stats questions.. please help.. (1 Viewer)

[nevereddy]

1.Let X represent the number of visits to a GP over the last 12 months. From past records of patients, the GP has constructed the following probability distribution for X.

x​

0
1
2
3
4​

P(X=x)​

0.5
0.2
0.1
0.1
0.1​

What is the expected number of visits given that a patient has had at least one visit? (Your answer should be correct to one decimal place.)

my answer was 0.5.. um P(x>1)= P(1)+P(2)+P(3)+P(4)= .2+.1+.1+.1

my answer was wrong


2.Consider a coin tossing game where you win 9 dollars if a head appears on a single toss of a fair coin but you lose 9 dollars if a tail appears. Let the random variable R represent the return on a single toss. What is the variance of R?
i got 0.25, it was wrong
i used the formula .. V= np(1-p) V= 1(0.5)(1-).5)

3.Let p be the probability of success in a binomial experiment. Of the following values of p, which one will produce a binomial probability distribution with the largest variance, given that the sample size (n) is the same in each case?
i got 0.3 . ( it was a guess)

4.It is known that 55% of your local electorate voted for the political party in government at a recent election. In a random sample of eight drawn from the local electorate, what is the probability that more than four voted for the party in government? (Your answer should be correct to four decimal places.)

i got .7396..

what i did was P(X>4)= P(4)+P(5)+P(6)+P(7)+P(8)

then, using P(x)= n!/x!(n-x)! * p^x(1-p)^n-x

OMG.. PLEASE HELP, im dieing in statistics..

 

[nevereddy]

oh.. all my answers were wrong .. -.-

 

[Kotchi]

Looks to me someone comes from UNSW
Just to know if your in my tute or not, what day's your tute?

 

[Timothy.Siu]

1. It asks for expected value, not probability.

2. It's not simply a binomial. It's a distribution f(x) for example, where f(9)=0.5 and f(-9)=0.5. So Var(X)=E(X^2)-[E(X)]^2. Now E(X)=0 and E(X^2)=(-9)^2*0.5+(9)^2*0.5=81. So Var(X)=81.

3. Largest variance should be probability closest to 0.5

4. It's binomial with p=55/100. It asks for MORE than 4. So 5,6,7,8.

Good luck.

 

[bro3]

asian knows maths

big deal

I am white and genius at maths XD

 

[Kotchi]

for the first question, my mate calculated the mean which was 1.1 and got it wrong. I got 1.2 (my x value were 01235, not 01234) and also got it wrong. Only thing i could think of was that you can't have 0.1 of a visit, so im considering putting 1.0

 

[deterministic]

Question 1 involves conditional probability, as it asks for the expected number of visits given that a patient has had at least one visit.
So P(X=x given X>0)= P(X=x and X>1)/P(X>0)
eg. P(X=1 given X>0)=P(X=1)/P(X>0)=0.2/0.5=2/5

Do the same for 2,3,4,5 and then find the expected value.