Probability of Consultation Time Differences at Hospital Clinics

[cloud360]

Homework Statement

The time X it takes a consultant to see a new patient at a hospital out-patient clinic may be assumed to be normally distributed with mean “16 minutes” and standard deviation “1.5 minutes”. The time Y it takes the same consultant to see a returning patient may be assumed to be normally distributed with mean “10 minutes” and standard deviation “1.2 minutes”. The consultation times of different patients mat be assumed to be independent.

(i)What is the probability the consultation time for the new patients is more than twice that of a returning patient

(ii) At the start of the Monday clinic, 5 new patients and 8 returning patients are waiting to be seen by the consultant. What is the probability that the consultant will see all of them within 3 hours?

(iii) At the start of the Tuesday clinic, 4 new patients and 6 returning patients are waiting to be seen. What is the probability that it will take the consultant, more time to see the retuning patients than to see the new patients?

Homework Equations

The Attempt at a Solution

(I) i think i must solve P(X>2Y)=P(X-2Y>0), does that mean where i see the random variable X in the normalization equation, i replace it with X-2Y

(ii) I was thinking this nee the bionomial distribution but am no sure

(iii) Is this P(X>Y)=P(X-Y>0)?

 

[Stephen Tashi]

(I) i think i must solve P(X>2Y)=P(X-2Y>0), does that mean where i see the random variable X in the normalization equation, i replace it with X-2Y

No. You probably mean "the normal probability density" instead of "the normalization equation" (whatever that would be), but the answer is still no. There are various ways to give an answer. One way is to say you want to integrate a bivariate normal density over the are of the x-y plane given by X > 2Y. Can you write the appropriate formula for the bivariate normal density that appies to this problem?

(ii) I was thinking this nee the bionomial distribution but am no sure.

I see no way to apply the binomial distribution.

Have you studied how to find the probability density function of the sum of independent normally distributed random variables?

(iii) Is this P(X>Y)=P(X-Y>0)?

Again this involves integrating a bivariate normal density over the area in the xy plane given by X > Y. Finding the formula for this bivariate normal density involves knowing about the density of the sum of independent normally distributed random variables.

 

[cloud360]

Isnt this the solution

Let W = X-2Y
Mean of W = mean of X - 2 Mean of Y = -4
Variance = standard deviation squared
Variance of W = V(X-2Y) = V(X)+4V(Y) = 2.25+4(1.44) =8.01
Standard deviation of Z = sqrt(8.01) = 2.8302

P( X > 2Y) = P( X-2Y > 0) = P(W > 0)

μ = -4
σ = 2.8302
standardize w to z = (x - μ) / σ
P(w > 0) = P( z > (0--4) / 2.8302)
= P(z > 1.4133) =0.0793
(From Normal probability table)

 

[Stephen Tashi]

Are you wanting to write a computer program to do the integration? Or use Matlab? Or do you merely need to write down the mathematical symbols for the integration? Explain what kind of course these problems came from.

 

[cloud360]

Are you wanting to write a computer program to do the integration? Or use Matlab? Or do you merely need to write down the mathematical symbols for the integration? Explain what kind of course these problems came from.

Simple statistics. I have not learned how to integrate the normal distribution function is my course. this is my course

I have learned how to prove to show, the integral of Hist(X) = 1

 

[Stephen Tashi]

Have you studied the Cauchy distribution?

 

[cloud360]

Have you studied the Cauchy distribution?

no, but i think this is a co variance question which does not requite knowledge of co variance

is this correct for q2?

Call the total time T

E[T] = 5E[X] + 8E[Y] = 5*16 + 8*10 = 160
Var[T] = Var[5X] + Var[8Y] = (5*1.5)^2 + (8*1.2)^2 = 148.41

To get the variance I used the fact that Cov[X,Y] = 0 since the times for all patients are independent.

So now you find the Z score.

Z = {T - E[T]}/StDev[T] = (T - 160)/sqrt(148.41)

So if T under 3 hours = 180 minutes is equivalent to

Z < 20/sqrt(148.41) = 1.6417

Looking up a standard normal table would give you a probability of 94.97%

 

[Stephen Tashi]

Var[T] = Var[5X] + Var[8Y] = (5*1.5)^2 + (8*1.2)^2 = 148.41

Shouldn't that be 5*(1.5)^2 + 8*(1.2)^2 ?

For example, Var[5X] would imply he saw a single patient whose distribution had a variance of five times the variance of X. Instead, he sees five patients independently, each with variance (1.5)^2 So the variance = sum of 5 variances, each of which is (1.5)^2.

 

[Stephen Tashi]

Isnt this the solution

Let W = X-2Y
Mean of W = mean of X - 2 Mean of Y = -4
Variance = standard deviation squared
Variance of W = V(X-2Y) = V(X)+4V(Y) = 2.25+4(1.44) =8.01
Standard deviation of Z = sqrt(8.01) = 2.8302

P( X > 2Y) = P( X-2Y > 0) = P(W > 0)

μ = -4
σ = 2.8302
standardize w to z = (x - μ) / σ
P(w > 0) = P( z > (0--4) / 2.8302)
= P(z > 1.4133) =0.0793

P(z > -1.4133) Otherwise, your method of working the problem is better than anything I was thinking about.

 

[cloud360]

Shouldn't that be 5*(1.5)^2 + 8*(1.2)^2 ?

For example, Var[5X] would imply he saw a single patient whose distribution had a variance of five times the variance of X. Instead, he sees five patients independently, each with variance (1.5)^2 So the variance = sum of 5 variances, each of which is (1.5)^2.

i think u take the n outside and don't square. only when we are taking sum of the random variables !

e.g if \Sigma(0 to n) X_i, then Var( \Sigma(0 to n)X_i)=nVar((0 to n)X_i). and not n²Var((0 to n)X_i)

i don't think we are taking sum of random variables here ,what is your opinion. i am not 100% sure

 

[Stephen Tashi]

i think u take the n outside and don't square. only when we are taking some of the random variables !

i don't think we are taking some of random variables here ,what is your opinion. i am not 100% sure

My opinion is that it should be 5*(1.5)^2 + 8*(1.2)^2 because there are 5 independent random variables of one kind plus 8 independent random variables of the other kind.

Consider the difference between 2X = on random variable multiplied by 2
vs (X + X) = the sum of two independent random variables

Var(X + X) = Var(X) + Var(X) = 2 Var X

vs

Var(2x) = 4 Var(X)