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March 11th, 2018, 03:43 PM   #1
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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 04:04 AM.
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March 12th, 2018, 06:21 PM   #2
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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 07:07 PM.
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March 12th, 2018, 10:51 PM   #3
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Quote:
Originally Posted by romsek View Post
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 10:57 PM.
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March 12th, 2018, 11:19 PM   #4
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Quote:
Originally Posted by Shanonhaliwell View Post
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

$P[X \leq 5] \approx 0.70293$
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March 12th, 2018, 11:43 PM   #5
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Quote:
Originally Posted by romsek View Post
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 12th, 2018 at 11:52 PM.
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March 16th, 2018, 07:53 AM   #6
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Can someone post the every step of this question please ? Thanks
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