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maths as level stats help

I have this 11 mark question for my stats homework but no idea how to do it. Can anyone explain or show me how?
The question is,

'two fair 5-sided spinners numbered 1-5 are spun. The score, s, is the difference between the two numbers showing.
a)find the probability distribution of s
b)find the mean and variance of s
c)a random variable T is defined by T = 1/2 s + 3, find the mean and variance of T

Reply 1

Damask-

The probability distribution looks like a table, with every possible outcome on one row, and the corresponding probability on the next row. Start off by listing the possible outcomes. You should have the formulae for the mean and variances, and with the distribution you have all you need for the last two questions.

Reply 2

cameron789

Okay so would S be 1 1 2 2 3 3 4 4 5 5 and P(S=s) 0.1 all the way across?

Reply 3

SamGrainger95

Original post by cameron789

Okay so would S be 1 1 2 2 3 3 4 4 5 5 and P(S=s) 0.1 all the way across?

no, because you are recording the differences between the two spinners then the table would have S being 0 (for example you could spin a 4 and another 4, the difference being zero), 1, 2, 3 and 4.

P(S=0) would be 5 x (1/5)^2 as to gain zero both the spinner's values must be the same, so 1 and 1, 2 and 2, 3 and 3, 4 and 4 or 5 and 5. The probability of scoring an individual value on Spinner 1 is 1/5, multiplied by 1/5 for Spinner 2 using AND rule and then summed for the five different possibilities using OR rule.

P(S=4) can only stand when a 1 and a 5 is spun. The probability of this is (1/5)^2, but as you could gain a 1 on S1 and a 5 on S2, or vice versa you must times this by 2 to get 2x(1/5)^2.

P(S=1) could be gained through 1 and 2, 2 and 1, 2 and 3, 3 and 2, 4 and 3, 3 and 4, 5 and 4 or 4 and 5. therefore it is 8x(1/5)^2

P(S=2) could be 1 and 3, 3 and 1, 2 and 4, 4 and 2, 3 and 5 or 5 and 3. therefore 6x(1/5)^2

P(S=3) could be 1 and 4, 4 and 1, 2 and 5 or 5 and 2. therefore 4x(1/5)^2

The table would therefore look like
1/5 8/25 6/25 4/25 2/25

You know you have done this correctly as all of the probabilities in the distribution table sum to 1.

I presume you know how to calculate E(S) and Var(S) from a probability distribution table for part (b)?

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