Normal Distrubition Question - How many wires will meet specifications?
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Wires manufactured for use in a certain computer system are specified to have resistances between 0.12 ohm and 0.14 ohm, the actual measured resistances of the wires produced by company A have a normal probability distribution with a mean of 0.13 ohm and a standard deviation of 0.005 ohm.
(i) What is the probability that a randomly selected wire from company A's production will meet the specifications?
(ii) If 4 such wires are used in the system and are selected from company A, what is the probability that all 4 will meet specifications?
(i) $$P(0.12≤X≤0.14)$$
$$P(frac{0.12-0.13}{0.005}≤Z≤frac{0.14-0.13}{0.005})$$
$$P(-2≤Z≤2)$$
$$P(Z≤2) - P(Z≤-2)$$
$$0.9772 - (1-0.9972)$$
$$0.9544$$
(ii) I'm stuck on this part, do you simply put $0.9544^4 = 0.8297?$
normal-distribution
asked May 23 '16 at 11:29
ModriscoModrisco
182110
$endgroup$
add a comment |
$begingroup$
Wires manufactured for use in a certain computer system are specified to have resistances between 0.12 ohm and 0.14 ohm, the actual measured resistances of the wires produced by company A have a normal probability distribution with a mean of 0.13 ohm and a standard deviation of 0.005 ohm.
(i) What is the probability that a randomly selected wire from company A's production will meet the specifications?
(ii) If 4 such wires are used in the system and are selected from company A, what is the probability that all 4 will meet specifications?
(i) $$P(0.12≤X≤0.14)$$
$$P(frac{0.12-0.13}{0.005}≤Z≤frac{0.14-0.13}{0.005})$$
$$P(-2≤Z≤2)$$
$$P(Z≤2) - P(Z≤-2)$$
$$0.9772 - (1-0.9972)$$
$$0.9544$$
(ii) I'm stuck on this part, do you simply put $0.9544^4 = 0.8297?$
normal-distribution
asked May 23 '16 at 11:29
ModriscoModrisco
182110
$endgroup$
1
$begingroup$
(ii) Yes, if we assume independence among the wires. Another way to see that your answer is correct is that the number of satisfactory wires follows a binomial distribution. (i) I didn't check the numbers but the process looks good.
$endgroup$
– Em.
May 23 '16 at 11:36
$begingroup$
Interesting use of binomial but wouldn't I need to know the total number of wires to do that? if everything I did is correct that's great
$endgroup$
– Modrisco
May 23 '16 at 11:39
1
$begingroup$
The total is 4, they told that. In other words, four wires is four trials where success is defined to be a wire that meets the specified requirements.
$endgroup$
– Em.
May 23 '16 at 11:41
add a comment |
$begingroup$
Wires manufactured for use in a certain computer system are specified to have resistances between 0.12 ohm and 0.14 ohm, the actual measured resistances of the wires produced by company A have a normal probability distribution with a mean of 0.13 ohm and a standard deviation of 0.005 ohm.
(i) What is the probability that a randomly selected wire from company A's production will meet the specifications?
(ii) If 4 such wires are used in the system and are selected from company A, what is the probability that all 4 will meet specifications?
(i) $$P(0.12≤X≤0.14)$$
$$P(frac{0.12-0.13}{0.005}≤Z≤frac{0.14-0.13}{0.005})$$
$$P(-2≤Z≤2)$$
$$P(Z≤2) - P(Z≤-2)$$
$$0.9772 - (1-0.9972)$$
$$0.9544$$
(ii) I'm stuck on this part, do you simply put $0.9544^4 = 0.8297?$
normal-distribution
asked May 23 '16 at 11:29
ModriscoModrisco
182110
$endgroup$
Wires manufactured for use in a certain computer system are specified to have resistances between 0.12 ohm and 0.14 ohm, the actual measured resistances of the wires produced by company A have a normal probability distribution with a mean of 0.13 ohm and a standard deviation of 0.005 ohm.
(i) What is the probability that a randomly selected wire from company A's production will meet the specifications?
(ii) If 4 such wires are used in the system and are selected from company A, what is the probability that all 4 will meet specifications?
