Probability “at-least” question with large numbers
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The mean monthly rent of all 1000 one-bedroom condos for rent is
1560 dollars, with a standard deviation of 224 dollars. 67.7% of these condos have
monthly rents lower than 1600 dollars
If 64 one-bedroom condos for rent are randomly selected from 1000
one-bedroom, what is the probability that at least 25 of the sampled
condos cost less than 1600 dollars per month to rent
I'm not entirely sure that I'm approaching this question in the right way. I've checked that it follows an approximately normal distribution (which it does) and I've calculated the standard deviation of the sample and found that the $1600 value is between 1 and 2 standard deviations of the mean (mean = 1560, mean + 1SD =1588, mean +2SD = 1616)
From there though, I'm not sure how to use the 68-95-99.7 rule to determine the probability (if its even possible, I honestly don't know if I'm on the right track with answering this question)
Thanks for any help anyone can give!
probability statistics
asked 2 hours ago
clum26
111
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clum26 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
up vote
2
down vote
favorite
The mean monthly rent of all 1000 one-bedroom condos for rent is
1560 dollars, with a standard deviation of 224 dollars. 67.7% of these condos have
monthly rents lower than 1600 dollars
If 64 one-bedroom condos for rent are randomly selected from 1000
one-bedroom, what is the probability that at least 25 of the sampled
condos cost less than 1600 dollars per month to rent
I'm not entirely sure that I'm approaching this question in the right way. I've checked that it follows an approximately normal distribution (which it does) and I've calculated the standard deviation of the sample and found that the $1600 value is between 1 and 2 standard deviations of the mean (mean = 1560, mean + 1SD =1588, mean +2SD = 1616)
From there though, I'm not sure how to use the 68-95-99.7 rule to determine the probability (if its even possible, I honestly don't know if I'm on the right track with answering this question)
Thanks for any help anyone can give!
probability statistics
asked 2 hours ago
clum26
111
New contributor
clum26 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If you approximate each room as independent, and a room can be cheap or expensive, the number of cheap bedrooms out of 64 is a binomial random variable. Did you learn about those?
– Mark
2 hours ago
I have learned about those. Would I be finding the probability then of when z > (25-np)/sqrt(nq*p)?
– clum26
1 hour ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The mean monthly rent of all 1000 one-bedroom condos for rent is
1560 dollars, with a standard deviation of 224 dollars. 67.7% of these condos have
monthly rents lower than 1600 dollars
If 64 one-bedroom condos for rent are randomly selected from 1000
one-bedroom, what is the probability that at least 25 of the sampled
condos cost less than 1600 dollars per month to rent
I'm not entirely sure that I'm approaching this question in the right way. I've checked that it follows an approximately normal distribution (which it does) and I've calculated the standard deviation of the sample and found that the $1600 value is between 1 and 2 standard deviations of the mean (mean = 1560, mean + 1SD =1588, mean +2SD = 1616)
From there though, I'm not sure how to use the 68-95-99.7 rule to determine the probability (if its even possible, I honestly don't know if I'm on the right track with answering this question)
Thanks for any help anyone can give!
probability statistics
asked 2 hours ago
clum26
111
New contributor
clum26 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The mean monthly rent of all 1000 one-bedroom condos for rent is
1560 dollars, with a standard deviation of 224 dollars. 67.7% of these condos have
monthly rents lower than 1600 dollars
If 64 one-bedroom condos for rent are randomly selected from 1000
one-bedroom, what is the probability that at least 25 of the sampled
condos cost less than 1600 dollars per month to rent
I'm not entirely sure that I'm approaching this question in the right way. I've checked that it follows an approximately normal distribution (which it does) and I've calculated the standard deviation of the sample and found that the $1600 value is between 1 and 2 standard deviations of the mean (mean = 1560, mean + 1SD =1588, mean +2SD = 1616)
From there though, I'm not sure how to use the 68-95-99.7 rule to determine the probability (if its even possible, I honestly don't know if I'm on the right track with answering this question)
Thanks for any help anyone can give!
probability statistics
probability statistics
asked 2 hours ago
clum26
111
New contributor
clum26 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
clum26
111
New contributor
clum26 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
clum26
111
New contributor
clum26 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
clum26
111
asked 2 hours ago
clum26
111
111
New contributor
clum26 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
clum26 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
clum26 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If you approximate each room as independent, and a room can be cheap or expensive, the number of cheap bedrooms out of 64 is a binomial random variable. Did you learn about those?
– Mark
2 hours ago
I have learned about those. Would I be finding the probability then of when z > (25-np)/sqrt(nq*p)?
– clum26
1 hour ago
add a comment |
If you approximate each room as independent, and a room can be cheap or expensive, the number of cheap bedrooms out of 64 is a binomial random variable. Did you learn about those?
– Mark
2 hours ago
I have learned about those. Would I be finding the probability then of when z > (25-np)/sqrt(nq*p)?
– clum26
1 hour ago
If you approximate each room as independent, and a room can be cheap or expensive, the number of cheap bedrooms out of 64 is a binomial random variable. Did you learn about those?
– Mark
2 hours ago
If you approximate each room as independent, and a room can be cheap or expensive, the number of cheap bedrooms out of 64 is a binomial random variable. Did you learn about those?
– Mark
2 hours ago
I have learned about those. Would I be finding the probability then of when z > (25-np)/sqrt(nq*p)?
– clum26
1 hour ago
I have learned about those. Would I be finding the probability then of when z > (25-np)/sqrt(nq*p)?
