Year 12s: what will be the first way you research your future options? >>

Normal approximation to Poisson

The number of telephone calls per day follows a Poisson distribution with mean 2

Use an approximate normal distribution to estimate the probability of at least 17 calls in in 8 days.

Newton.

Reply 1

Jonny W

X
= Number of calls in 8 days
~ Po(16)

So X ~ N(16, 16) approximately.

P(X >= 17)
~= P(N(16, 16) >= 16.5) . . . . . continuity correction
= 1 - Phi(0.5/4)
= 0.4503

Reply 2

RichE

I got the same answer as the book arrived at as P(N(0,1)>0.125)

Reply 3

Jonny W

Why phi(0.5/4)? Should it not be Phi((1/4)(17.5-16))?

Yes, if we wanted P(X > 17), ie, P(X >= 18). But the question asks for the probability of "at least 17 calls" not "more than 17 calls"

Reply 4

RichE

Newton

Why does P(X>17)=P(X>=18)?

Newton.

P(X>17) = P(X>=18) for a discrete integer-valued variable

Reply 5

Jonny W

X counts the number of calls so it has to be an integer.

P(X > 17)
= P(17 < X < 18) + P(X >= 18)
= 0 + P(X >= 18)
= P(X >= 18)

Reply 6

RichE

Newton

Ok. Please tell Me if my understanding is then correct.

It is asking for the probability of at least 17 calls i. e. P(X>=17).

By continuity correction this transforms to P(X>=17.5).

But since X must take an integer value we are looking for P(X>=18).

Newton.

No we end up working out

P(N(16,16)>16.5)

because each interval

P( n-1/2 < N(16,16) < n+1/2)

is a good approximation for P(X=n)

Reply 7

RichE

Newton

So you minus a half?

Newton.

yes - well as appropriate

if we'd been asked what is P(X <= 17) we'd have worked out P(N(16,16)<17.5)

Reply 8

RichE

Newton

But that is what we are being asked for, hence the 'at least'.

Newton.

no that would be phrased as "at most 17"

just remember

P(X=n) is approximately P(n-1/2 < N(16,16) < n+1/2)

and add up whichever of these you need

Reply 9

RichE

Newton

I mean P(X>=17).

Newton.

yes so you need to add

P(X=17)+P(X=18)+P(X=19)+...

and by the approx I gave you this is

P(16.5<N<17.5)+P(17.5<N<18.5)+...

= P(16.5<N)

Reply 10

yazan_l

Newton

tell Me if my understanding is then correct

why Me?

Reply 11

Newton

yazan_l

why Me?

Obsessive Compulsive Behaviour.

Newton.

Reply 12

confused?

/Ok. Please tell Me if my understanding is then correct.

It is asking for the probability of at least 17 calls i. e. P(X>=17).

By continuity correction this transforms to P(X>=17.5).

But since X must take an integer value we are looking for P(X>=18).

Newton./

If its P(X>=17) it must include the integer 17 therefore with the continuity correction it becomes P(X>=16.5).

Just remember what integers you have to include!