Question

In: Statistics and Probability

Suppose that in West Campus in September of 2019, out of 12 accidents that resulted in...

Suppose that in West Campus in September of 2019, out of 12 accidents that resulted in trips to Health Services or the emergency room, 4 happened on Friday the 13th. Is this a good reason for my GF Liz, who is very superstitious, to worry that next Friday, December the 13th, 2019, I am particularly in danger in my dorm room in Rich Hall? Just for the sake of this problem, let's pretend that December has only 30 days, so we can compare it with September.

Hint: This is similar to the last one, but the model in question is that we know that 12 accidents occurred in September of 2019, and the probability that any one of them occured on Friday the 13th is the same as any other day, namely 1/30. Therefore the binomial model is X = "how many of the 12 accidents occurred on Friday the 13th?" You want to consider the likelihood of at least 4 accidents on that date.

Solutions

Expert Solution

Solution

Back-up Theory

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and

p = probability of one success, then, probability mass function (pmf) of X is given by

p(x) = P(X = x) = (nCx)(px)(1 - p)n – x, x = 0, 1, 2, ……. , n ………….........................................................................………..(1)

[This probability can also be directly obtained using Excel Function: Statistical, BINOMDIST]....................…........……….(1a)

Now, to work out the solution,

Let X = number of accidents out of 12 accidents that occurred on Friday the 13th.

Then, under the hypothesis that any one of the accidents occurring on Friday the 13th is the same

as any other day, namely 1/30;

X ~ B(12, 1/30) ......................................................................................................................................................................(2)

Now, the likelihood of at least 4 accidents on that date

= P(X ≥ 4)

= 1 – P(X ≤ 3)

= 1 – Σ(x = 0, 1, 2, 3){(12Cx)(1/30)x)(29/30)12 – x } [vide (1) and (2)]

= 1 – 0.9995 [vide (1a)]

= 0.0005.

This probability is incredibly low. But, on Friday the 13th September of 2019, 4 accidents did occur. Hence, that event cannot be due to chance.

Thus, there is evidence to suggest that the superstition could be genuine. Answer

DONE


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