Statistic question?

Question

Support whenever invited to a party, the probability that person arrives with a guest is 1/3, arrives alone is 1/3, and doesn't attend is 1/3. If the company invites all of its 300 employees and their guests, what is the probability that at least 320 arrives?

Answer 1

The expected number is defined to be E(s) = P(s)N; where P(s) is the probability of success s and N is the total number of possible outcomes.

Thus, with N = 300 employees invited, the E(s = with guest) = P(s = with guest)300 = 1/3 (300) = 100 employees + (presumed) one guest each = 200 attendees.

Similarly, P(s = alone)300 = 100 employees
P(s = decline)300 = 100 employees.

So the expected number of people attending is 300, 200 employees plus 100 guests.

Expected number is the same as mean or average. If we assume a normal distribution, that means a standard deviation for this party is about SD = Ave/3 = 300/3 = 100. This is based on the standard z curve that has about three standard deviations from the tip of one tail to the mean.

So the probability that P(s <= 300) = 1/2; the probability that up to and including 300 people (including guests) will attend. The probability that P(SD) = P(300 < s <= 400) ~ .34, which is the probability under the curve between mean and one standard deviation from the mean.

If we assume linearity, we can write s/SD = P(s)/P(SD); where s = 20 attendees above the average of 300, SD = 100 attendees, and P(SD) = .34

We then solve for P(300 < s <= 320) = (s/SD)P(SD) = (20/100)*.34 = .068 or about .07. Thus, P(s <= 320) = P(s <= 300) + P(300 < s <= 320) = .50 + .07 = .57 so that P(s >= 320) = 1 - P(s <= 320) = 1 - .57 = .43.

These are all approximations based on the assumption that there are three standard deviations from tip to mean of a Normal curve and that we can assume s/SD = P(s)/P(SD) to estimate that bit of probability above the mean and below 320. That gets us P(s < 320); so that P(s >= 320) = 1 - P(s < 320), which is the probability at least 320 will attend the gala.

Clearly, the probability under a Normal curve is not linear with the amount of standard deviation. But the amount of error incurred by that assumption should be acceptable in most cases where extreme accuracy is not required...like planning for a party.