Resistors to be used in a circuit have average resistance 200 ohms and standard deviation 10 ohms…












2












$begingroup$



Resistors to be used in a circuit have average resistance 200 ohms and standard deviation 10 ohms. Suppose 25 of these resistors are randomly selected to be used in a circuit.
a) What is the probability that the average resistance for the 25 resistors is between 199 and 202 ohms?
b) Find the probability that the total resistance does not exceed 5100 ohms.




I keep getting a negative result in part a. This follows the central limit theorem. I calculated my interval to be:



$$P((199-200)/10/(sqrt{25}) < (X-200)/10/(sqrt{25}) < (202-200)/10/(sqrt{25})$$



$$P(-0.5 < Z < 1)$$



Wouldn't this be $$Phi(1)-[1-Phi(0.5)]?$$










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$endgroup$












  • $begingroup$
    $0.8413447 - (1-0.6914625)= 0.8413447-0.3085375=0.5328072gt 0$
    $endgroup$
    – Henry
    Jul 7 '16 at 7:32










  • $begingroup$
    @Henry's comment - adding some detail: average will be 200 with std dev of 2, so 199 is 1/2 std dev below the mean and 202 is 1 std dev above. 1/2 std dev is 19.15% (from a table) and 1 std dev is 34.13% (using a table for the .13% part).
    $endgroup$
    – stretch
    Jul 8 '17 at 13:37
















2












$begingroup$



Resistors to be used in a circuit have average resistance 200 ohms and standard deviation 10 ohms. Suppose 25 of these resistors are randomly selected to be used in a circuit.
a) What is the probability that the average resistance for the 25 resistors is between 199 and 202 ohms?
b) Find the probability that the total resistance does not exceed 5100 ohms.




I keep getting a negative result in part a. This follows the central limit theorem. I calculated my interval to be:



$$P((199-200)/10/(sqrt{25}) < (X-200)/10/(sqrt{25}) < (202-200)/10/(sqrt{25})$$



$$P(-0.5 < Z < 1)$$



Wouldn't this be $$Phi(1)-[1-Phi(0.5)]?$$










share|cite|improve this question











$endgroup$












  • $begingroup$
    $0.8413447 - (1-0.6914625)= 0.8413447-0.3085375=0.5328072gt 0$
    $endgroup$
    – Henry
    Jul 7 '16 at 7:32










  • $begingroup$
    @Henry's comment - adding some detail: average will be 200 with std dev of 2, so 199 is 1/2 std dev below the mean and 202 is 1 std dev above. 1/2 std dev is 19.15% (from a table) and 1 std dev is 34.13% (using a table for the .13% part).
    $endgroup$
    – stretch
    Jul 8 '17 at 13:37














2












2








2





$begingroup$



Resistors to be used in a circuit have average resistance 200 ohms and standard deviation 10 ohms. Suppose 25 of these resistors are randomly selected to be used in a circuit.
a) What is the probability that the average resistance for the 25 resistors is between 199 and 202 ohms?
b) Find the probability that the total resistance does not exceed 5100 ohms.




I keep getting a negative result in part a. This follows the central limit theorem. I calculated my interval to be:



$$P((199-200)/10/(sqrt{25}) < (X-200)/10/(sqrt{25}) < (202-200)/10/(sqrt{25})$$



$$P(-0.5 < Z < 1)$$



Wouldn't this be $$Phi(1)-[1-Phi(0.5)]?$$










share|cite|improve this question











$endgroup$





Resistors to be used in a circuit have average resistance 200 ohms and standard deviation 10 ohms. Suppose 25 of these resistors are randomly selected to be used in a circuit.
a) What is the probability that the average resistance for the 25 resistors is between 199 and 202 ohms?
b) Find the probability that the total resistance does not exceed 5100 ohms.




