Alice has noticed that the probability of any browsing customers eventually buying sth in the store is 18%, assume that the decisions of buying are independent of one another. Thus, what is the expected number of browsing customers who would buy sth in a group of 50 people ?
Asked
Active
Viewed 37 times
0
-
This sounds like a binomial distribution problem to me. Can you see why? What would $n$ be? What would $p$ be? What do you know about the expected number of a binomial distribution? – JMoravitz Jun 17 '17 at 14:31
-
What would you think the answer was? – lulu Jun 17 '17 at 14:31
1 Answers
1
We note that $P(X_i=1) = 0.18$, where $X_i$ is an indicator that the customer $i$ buys smh, therefore $\mathbb E[X_i]=0.18$ for every $i$.
Therefore, We can write $S = \sum_{i=1}^{50} X_i$.
From the linearity of the expected value we get $\mathbb E[S] = \sum_{i=1}^{50} \mathbb E[X_i] = 50*\mathbb E[X_i] = 50*0.18 = 9 $.
Mickey
- 937