Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    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
    Offline

    14
    ReputationRep:
    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.
    • Thread Starter
    Offline

    0
    ReputationRep:
    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?
    Offline

    0
    ReputationRep:
    (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)?



    Posted from TSR Mobile
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.