– clum26
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
We know that the probability of having a condo be cheaper than $$1600$ is $66.7%$, and for it being more expensive it is $33.3%$. The probability of getting $k$ condos costing less than $$1600$ is $binom{64}kcdot0.333^{64-k}cdot0.667^k$ because the $0.333^{64-k}$ is to show the probability of getting $64-k$ condos costing over $$1600$, $0.667^k$ is used to find the probability of getting $k$ condos that cost under $$1600$, and the $binom{64}k$ is to find the number of possibilities of which these conditions are met (i.e. HHHHHTTTTT and TTTTTHHHHH all have 5 heads, so they would both be counted if we were finding the number of possibilities of getting 5 heads in 10 coin flips). Using Wolfram Alpha, we can show the probability of getting $25$ or over condos costing less than $$1600$ is $0.9999983$.
Source: https://www.wolframalpha.com/input/?i=Sum(binom(64,k)*0.333%5E(64-k)*0.667%5Ek)+from+25+to+64
answered 2 mins ago
Kyky
387113
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
We know that the probability of having a condo be cheaper than $$1600$ is $66.7%$, and for it being more expensive it is $33.3%$. The probability of getting $k$ condos costing less than $$1600$ is $binom{64}kcdot0.333^{64-k}cdot0.667^k$ because the $0.333^{64-k}$ is to show the probability of getting $64-k$ condos costing over $$1600$, $0.667^k$ is used to find the probability of getting $k$ condos that cost under $$1600$, and the $binom{64}k$ is to find the number of possibilities of which these conditions are met (i.e. HHHHHTTTTT and TTTTTHHHHH all have 5 heads, so they would both be counted if we were finding the number of possibilities of getting 5 heads in 10 coin flips). Using Wolfram Alpha, we can show the probability of getting $25$ or over condos costing less than $$1600$ is $0.9999983$.
Source: https://www.wolframalpha.com/input/?i=Sum(binom(64,k)*0.333%5E(64-k)*0.667%5Ek)+from+25+to+64
answered 2 mins ago
Kyky
387113
add a comment |
up vote
0
down vote
We know that the probability of having a condo be cheaper than $$1600$ is $66.7%$, and for it being more expensive it is $33.3%$. The probability of getting $k$ condos costing less than $$1600$ is $binom{64}kcdot0.333^{64-k}cdot0.667^k$ because the $0.333^{64-k}$ is to show the probability of getting $64-k$ condos costing over $$1600$, $0.667^k$ is used to find the probability of getting $k$ condos that cost under $$1600$, and the $binom{64}k$ is to find the number of possibilities of which these conditions are met (i.e. HHHHHTTTTT and TTTTTHHHHH all have 5 heads, so they would both be counted if we were finding the number of possibilities of getting 5 heads in 10 coin flips). Using Wolfram Alpha, we can show the probability of getting $25$ or over condos costing less than $$1600$ is $0.9999983$.
Source: https://www.wolframalpha.com/input/?i=Sum(binom(64,k)*0.333%5E(64-k)*0.667%5Ek)+from+25+to+64
answered 2 mins ago
Kyky
387113
add a comment |
up vote
0
down vote
up vote
0
down vote
We know that the probability of having a condo be cheaper than $$1600$ is $66.7%$, and for it being more expensive it is $33.3%$. The probability of getting $k$ condos costing less than $$1600$ is $binom{64}kcdot0.333^{64-k}cdot0.667^k$ because the $0.333^{64-k}$ is to show the probability of getting $64-k$ condos costing over $$1600$, $0.667^k$ is used to find the probability of getting $k$ condos that cost under $$1600$, and the $binom{64}k$ is to find the number of possibilities of which these conditions are met (i.e. HHHHHTTTTT and TTTTTHHHHH all have 5 heads, so they would both be counted if we were finding the number of possibilities of getting 5 heads in 10 coin flips). Using Wolfram Alpha, we can show the probability of getting $25$ or over condos costing less than $$1600$ is $0.9999983$.
Source: https://www.wolframalpha.com/input/?i=Sum(binom(64,k)*0.333%5E(64-k)*0.667%5Ek)+from+25+to+64
answered 2 mins ago
Kyky
387113
We know that the probability of having a condo be cheaper than $$1600$ is $66.7%$, and for it being more expensive it is $33.3%$. The probability of getting $k$ condos costing less than $$1600$ is $binom{64}kcdot0.333^{64-k}cdot0.667^k$ because the $0.333^{64-k}$ is to show the probability of getting $64-k$ condos costing over $$1600$, $0.667^k$ is used to find the probability of getting $k$ condos that cost under $$1600$, and the $binom{64}k$ is to find the number of possibilities of which these conditions are met (i.e. HHHHHTTTTT and TTTTTHHHHH all have 5 heads, so they would both be counted if we were finding the number of possibilities of getting 5 heads in 10 coin flips). Using Wolfram Alpha, we can show the probability of getting $25$ or over condos costing less than $$1600$ is $0.9999983$.
Source: https://www.wolframalpha.com/input/?i=Sum(binom(64,k)*0.333%5E(64-k)*0.667%5Ek)+from+25+to+64
answered 2 mins ago
Kyky
387113
answered 2 mins ago
Kyky
387113
answered 2 mins ago
Kyky
387113
answered 2 mins ago
Kyky
387113
387113
add a comment |
add a comment |
clum26 is a new contributor. Be nice, and check out our Code of Conduct.
clum26 is a new contributor. Be nice, and check out our Code of Conduct.
clum26 is a new contributor. Be nice, and check out our Code of Conduct.
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If you approximate each room as independent, and a room can be cheap or expensive, the number of cheap bedrooms out of 64 is a binomial random variable. Did you learn about those?
– Mark
2 hours ago
I have learned about those. Would I be finding the probability then of when z > (25-np)/sqrt(nq*p)?
– clum26
1 hour ago