Can someone please check if my procedure to solve this problem is correct? I don't have anywhere to check if my answer is right.
Thank you in advance!
Computer manufacturer has launched a new product, the laptop 2017. The daily sales volume of the laptop is assumed to be Poisson distributed with parameter $\lambda=4.$
a) Determine the probability that at most 3 laptops are sold on a given day.
By rule: $P(k)=\cfrac{\lambda^{k}}{k!} e^{-\lambda}$
Hence,
$P(0)=0.01831$
$P(1)= 0.07326$
$P(2) = 0.1465$
$P(3) = 0.1953$
$P(X≤3)=P(0) + P(1) + P(2) +P(3)=0.01831 + 0.07326 + 0.1465 +0.1953 = > 0.43337 $
b) To make a profit with the new product, the company needs a daily turnover of at least 2000 euros. Suppose that the price of the laptop 2017 is equal to 500 euros. What is the probability of making a profit on a given day?
Given the conditions, they need to sell at least 4 laptops on a given day to make a profit. Hence, we want to find $P(X≥4)$ or in other words $1- P(X≤3)$, so from above: $1 - 0.43337 = 0.5666$
It is correct. But un the last row of item a) just use =.