At four decimal places, the difference is a rather noticable 0.0087. Anyone who does this problem correctly (to more than one decimal place) will be awarded "Wrong Answer"!

As a former teacher of statistics, this error, which I hope none of MY former students would make, is decidedly annoying.

[[Teaching Mode ON]]

There is some discrete probability distribution for the number of tickets which these 100 students will wish to buy. That is, there is a number representing the probability that 100 students will want 100 tickets, and another number for 101 tickets, yet another number (probably very close to zero) for zero tickets. But no number for 100.2 tickets. - Noone will ask to buy one fifth of a ticket! (Or rather the probability of that is so small that we will ignore it.)

The central limit theorem tells us that the average number of tickets demanded will be 2.4×100 = 240, with a standard deviation of 2.0×√100 = 20. It also tells us that the actual distribution will be "approximately" normal. But the normal distribution is continuous, not discrete. A discrete distribution which closely approximates the normal distribution is obtained by taking the normal probability of a value between N-0.5 and N+0.5 as the discrete probability of a value exactly equal to N. (Many discrete distributions arise in exactly this way when observed values are rounded off for reporting.) The probability that exactly 250 tickets are demanded would then be approximately

    Norm.CDF((250.5-240)/20) - Norm.CDF((249.5-250)/20)

while the probability of exactly 249 tickets is approximately

    Norm.CDF((249.5-240)/20) - Norm.CDF((248.5-250)/20)

et cetera. So the correct approximation for the probability that 250 or fewer tickets are demanded telescopes to: Norm.CDF((250.5-240)/20) rather that the the mistake which is currently being accepted.

[[Teaching Mode OFF]]