Normal Distrubition Question - How many wires will meet specifications?












1












$begingroup$



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







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$endgroup$








  • 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
















1












$begingroup$



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







share|cite|improve this question









$endgroup$








  • 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














1












1








1





$begingroup$



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







share|cite|improve this question









$endgroup$





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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
1






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oldest

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1












$begingroup$

Using binomial distribution to solve:



$n=4$,
$r=4$,
$p=0.954$



$p(4)= [4!/(0!4!)] [0.954^4] [(1-0.954)^{4-4}]
= [4!/4!] [0.954^4] [0.046^0]
= 0.954^4
= 0.83$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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/…
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    – Sauhard Sharma
    Jan 3 at 6:34











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Using binomial distribution to solve:



$n=4$,
$r=4$,
$p=0.954$



$p(4)= [4!/(0!4!)] [0.954^4] [(1-0.954)^{4-4}]
= [4!/4!] [0.954^4] [0.046^0]
= 0.954^4
= 0.83$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












$begingroup$

Using binomial distribution to solve:



$n=4$,
$r=4$,
$p=0.954$



$p(4)= [4!/(0!4!)] [0.954^4] [(1-0.954)^{4-4}]
= [4!/4!] [0.954^4] [0.046^0]
= 0.954^4
= 0.83$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

Using binomial distribution to solve:



$n=4$,
$r=4$,
$p=0.954$



$p(4)= [4!/(0!4!)] [0.954^4] [(1-0.954)^{4-4}]
= [4!/4!] [0.954^4] [0.046^0]
= 0.954^4
= 0.83$






share|cite|improve this answer











$endgroup$



Using binomial distribution to solve:



$n=4$,
$r=4$,
$p=0.954$



$p(4)= [4!/(0!4!)] [0.954^4] [(1-0.954)^{4-4}]
= [4!/4!] [0.954^4] [0.046^0]
= 0.954^4
= 0.83$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 3 at 7:10









Sauhard Sharma

953318




953318










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


















  • $begingroup$
    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/…
$endgroup$
– Sauhard Sharma
Jan 3 at 6:34


















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