Question-13
A shopkeeper sells mobile phones. The demand for mobile phone follows a Poisson distribution with mean 4.6 per week. The shopkeeper has 5 mobile phones in his shop at the beginning of a week. Find the probability that this will not be enough to satisfy the demand for mobile phones in that week. Enter your answer correct up to two decimals accuracy.
Define random variable and find the condition for “not satisfying the demand” and apply the PMF formula of Poisson Distribution.
0.32
Let \(X\) be a random variable which denotes the demand for mobile phones. Shopkeeper has 5 phones in his shop at the beginning of the week. So, this will not be enough to satisfy the demand for mobile phones in that week when \(X > 5\). From the question, it is given that \(\lambda = 4.6\). Thus, the required probability, \(P(X > 5) = 1-P(X \leq 5)\)
\(= 1- \left(\dfrac{e^{-4.6}\times 4.6^0}{0!} + \dfrac{e^{-4.6}\times 4.6^1}{1!} + \dfrac{e^{-4.6}\times 4.6^2}{2!} + \dfrac{e^{-4.6}\times 4.6^3}{3!} + \dfrac{e^{-4.6}\times 4.6^4}{4!} + \dfrac{e^{-4.6}\times 4.6^5}{5!} \right )\)