Mixing
Poisson Distribution Probability - Is my solution correct?
$begingroup$
Can someone please check if my procedure to solve this problem is correct? I don't have anywhere to check if my answer is right.
Thank you in advance!
Computer manufacturer has launched a new product, the laptop 2017. The daily sales volume of the laptop is assumed to be Poisson distributed with parameter $lambda=4.$
a) Determine the probability that at most 3 laptops are sold on a given day.
By rule: $P(k)=cfrac{lambda^{k}}{k!} e^{-lambda}$
Hence,
$P(0)=0.01831$
$P(1)= 0.07326$
$P(2) = 0.1465$
$P(3) = 0.1953$
$P(X≤3)=P(0) + P(1) + P(2) +P(3)=0.01831 + 0.07326 + 0.1465 +0.1953 =
> 0.43337 $
b) To make a profit with the new product, the company needs a daily turnover of at least 2000 euros. Suppose that the price of the laptop 2017 is equal to 500 euros. What is the probability of making a profit on a given day?
Given the conditions, they need to sell at least 4 laptops on a given
day to make a profit. Hence, we want to find $P(X≥4)$ or in other words
$1- P(X≤3)$, so from above: $1 - 0.43337 = 0.5666$
probability statistics poisson-distribution
edited Feb 3 at 13:30
Fozoro
1266
asked Feb 3 at 12:45
VRTVRT
957
$endgroup$
add a comment |
$begingroup$
Can someone please check if my procedure to solve this problem is correct? I don't have anywhere to check if my answer is right.
Thank you in advance!
Computer manufacturer has launched a new product, the laptop 2017. The daily sales volume of the laptop is assumed to be Poisson distributed with parameter $lambda=4.$
a) Determine the probability that at most 3 laptops are sold on a given day.
By rule: $P(k)=cfrac{lambda^{k}}{k!} e^{-lambda}$
Hence,
$P(0)=0.01831$
$P(1)= 0.07326$
$P(2) = 0.1465$
$P(3) = 0.1953$
$P(X≤3)=P(0) + P(1) + P(2) +P(3)=0.01831 + 0.07326 + 0.1465 +0.1953 =
> 0.43337 $
b) To make a profit with the new product, the company needs a daily turnover of at least 2000 euros. Suppose that the price of the laptop 2017 is equal to 500 euros. What is the probability of making a profit on a given day?
Given the conditions, they need to sell at least 4 laptops on a given
day to make a profit. Hence, we want to find $P(X≥4)$ or in other words
$1- P(X≤3)$, so from above: $1 - 0.43337 = 0.5666$
probability statistics poisson-distribution
edited Feb 3 at 13:30
Fozoro
1266
asked Feb 3 at 12:45
VRTVRT
957
$endgroup$
add a comment |
$begingroup$
Can someone please check if my procedure to solve this problem is correct? I don't have anywhere to check if my answer is right.
Thank you in advance!
Computer manufacturer has launched a new product, the laptop 2017. The daily sales volume of the laptop is assumed to be Poisson distributed with parameter $lambda=4.$
a) Determine the probability that at most 3 laptops are sold on a given day.
By rule: $P(k)=cfrac{lambda^{k}}{k!} e^{-lambda}$
Hence,
$P(0)=0.01831$
$P(1)= 0.07326$
$P(2) = 0.1465$
$P(3) = 0.1953$
$P(X≤3)=P(0) + P(1) + P(2) +P(3)=0.01831 + 0.07326 + 0.1465 +0.1953 =
> 0.43337 $
b) To make a profit with the new product, the company needs a daily turnover of at least 2000 euros. Suppose that the price of the laptop 2017 is equal to 500 euros. What is the probability of making a profit on a given day?
Given the conditions, they need to sell at least 4 laptops on a given
day to make a profit. Hence, we want to find $P(X≥4)$ or in other words
$1- P(X≤3)$, so from above: $1 - 0.43337 = 0.5666$
probability statistics poisson-distribution
edited Feb 3 at 13:30
Fozoro
1266
asked Feb 3 at 12:45
VRTVRT
957
$endgroup$
Can someone please check if my procedure to solve this problem is correct? I don't have anywhere to check if my answer is right.
Thank you in advance!
Computer manufacturer has launched a new product, the laptop 2017. The daily sales volume of the laptop is assumed to be Poisson distributed with parameter $lambda=4.$
a) Determine the probability that at most 3 laptops are sold on a given day.
By rule: $P(k)=cfrac{lambda^{k}}{k!} e^{-lambda}$
Hence,
$P(0)=0.01831$
$P(1)= 0.07326$
$P(2) = 0.1465$
$P(3) = 0.1953$
$P(X≤3)=P(0) + P(1) + P(2) +P(3)=0.01831 + 0.07326 + 0.1465 +0.1953 =
> 0.43337 $
b) To make a profit with the new product, the company needs a daily turnover of at least 2000 euros. Suppose that the price of the laptop 2017 is equal to 500 euros. What is the probability of making a profit on a given day?
Given the conditions, they need to sell at least 4 laptops on a given
day to make a profit. Hence, we want to find $P(X≥4)$ or in other words
$1- P(X≤3)$, so from above: $1 - 0.43337 = 0.5666$
probability statistics poisson-distribution
probability statistics poisson-distribution
edited Feb 3 at 13:30
Fozoro
1266
asked Feb 3 at 12:45
VRTVRT
957
edited Feb 3 at 13:30
Fozoro
1266
asked Feb 3 at 12:45
VRTVRT
957
edited Feb 3 at 13:30
Fozoro
1266
edited Feb 3 at 13:30
Fozoro
1266
edited Feb 3 at 13:30
Fozoro
1266
1266
asked Feb 3 at 12:45
VRTVRT
957
asked Feb 3 at 12:45
VRTVRT
957
asked Feb 3 at 12:45
VRTVRT
957
957
add a comment |
add a comment |
1 Answer
1
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$begingroup$
It is correct. But un the last row of item a) just use =.
answered Feb 3 at 13:18
Gabriel PalauGabriel Palau
1126
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
It is correct. But un the last row of item a) just use =.
answered Feb 3 at 13:18
Gabriel PalauGabriel Palau
1126
$endgroup$
add a comment |
$begingroup$
It is correct. But un the last row of item a) just use =.
answered Feb 3 at 13:18
Gabriel PalauGabriel Palau
1126
$endgroup$
add a comment |
$begingroup$
It is correct. But un the last row of item a) just use =.
answered Feb 3 at 13:18
Gabriel PalauGabriel Palau
1126
$endgroup$
It is correct. But un the last row of item a) just use =.
answered Feb 3 at 13:18
Gabriel PalauGabriel Palau
1126
answered Feb 3 at 13:18
Gabriel PalauGabriel Palau
1126
answered Feb 3 at 13:18
Gabriel PalauGabriel Palau
1126
answered Feb 3 at 13:18
Gabriel PalauGabriel Palau
1126
1126
add a comment |
add a comment |
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