distributed with a mean of 50 and a standard deviation of 2. He finds more than 45.D5 geocaches in 99% of months. Find D.
5. A group of geocachers were asked about their caching habits. Two fifths of them liked puzzle caches. Two thirds of the group went to event caches. Of those who went to event caches, nine twenty-fifths also liked puzzle caches. Find the probability that a randomly chosen member of the group does not like puzzle caches and does not go to event caches. If you write your answer as a fraction in the simplest form possible, E= the numerator of the fraction.
You're correct, the setter has been a little wayward here and there. There was an earlier question:-
(8-sqrt(18))/sqrt(2) = f + g x sqrt(2), b= f + g. (Where ‘x’ means times, and sqrt(18) means the square root of 18).
I believe this to be insoluble. B=1 is what is expected, though I only got that by solving the other 4 questions and a little trial and error.
Answer#1 Spearman Rank
This is a standard, rote sort of computation.
Using this calculator
you quickly get that ρ = 0.5, so A=5.
#2 Standard Deviation
This is an even more standard computation. Using this calculator
. You get 6.71118, so B=1.
Note that with Wolfram, it suggests you try this as a population standard deviation instead of a sample s.d. -- it's not clear which one it is, according to the question, although it is again clear that the idea of "digits" is incorrect (the question clearly uses four digits).
This one requires some work. First of all, the expectation is 2.2 which means:
1*P(1) + 2*P(2) + 3*P(3) = 2.2
0.3 + 2f + 3g = 2.2
But also we know 0.3 + f + g = 1, since these are the only possible outcomes.
Solving for f and g gives us
P(2) = 0.2
P(3) = 0.5
The variance is:
0.3*1^2 + 0.2*2^2 + 0.5*3^2 - 2.2^2 = 0.76
so it looks like C=7.
#4 Normal Distribution
This is something you just look up in a table. A normal distribution tends to be close to the mean, so being greater than 45 is reasonable when the mean is 50. Checking any table or calculator can give you that 45.3473 is the number you want... I suppose if we round that off, everything works out and we get D=3.
#5 Venn Diagrams
This is just a Venn diagram sort of problem. It can be understood by considering two properties: P could be puzzle caches and E event caches.
The information we have is:
Pr(P) = 2/5
Pr(E) = 2/3
Pr(P|E) = 9/25
That last one is a conditional probability
. It can be rewritten:
Pr(P & E) / Pr(E) = 9/25
That gives you Pr(P&E) = (9/25)*(2/3) = 6/25
Now, what you want is to exclude P and E. You could simply subtract 1 - Pr(P) - Pr(E) = 1 - 2/3 - 2/5 = -1/15.
That is obviously wrong because it is negative. The problem is you have subtracted twice anyone who falls into both categories. For the purposes of excluding them from the total, you only want to subtract them out once. The easiest way to compensate is to add them back in:
1 - 2/3 - 2/5 + 6/25 = 13/75
This is a very small version of the principle of inclusion/exclusion
A = 5
B = 1
C = 7
D = 3
F = 13
Note that there are ongoing issues with "n decimal digits" often meaning something incorrect, but otherwise this is fine. For example, (8 - √18) / √2 = -3 + 4√2