From past experience, it is known that the number of tickets purchased by studen

Question

From past experience, it is known that the number of tickets purchased by students standing in line at the ticket window for the football match of Alabama against Auburn follows a distribution that has mean equal to 240 and standard deviation equal to 200. Suppose that few hours before the start of one of these matches there are 100 eager students standing in line to purchase tickets. If only 250 tickets remain, what is the probability that all 100 students will be able to purchase the tickets they desire?

Explanation / Answer

Solution:-

We are given that µ = 240, = 200 and n = 100.

There are 250 tickets available, so the 100 students will be able to purchase the tickets they want if all together ask for less than 250 tickets.

The probability for that is P(T < 250) = P(z < (250100(240)) / (sqrt(100) * 200) ) = P(z < -11.875) < 0.00001.

NOTE:- Here, the mean and standard distribution is quite large comparitively, thus the calculated value is also large for z, because usually the value of z is between -6 and 6, however here it is -11.875. If there is any change in the question, just plug the values of mean and SD.

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