(i) $$P(0.12≤X≤0.14)$$
$$P(frac{0.12-0.13}{0.005}≤Z≤frac{0.14-0.13}{0.005})$$
$$P(-2≤Z≤2)$$
$$P(Z≤2) - P(Z≤-2)$$
$$0.9772 - (1-0.9972)$$
$$0.9544$$
(ii) I'm stuck on this part, do you simply put $0.9544^4 = 0.8297?$
normal-distribution
normal-distribution
asked May 23 '16 at 11:29
ModriscoModrisco
182110
asked May 23 '16 at 11:29
ModriscoModrisco
182110
asked May 23 '16 at 11:29
ModriscoModrisco
182110
asked May 23 '16 at 11:29
ModriscoModrisco
182110
asked May 23 '16 at 11:29
ModriscoModrisco
182110
182110
1
$begingroup$
(ii) Yes, if we assume independence among the wires. Another way to see that your answer is correct is that the number of satisfactory wires follows a binomial distribution. (i) I didn't check the numbers but the process looks good.
$endgroup$
– Em.
May 23 '16 at 11:36
$begingroup$
Interesting use of binomial but wouldn't I need to know the total number of wires to do that? if everything I did is correct that's great
$endgroup$
– Modrisco
May 23 '16 at 11:39
1
$begingroup$
The total is 4, they told that. In other words, four wires is four trials where success is defined to be a wire that meets the specified requirements.
$endgroup$
– Em.
May 23 '16 at 11:41
add a comment |
1
$begingroup$
(ii) Yes, if we assume independence among the wires. Another way to see that your answer is correct is that the number of satisfactory wires follows a binomial distribution. (i) I didn't check the numbers but the process looks good.
$endgroup$
– Em.
May 23 '16 at 11:36
$begingroup$
Interesting use of binomial but wouldn't I need to know the total number of wires to do that? if everything I did is correct that's great
$endgroup$
– Modrisco
May 23 '16 at 11:39
1
$begingroup$
The total is 4, they told that. In other words, four wires is four trials where success is defined to be a wire that meets the specified requirements.
$endgroup$
– Em.
May 23 '16 at 11:41
1
1
$begingroup$
(ii) Yes, if we assume independence among the wires. Another way to see that your answer is correct is that the number of satisfactory wires follows a binomial distribution. (i) I didn't check the numbers but the process looks good.
$endgroup$
– Em.
May 23 '16 at 11:36
$begingroup$
(ii) Yes, if we assume independence among the wires. Another way to see that your answer is correct is that the number of satisfactory wires follows a binomial distribution. (i) I didn't check the numbers but the process looks good.
$endgroup$
– Em.
May 23 '16 at 11:36
$begingroup$
Interesting use of binomial but wouldn't I need to know the total number of wires to do that? if everything I did is correct that's great
$endgroup$
– Modrisco
May 23 '16 at 11:39
$begingroup$
Interesting use of binomial but wouldn't I need to know the total number of wires to do that? if everything I did is correct that's great
$endgroup$
– Modrisco
May 23 '16 at 11:39
1
1
$begingroup$
The total is 4, they told that. In other words, four wires is four trials where success is defined to be a wire that meets the specified requirements.
$endgroup$
– Em.
May 23 '16 at 11:41
$begingroup$
The total is 4, they told that. In other words, four wires is four trials where success is defined to be a wire that meets the specified requirements.
$endgroup$
– Em.
