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March 11th, 2018, 04:43 PM  #1 
Member Joined: Feb 2018 From: Canada Posts: 37 Thanks: 2 
For a shirt ironing service at the dry cleaners, they charge a base price of \$3 plus an additional \$1.50 per shirt to be ironed. Suppose the number of shirts customers at the dry cleaners require to be ironed follows a Poisson distribution with mean 4.5. What is the probability a random customer in the store will use the dry cleaning service and spend no more than \$10 on the dry cleaning services? I am having a problem with this question, for mean 4.5 shirts, I think the price would be \$3 + \$1.5*3.5 = \$8.25. If X be the number of shirts customers at the dry cleaners require to be ironed, then Y be an amount of money customer have to pay for the services. Then Y = \$3 + \$1.5*(X  1). Since X follows a Poisson distribution with mean 4.5, Y would follow a Poisson distribution with $\lambda$ = 8.25. Am I right? Last edited by skipjack; March 20th, 2018 at 05:04 AM. 
March 12th, 2018, 07:21 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,174 Thanks: 1143 
I would convert things the other way. Spending no more than \$10 is the same as having no more than 4 shirts ironed. So we evaluate $P[X \leq 4]$, where $X$ is a Poisson distributed rv with $\lambda = 4.5$ $P[X \leq 4] \approx 0.5321$ Last edited by romsek; March 12th, 2018 at 08:07 PM. 
March 12th, 2018, 11:51 PM  #3  
Member Joined: Feb 2018 From: Canada Posts: 37 Thanks: 2  Quote:
How would you come up with 4 shirts. If X is the number of shirts customers at the dry cleaners require to be ironed then Y be a amount of money customer have to pay for the services. So, $Y=3+1.5*(X1)$ $10=3+1.5*(X1) \text{ then } X \approx 5$ $P[X \leq 5]$, where $X$ is a Poisson distributed rv with $\lambda = 4.5$ $P[X \leq 5] \approx 0.7029304$ Last edited by Shanonhaliwell; March 12th, 2018 at 11:57 PM.  
March 13th, 2018, 12:19 AM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 2,174 Thanks: 1143  Quote:
5 shirts cost \$3 + 5(\$1.5) = \$10.5 > 10 oh, are you saying the base \$3 gets you 1 shirt ironed? In that case you're right. You can get 5 shirts ironed < \$10 $P[X \leq 5] \approx 0.70293$  
March 13th, 2018, 12:43 AM  #5  
Member Joined: Feb 2018 From: Canada Posts: 37 Thanks: 2  Quote:
Thanks. Last edited by Shanonhaliwell; March 13th, 2018 at 12:52 AM.  
March 16th, 2018, 08:53 AM  #6 
Newbie Joined: Mar 2018 From: Canada Posts: 1 Thanks: 0 
Can someone post the every step of this question please ? Thanks


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distribution, poisson, problem 
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