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 March 11th, 2018, 04:43 PM #1 Member   Joined: Feb 2018 From: Canada Posts: 42 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: 2,324 Thanks: 1233 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$\lambda = 4.5P[X \leq 4] \approx 0.5321$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: 42 Thanks: 2 Quote:  Originally Posted by romsek 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 $\lambda = 4.5$ $P[X \leq 4] \approx 0.5321$

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, $Y=3+1.5*(X-1)$
$10=3+1.5*(X-1) \text{ then } X \approx 5$

$P[X \leq 5]$, 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
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Quote:
 Originally Posted by Shanonhaliwell 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, $Y=3+1.5*(X-1)$ $10=3+1.5*(X-1) \text{ then } X \approx 5$ $P[X \leq 5]$, where $X$ is a Poisson distributed rv with $\lambda = 4.5$ $P[X \leq 5] \approx 0.7029304$
4 shirts costs \$3+4(\$1.5) = \$9 < 10 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
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Quote:
 Originally Posted by romsek 4 shirts costs \$3+4(\$1.5) = \$9 < 10 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 $P[X \leq 5] \approx 0.70293$
I am not sure if the base price \\$3 can get a first shirt ironed. Let me check again with my professor.
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

 Tags distribution, poisson, problem

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