I keep getting a negative result in part a. This follows the central limit theorem. I calculated my interval to be:



$$P((199-200)/10/(sqrt{25}) < (X-200)/10/(sqrt{25}) < (202-200)/10/(sqrt{25})$$



$$P(-0.5 < Z < 1)$$



Wouldn't this be $$Phi(1)-[1-Phi(0.5)]?$$







probability statistics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 3 '15 at 16:03









Empty

8,12152661




8,12152661










asked Dec 9 '13 at 19:12









user108626user108626

497




497












  • $begingroup$
    $0.8413447 - (1-0.6914625)= 0.8413447-0.3085375=0.5328072gt 0$
    $endgroup$
    – Henry
    Jul 7 '16 at 7:32










  • $begingroup$
    @Henry's comment - adding some detail: average will be 200 with std dev of 2, so 199 is 1/2 std dev below the mean and 202 is 1 std dev above. 1/2 std dev is 19.15% (from a table) and 1 std dev is 34.13% (using a table for the .13% part).
    $endgroup$
    – stretch
    Jul 8 '17 at 13:37


















  • $begingroup$
    $0.8413447 - (1-0.6914625)= 0.8413447-0.3085375=0.5328072gt 0$
    $endgroup$
    – Henry
    Jul 7 '16 at 7:32










  • $begingroup$
    @Henry's comment - adding some detail: average will be 200 with std dev of 2, so 199 is 1/2 std dev below the mean and 202 is 1 std dev above. 1/2 std dev is 19.15% (from a table) and 1 std dev is 34.13% (using a table for the .13% part).
    $endgroup$
    – stretch
    Jul 8 '17 at 13:37
















$begingroup$
$0.8413447 - (1-0.6914625)= 0.8413447-0.3085375=0.5328072gt 0$
$endgroup$
– Henry
Jul 7 '16 at 7:32




$begingroup$
$0.8413447 - (1-0.6914625)= 0.8413447-0.3085375=0.5328072gt 0$
$endgroup$
– Henry
Jul 7 '16 at 7:32












$begingroup$
@Henry's comment - adding some detail: average will be 200 with std dev of 2, so 199 is 1/2 std dev below the mean and 202 is 1 std dev above. 1/2 std dev is 19.15% (from a table) and 1 std dev is 34.13% (using a table for the .13% part).
$endgroup$
– stretch
Jul 8 '17 at 13:37




$begingroup$
@Henry's comment - adding some detail: average will be 200 with std dev of 2, so 199 is 1/2 std dev below the mean and 202 is 1 std dev above. 1/2 std dev is 19.15% (from a table) and 1 std dev is 34.13% (using a table for the .13% part).
$endgroup$
– stretch
Jul 8 '17 at 13:37










1 Answer
1






active

oldest

votes


















0












$begingroup$

This is not negative.



$$Phi(1) + Phi(0.5)-1=0.5318$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What values are you using for your phi's?
    $endgroup$
    – user108626
    Dec 9 '13 at 19:21










  • $begingroup$
    @user108626 The values from a (standard) Normal Distribution Table.
    $endgroup$
    – JohnK
    Dec 9 '13 at 19:22












  • $begingroup$
    That's what I thought! I just realized our homework gave us a different table where the shaded area is greater than Z as opposed to being less than Z on the standard normal distribution table. Thanks for your help!
    $endgroup$
    – user108626
    Dec 9 '13 at 19:30








  • 1




    $begingroup$
    @user108626 And now you can do part(b) as well without problems.
    $endgroup$
    – JohnK
    Dec 9 '13 at 19:41











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









0












$begingroup$

This is not negative.



$$Phi(1) + Phi(0.5)-1=0.5318$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What values are you using for your phi's?
    $endgroup$
    – user108626
    Dec 9 '13 at 19:21










  • $begingroup$
    @user108626 The values from a (standard) Normal Distribution Table.
    $endgroup$
    – JohnK
    Dec 9 '13 at 19:22












  • $begingroup$
    That's what I thought! I just realized our homework gave us a different table where the shaded area is greater than Z as opposed to being less than Z on the standard normal distribution table. Thanks for your help!
    $endgroup$
    – user108626
    Dec 9 '13 at 19:30








