Based on recent records, the manager of a car painting center has determined the following probability distribution for the number of customers per day. If the center has the capacity to serve two customers per day,
$$\begin{array}{ccccccc} x&0&1&2&3&4&5\\ f(x)&0.06&0.17&0.24&0.21&0.12&0.2\\ \end{array}$$
By how much must the capacity be increased so the probability of turning a customer away is less than $0.10$?
I have no idea how to answer this question. Can you please help? Thanks!
The question is not really clear about what a capacity is and what it is now (so that it must be increased). Taking my interpretation of the question, if you have capacity to take care of at most $4$ customers, you turn a customer away when $5$ of them shows up, which happens with probability $0.2$. When your capacity is $3$, you turn away customers away with probability $0.12+0.2$. Hence you need capacity to service $5$ customers.
I agree with @Jan that there isn't enough information to make sense of this problem.
If the number of customers $X$ in a day is Poisson, then $P(X = 0) \approx .06$ implies the rate per day is $\lambda \approx 2.8,$ and the distribution $\mathsf{Pois}(\lambda = 2.8)$ is not a bad fit to the probabilities in the image. That would imply $P(X > 5) \approx 0.065 < 0.10.$
There must be capacity for 5 per day or 5 wouldn't show up in the 'recent' records. So everything should be fine without increasing capacity.
Another try was to use $\lambda \approx \bar X = 1.86,$ but $\mathsf{Pois}(\lambda = 1.86)$ doesn't fit the 'recent data' so well. And it would imply an even smaller $P(X > 5),$ hence even less urgency to expand. A binomial distribution with $n = 5$ and $\mu \approx 1.86$ looks nothing like the distribution given.
So my well of "plausible speculation" has run dry. It is not clear what is meant by 'capacity' or under what circumstances customers are 'turned away'. As always, it might have helped to know what probability topics you have been studying lately.