Calculating Mean and Variance for Graduation Ceremony Guest Tickets

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Homework Help Overview

The problem involves calculating the mean and variance of guest tickets requested by graduating students, based on given probabilities for different ticket requests. It also includes a probability question regarding the availability of tickets for a graduation ceremony.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the mean and variance, with some questioning the initial methods used. There is a focus on the interpretation of probabilities and the correct application of formulas. The Central Limit Theorem is mentioned as a potential approach for the second part of the problem.

Discussion Status

Participants are actively revising their calculations and reasoning. Some have expressed confusion about the mean and variance, while others are considering different interpretations of the problem. There is no explicit consensus on the correct approach yet, but there is movement towards clarifying misunderstandings.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an ongoing discussion about the assumptions related to the distribution of ticket requests.

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Homework Statement


Suppose that, on average, 70% of graduating students want 2 guest tickets for a graduation ceremony, 20% want 1 guest ticket and the remaining10% don't want any guest tickets.

(a) Let X be the number of tickets required by a randomly chosen student. Find the mean and variance of X.

(b) If 500 guest tickets are available for a ceremony at which 300 students are graduating, what is the probability that there will not be enough tickets to satisfy demand?


Homework Equations


Variance = E(X2) - E2(X)


The Attempt at a Solution


I've got the mean to be ((0.7*2) + (0.2*1) + (0.1*0))/3 = 0.53and the variance to be ((0.72*2) + (0.22*1) + (0.12*0))/3 - 0.532 = 0.7391 but this seems wrong because the variance is then bigger than the mean?!
 
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Okay I have since realized that for part a) I think i was doing it wrong so now for the mean I have:

((0.7*2) + (0.2*1) + (0.1*0))/3 = 0.53

But for the Variance I have:

((0.7 - 0.53)2+ (0.2 - 0.53)2 + (0.1 - 0.53)2)/3 = 0.1076 which makes far more sense!

Now I'm thinking of using the Central Limit Theorem to solve b)??!
 
vikkisut88 said:
Okay I have since realized that for part a) I think i was doing it wrong so now for the mean I have:

((0.7*2) + (0.2*1) + (0.1*0))/3 = 0.53
This is still incorrect. Suppose every student wanted one ticket. Would the average number be ((0.7*1) + (0.2*1) + (0.1*1))/3 = 1/3? Of course not. The ".7", ".2", and ".1" are already "averages". Don't divide by 3: 0.6*2+ 0.2*1+ 0.1*0= 1.4.

But for the Variance I have:

((0.7 - 0.53)2+ (0.2 - 0.53)2 + (0.1 - 0.53)2)/3 = 0.1076 which makes far more sense!
There is no reason in the world why the variance can't be larger than the mean.

Now I'm thinking of using the Central Limit Theorem to solve b)??!
 
Right okay, so now for the variance do I use the same formula but use 1.6 (I presume it was just a typo and it should have been 0.7 *2) instead of 0.53?
Do I then use the Central Limit Theorem for part b?
 

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