The Student Room Group

Poisson distribution practice question

When pond water is examined under the microscope there are nasty bugs to be seen. The average concentration of these bugs is three per millilitre. Determine the probability that a random sample of 5 millilitres of pond water contains exactly 14 bugs
v)(b) using the normal approximation to the Poisson distribution.
(edited 11 years ago)
Original post by sammy3000
When pond water is examined under the microscope there are nasty bugs to be seen. The average concentration of these bugs is three per millilitre. Determine the probability that a random sample of 5 millilitres of pond water contains exactly 14 bugs
v)(b) using the normal approximation to the Poisson distribution.


If there are 3/ml, then what's the average no. per 5ml?

Hence the Poisson distribution of bugs per 5ml is?
Reply 2
15
Reply 3
I done 14-15/sqrt(15) but i dnt kno what to do after that
Original post by sammy3000
I done 14-15/sqrt(15) but i dnt kno what to do after that


The normal approximation to the poission will use a continuity correction.

So exactly 14, translates to 13.5 to 14.5.

The mean and the variance will both be 15, so your distribution is N(15,15)\sim N(15,15)
Reply 5
i still dont get it, would the answer be 13.5-15/sqrt(15)<14.4-15/sqrt(15)
Original post by sammy3000
i still dont get it, would the answer be 13.5-15/sqrt(15)<14.4-15/sqrt(15)


Those two values would be the standardized normal distribution values, which you'd need to look up in the Normal tables.

Which may be tricky if you've not been taught how to do it. For negative z values, you need to look up the corresponding positive value and subtract the result from 1.
Reply 7
i keep getting 0.1002 but the answer is 0.092
Original post by sammy3000
i keep getting 0.1002 but the answer is 0.092


I get 0.0993, which allowing for 4th digit approximations on the Normal distribution tables is probably the same as yours.

I'm inclined to think the given answer is wrong.
Reply 9
how did you get that
Reply 10
i got -0.387 and -0.129 and from the values in the table i got 0.6480 and 0.5517 I then minus it from 1 which gave 0.4483-0.352=0.0963
Original post by sammy3000
i got -0.387 and -0.129 and from the values in the table i got 0.6480 and 0.5517 I then minus it from 1 which gave 0.4483-0.352=0.0963


I got -0.3873 giving 0.6507

and -0.1291 giving 0.5516 (corrected)

giving 0.0991

Your 0.6480 corresponds to 0.38 (you've only read the table to 2 sig figs.)

Quick Reply

Latest