VCE Methods Question Thread!

[TrueTears]

For any normal random variable , then any linear transformation of it will still be normal. More precisely, if then .

What if it is a discrete probability distribution? is it the same? (like the probability tables)

A normal random variable is characterised by a pdf not a pmf. The closest discrete 'approximation' to a normal distribution is the (limiting distribution) binomial random variable (which is discrete), but by the central limit theorem, it's limiting (non-degenerate) distribution is normal.

In general, if Y is a discrete random variable, then any real function of Y, g(Y), is also a discrete random variable, then the pmf of g(Y)=X is given by:

[Jaswinder]

with continuous probability distributions question which require you to calculate suppose the height of a plant which is less than 10cm. why do we put (-)infinty as the lower limit? shouldnt it be from 10 to 0 instead of 10 to (-)infinty, when they ask for proportion of plants whose height is less than 10cm?

[Zealous]

with continuous probability distributions question which require you to calculate suppose the height of a plant which is less than 10cm. why do we put (-)infinty as the lower limit? shouldnt it be from 10 to 0 instead of 10 to (-)infinty, when they ask for proportion of plants whose height is less than 10cm?

In my experience, I've found that when the question asks to round, this will take care of that. The probability of something being lower than 0 would be negligible, in a lot of cases due to small standard deviations.

So personally I use negative infinity when calculating the continuous distribution (unless there's a lower bound), but I'm interested as to what other's have to say.

[TrueTears]

with continuous probability distributions question which require you to calculate suppose the height of a plant which is less than 10cm. why do we put (-)infinty as the lower limit? shouldnt it be from 10 to 0 instead of 10 to (-)infinty, when they ask for proportion of plants whose height is less than 10cm?

All depends on the support for the associated pdf.

[Jaswinder]

All depends on the support for the associated pdf.

not sure what that means?

[TrueTears]

http://en.wikipedia.org/wiki/Support_%28mathematics%29#In_probability_and_measure_theory

[Jaswinder]

Hannah determines that the number of minutes, she uses her mobile phone in a randomly chosen moth is normally distributed with mean of 120 minutes and standard deviation of 7 minutes. Hannah rents her mobile phone with calls charged at $1.25 per minute and a fixed charge of $34 per month. Given C, the monthly cost of the mobile phone, is a random variable with a normal distribution find:

Find Var(c), the variance of C and hence SD(C), the standard deviation of C.

[abcdqd]

Hey guys, I'm struggling to wrap my head around this question, any help would be appreciated thanks.
There are two lines at a supermarket checkout: regular and express. It is known that 88% of people go to the regular line. The time spent in the regular line is normally distributed with a mean of 6.6 mins and st.dev. of 2.2 1.2 mins. For the express line, the mean is 4.2290 mins and st.dev. is 0.8 mins.
Two people are chosen at random. Find the probability that exactly one person is from the regular line, given that both people have been waiting for longer than 4.2290 minutes. Pr (time spent in line) of each person is independent.

[09Ti08]

Probability that a person from the regular line has been waiting for longer than 4.229 mins: normCdf(4.229,+infinity,6.6,2.2)=0.8594. But since 88% of people go to regular line, the "overall" probability is: 0.8594*88%=0.7563

Probability that a person from the express line has been waiting for longer than 4.229 mins: 0.5. Again, the overall probability is: 0.5*12%=0.06

I think the answer to your question is: 0.7563/(0.7563+0.06)=0.9265

[friedchicken]

I got 2*(0.9759*0.88/(0.9759*0.88+0.5*0.12))*(0.50*0.12/(0.9759*0.88+0.5*0.12))=0.1221
I think it's binomial with p=0.97598*0.88/(0.9759*0.88+0.5*0.12).

[Hancock]

Hey guys, I'm struggling to wrap my head around this question, any help would be appreciated thanks.
There are two lines at a supermarket checkout: regular and express. It is known that 88% of people go to the regular line. The time spent in the regular line is normally distributed with a mean of 6.6 mins and st.dev. of 2.2 1.2 mins. For the express line, the mean is 4.2290 mins and st.dev. is 0.8 mins.
Two people are chosen at random. Find the probability that exactly one person is from the regular line, given that both people have been waiting for longer than 4.2290 minutes. Pr (time spent in line) of each person is independent.

Let's define two people, X and Y.

Pr(X == R | X, Y > 4.2290) = Pr(X == R intersect X,Y > 4.2290) / Pr(X, Y > 4.2290)

Pr(X, Y > 4.2290) = 0.88*Pr(Regular > 4.2290) + 0.12*Pr(Express > 4.2290)
Pr(X, Y > 4.2290) = 0.88*normcdf(mean = 6.6, sd = 1.2, lower = 4.2290, upper = inf) + 0.12*normcdf(mean = 4.2290, sd = 0.8, lower = 4.2290, upper = inf)
Pr(X, Y > 4.2290) = 0.88*0.9759 + 0.12*0.5 = 0.9188

Now, the probability that the chosen person X is from the regular line intersecting both X, Y > 4.2290 is (since regular line):

Pr(X == R intersect X,Y > 4.2290) = Pr(X > 4.2290) = 0.9759*0.88 (since we chosen randomly between the two lines).

Pr(X == R | X, Y > 4.2290) = 0.9759*0.88/0.9188 = 0.935

[abcdqd]

but isn't 0.9188 the probability of one person waiting more than 4.2290 minutes?

could you also do it this way:
pr (x>4.229) in Regular = 0.975913, pr(x>4.2990) in Express = 0.5
Want to find: (Pr exactly 1 from regular | both>4.229)
Possible combinations
R(>4.229) R(>4.229)=0.88*0.975913*0.88*0.975913=0.737544
R(>4.229) E(>4.229)=0.88*0.975913*0.12*0.5=0.051528
E(>4.229) R(>4.229)=0.12*0.5*0.88*0.975913=0.051528
E(>4.229) E(>4.229) =0.12*0.5*0.12*0.5=0.0036

So Pr(exactly 1 regular intersect both>4.229)= 0.051528*2=0.103056 and Pr(both> 4.229)= 0.737544+0.103056+0.0036=0.8442, which would give (Pr exactly 1 from regular | both>4.229)=0.1221, which is the same as friedchicken's answer?

[Alwin]

Hannah determines that the number of minutes, she uses her mobile phone in a randomly chosen moth is normally distributed with mean of 120 minutes and standard deviation of 7 minutes. Hannah rents her mobile phone with calls charged at $1.25 per minute and a fixed charge of $34 per month. Given C, the monthly cost of the mobile phone, is a random variable with a normal distribution find:

Find Var(c), the variance of C and hence SD(C), the standard deviation of C.

All you need to know is:



I think you can take it from here

[TrueTears]

For those that are interested, more generally, define:

and

where for are iid RV and for are iid RV, then:



and



Exercise: Try prove and by expanding out the sum and apply definition of variance and covariance.

[Alwin]

CRAP! Thanks TT for subtly pointing out my massive mistake.. mixed up var(ax+b) and E(ax+b). Gee its been a while

For anyone wondering how to prove it using the definition of variance that TT mentioned (also method of proof i prefer):
Var[aX + b] = E[ (aX + b)²] - (E [aX + b])² .
= E[ a²X² + 2abX + b²] - (aE(X) + b)²
= a²E(X2) + 2abE(X) + b² - a²E²(X) - 2abE(X) - b²
= a²E(X²) - a²E²(X) = a²Var(X)

forgot that the b² cancels in my previous post..