  • 1




    $begingroup$
    @user108626 And now you can do part(b) as well without problems.
    $endgroup$
    – JohnK
    Dec 9 '13 at 19:41
















0












$begingroup$

This is not negative.



$$Phi(1) + Phi(0.5)-1=0.5318$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What values are you using for your phi's?
    $endgroup$
    – user108626
    Dec 9 '13 at 19:21










  • $begingroup$
    @user108626 The values from a (standard) Normal Distribution Table.
    $endgroup$
    – JohnK
    Dec 9 '13 at 19:22












  • $begingroup$
    That's what I thought! I just realized our homework gave us a different table where the shaded area is greater than Z as opposed to being less than Z on the standard normal distribution table. Thanks for your help!
    $endgroup$
    – user108626
    Dec 9 '13 at 19:30








  • 1




    $begingroup$
    @user108626 And now you can do part(b) as well without problems.
    $endgroup$
    – JohnK
    Dec 9 '13 at 19:41














0












0








0





$begingroup$

This is not negative.



$$Phi(1) + Phi(0.5)-1=0.5318$$






share|cite|improve this answer









$endgroup$



This is not negative.



$$Phi(1) + Phi(0.5)-1=0.5318$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '13 at 19:19









JohnKJohnK

2,83011738




2,83011738












  • $begingroup$
    What values are you using for your phi's?
    $endgroup$
    – user108626
    Dec 9 '13 at 19:21










  • $begingroup$
    @user108626 The values from a (standard) Normal Distribution Table.
    $endgroup$
    – JohnK
    Dec 9 '13 at 19:22












  • $begingroup$
    That's what I thought! I just realized our homework gave us a different table where the shaded area is greater than Z as opposed to being less than Z on the standard normal distribution table. Thanks for your help!
    $endgroup$
    – user108626
    Dec 9 '13 at 19:30








  • 1




    $begingroup$
    @user108626 And now you can do part(b) as well without problems.
    $endgroup$
    – JohnK
    Dec 9 '13 at 19:41


















  • $begingroup$
    What values are you using for your phi's?
    $endgroup$
    – user108626
    Dec 9 '13 at 19:21










  • $begingroup$
    @user108626 The values from a (standard) Normal Distribution Table.
    $endgroup$
    – JohnK
    Dec 9 '13 at 19:22












  • $begingroup$
    That's what I thought! I just realized our homework gave us a different table where the shaded area is greater than Z as opposed to being less than Z on the standard normal distribution table. Thanks for your help!
    $endgroup$
    – user108626
    Dec 9 '13 at 19:30








  • 1




    $begingroup$
    @user108626 And now you can do part(b) as well without problems.
    $endgroup$
    – JohnK
    Dec 9 '13 at 19:41
















$begingroup$
What values are you using for your phi's?
$endgroup$
– user108626
Dec 9 '13 at 19:21




$begingroup$
What values are you using for your phi's?
$endgroup$
– user108626
Dec 9 '13 at 19:21












$begingroup$
@user108626 The values from a (standard) Normal Distribution Table.
$endgroup$
– JohnK
Dec 9 '13 at 19:22






$begingroup$
@user108626 The values from a (standard) Normal Distribution Table.
$endgroup$
– JohnK
Dec 9 '13 at 19:22














$begingroup$
That's what I thought! I just realized our homework gave us a different table where the shaded area is greater than Z as opposed to being less than Z on the standard normal distribution table. Thanks for your help!
$endgroup$
– user108626
Dec 9 '13 at 19:30






$begingroup$
That's what I thought! I just realized our homework gave us a different table where the shaded area is greater than Z as opposed to being less than Z on the standard normal distribution table. Thanks for your help!
$endgroup$
– user108626
Dec 9 '13 at 19:30






1




1




$begingroup$
@user108626 And now you can do part(b) as well without problems.
$endgroup$
– JohnK
Dec 9 '13 at 19:41




$begingroup$
@user108626 And now you can do part(b) as well without problems.
$endgroup$
– JohnK
Dec 9 '13 at 19:41


















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