Question

Question closed.

Answer 1

Do you mean: "with mean 0.8" and not "0.8n"?

Answer 2

0.8n. Correct.

Answer 3

o.O

Answer 4

For some reason, I do not see your question any more. :(

If mean = 0.8n for n samples, that would imply that the mean increases as sample size increases. That contradicts the fat that the mean \(\lambda\) of a Poisson distribution should be a constant. Could you check the question again?

Answer 5

If we fix \(n=50\), then we fix \(\lambda=0.8(50)=40\).

You want to find \(P(X>45)\), which is given by
\[\large\sum_{x=45}^{50}\frac{e^{-40}40^x}{x!}\]

Answer 6

For those just passing by, the question was:

"A biologist gathers leaves of a certain plant in order to collect insects of a particular type. From past experience she knows that the distribution of the number of insects on n leaves may be modeled by a Poisson distribution with mean 0.8n. use a distributional approximation to find, to two decimal places, the probability that the total number of insects on the next 50 leaves to be examined will exceed 45."

Answer 7

Slight correction:
\[\large P(X>45)=\sum_{x=\color{red}{46}}^{50}\frac{e^{-40}40^x}{x!}\]

Answer 8

If you need to approximate using the normal distribution, you can use \(N(\lambda, \sqrt\lambda)\) and calculate the difference of the two tails (45 and 50). The value will be about 14.5% too high, since \(\lambda=40\) is relatively small for a good approximation using the normal distribution.

Answer 9

The answer is supposed to be 0.19

Answer 10

Using the normal distribution approximation?

Answer 11

Yup

Answer 12

0.19 is correct.
In fact, the question asks for >45 bugs, so the upper bound for summation should be \(\infty\), which brings P(x>45) to 0.19035 approx.
The corresponding normal approximation (using the right tail of 45) would give 0.192 if you apply the continuity correction.