I think this question is less complicated than I'm making it in my head but I can't figure it out, pls could someone explain?

The average number of misprints on a page is 1.8. What is the most likely

number of misprints in a four sided leaflet?

this is as far as i got and now im stuck:

Average no. of misprints on page = 1.8

Average no. of misprints on four pages = 7.2

X is no. of misprints in a four sided leaflet

X ~ Po (7.2)

The average number of misprints on a page is 1.8. What is the most likely

number of misprints in a four sided leaflet?

this is as far as i got and now im stuck:

Average no. of misprints on page = 1.8

Average no. of misprints on four pages = 7.2

X is no. of misprints in a four sided leaflet

X ~ Po (7.2)

Assuming the number of misprints does indeed follow Poisson distribution...

Unless there are tricks on your calculator, you kind of need to just "guess", especially with the pmf of Poisson not being nice at all to find the maximum of using calculus.

The really dumb but surely correct way is to check manually from P(X=0), P(X=1),... until the probability drops.

In theory, it should be less than 7 (Poisson is skewed right).

Unless there are tricks on your calculator, you kind of need to just "guess", especially with the pmf of Poisson not being nice at all to find the maximum of using calculus.

The really dumb but surely correct way is to check manually from P(X=0), P(X=1),... until the probability drops.

In theory, it should be less than 7 (Poisson is skewed right).

(edited 1 month ago)

Original post by tonyiptony

Assuming the number of misprints does indeed follow Poisson distribution...

Unless there are tricks on your calculator, you kind of need to just "guess", especially with the pmf of Poisson not being nice at all to find the maximum of using calculus.

The really dumb but surely correct way is to check manually from P(X=0), P(X=1),... until the probability drops.

In theory, it should be less than 7 (Poisson is skewed right).

Unless there are tricks on your calculator, you kind of need to just "guess", especially with the pmf of Poisson not being nice at all to find the maximum of using calculus.

The really dumb but surely correct way is to check manually from P(X=0), P(X=1),... until the probability drops.

In theory, it should be less than 7 (Poisson is skewed right).

tysm 🙂 i gave it a shot and got the right answer

Original post by ferdi295

tysm 🙂 i gave it a shot and got the right answer

In this question you're being asked to find the mode (the value most likely to occur) of the distribution, and for a Poisson it is:

If lambda is not an integer, then it is the largest integer less than lambda.

If lambda is an integer, then there are two modes, lamdba, and lambda minus one.

Original post by tonyiptony

Assuming the number of misprints does indeed follow Poisson distribution...

Unless there are tricks on your calculator, you kind of need to just "guess", especially with the pmf of Poisson not being nice at all to find the maximum of using calculus.

The really dumb but surely correct way is to check manually from P(X=0), P(X=1),... until the probability drops.

In theory, it should be less than 7 (Poisson is skewed right).

Unless there are tricks on your calculator, you kind of need to just "guess", especially with the pmf of Poisson not being nice at all to find the maximum of using calculus.

The really dumb but surely correct way is to check manually from P(X=0), P(X=1),... until the probability drops.

In theory, it should be less than 7 (Poisson is skewed right).

It doesn't need calculus:

Poisson is $P(X=k) = e^{-\lambda} \dfrac{\lambda^k}{k!}$

So $\dfrac{P(X=k+1)}{P(X=k)} = \dfrac{\lambda}{k+1}$

So P(X=k+1) is bigger than P(X=k) until $k+1 \ge \lambda$.

(edited 1 month ago)

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