May 23 '16 at 11:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Using binomial distribution to solve:
$n=4$,
$r=4$,
$p=0.954$
$p(4)= [4!/(0!4!)] [0.954^4] [(1-0.954)^{4-4}]
= [4!/4!] [0.954^4] [0.046^0]
= 0.954^4
= 0.83$
edited Jan 3 at 7:10
Sauhard Sharma
953318
answered Jan 3 at 5:51
benben
112
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Hi ! Welcome to MSE. Please use Mathjax to post equations as that makes it easier to actually see the information. Here's a reference link to Mathjax math.meta.stackexchange.com/questions/5020/…
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– Sauhard Sharma
Jan 3 at 6:34
add a comment |
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1 Answer
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$begingroup$
Using binomial distribution to solve:
$n=4$,
$r=4$,
$p=0.954$
$p(4)= [4!/(0!4!)] [0.954^4] [(1-0.954)^{4-4}]
= [4!/4!] [0.954^4] [0.046^0]
= 0.954^4
= 0.83$
edited Jan 3 at 7:10
Sauhard Sharma
953318
answered Jan 3 at 5:51
benben
112
$endgroup$
$begingroup$
Hi ! Welcome to MSE. Please use Mathjax to post equations as that makes it easier to actually see the information. Here's a reference link to Mathjax math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Sauhard Sharma
Jan 3 at 6:34
add a comment |
$begingroup$
Using binomial distribution to solve:
$n=4$,
$r=4$,
$p=0.954$
$p(4)= [4!/(0!4!)] [0.954^4] [(1-0.954)^{4-4}]
= [4!/4!] [0.954^4] [0.046^0]
= 0.954^4
= 0.83$
edited Jan 3 at 7:10
Sauhard Sharma
953318
answered Jan 3 at 5:51
benben
112
$endgroup$
$begingroup$
Hi ! Welcome to MSE. Please use Mathjax to post equations as that makes it easier to actually see the information. Here's a reference link to Mathjax math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Sauhard Sharma
Jan 3 at 6:34
add a comment |
$begingroup$
Using binomial distribution to solve:
$n=4$,
$r=4$,
$p=0.954$
$p(4)= [4!/(0!4!)] [0.954^4] [(1-0.954)^{4-4}]
= [4!/4!] [0.954^4] [0.046^0]
= 0.954^4
= 0.83$
edited Jan 3 at 7:10
Sauhard Sharma
953318
answered Jan 3 at 5:51
benben
112
$endgroup$
Using binomial distribution to solve:
$n=4$,
$r=4$,
$p=0.954$
$p(4)= [4!/(0!4!)] [0.954^4] [(1-0.954)^{4-4}]
= [4!/4!] [0.954^4] [0.046^0]
= 0.954^4
= 0.83$
edited Jan 3 at 7:10
Sauhard Sharma
953318
answered Jan 3 at 5:51
benben
112
edited Jan 3 at 7:10
Sauhard Sharma
953318
edited Jan 3 at 7:10
Sauhard Sharma
953318
edited Jan 3 at 7:10
Sauhard Sharma
953318
953318
answered Jan 3 at 5:51
benben
112
answered Jan 3 at 5:51
benben
112
answered Jan 3 at 5:51
benben
112
112
$begingroup$
Hi ! Welcome to MSE. Please use Mathjax to post equations as that makes it easier to actually see the information. Here's a reference link to Mathjax math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Sauhard Sharma
Jan 3 at 6:34
add a comment |
$begingroup$
Hi ! Welcome to MSE. Please use Mathjax to post equations as that makes it easier to actually see the information. Here's a reference link to Mathjax math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Sauhard Sharma
Jan 3 at 6:34
$begingroup$
Hi ! Welcome to MSE. Please use Mathjax to post equations as that makes it easier to actually see the information. Here's a reference link to Mathjax math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Sauhard Sharma
Jan 3 at 6:34
$begingroup$
Hi ! Welcome to MSE. Please use Mathjax to post equations as that makes it easier to actually see the information. Here's a reference link to Mathjax math.meta.stackexchange.com/questions/5020/…
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– Sauhard Sharma
Jan 3 at 6:34
add a comment |
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(ii) Yes, if we assume independence among the wires. Another way to see that your answer is correct is that the number of satisfactory wires follows a binomial distribution. (i) I didn't check the numbers but the process looks good.
$endgroup$
– Em.
May 23 '16 at 11:36
$begingroup$
Interesting use of binomial but wouldn't I need to know the total number of wires to do that? if everything I did is correct that's great
$endgroup$
– Modrisco
May 23 '16 at 11:39
1
$begingroup$
The total is 4, they told that. In other words, four wires is four trials where success is defined to be a wire that meets the specified requirements.
$endgroup$
– Em.
May 23 '16 at 11:41