Resistors to be used in a circuit have average resistance 200 ohms and standard deviation 10 ohms…
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
edited Nov 3 '15 at 16:03
Empty
8,07252559
asked Dec 9 '13 at 19:12
user108626
497
add a comment |
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
edited Nov 3 '15 at 16:03
Empty
8,07252559
asked Dec 9 '13 at 19:12
user108626
497
$0.8413447 - (1-0.6914625)= 0.8413447-0.3085375=0.5328072gt 0$
– Henry
Jul 7 '16 at 7:32
@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).
– stretch
Jul 8 '17 at 13:37
add a comment |
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
edited Nov 3 '15 at 16:03
Empty
8,07252559
asked Dec 9 '13 at 19:12
user108626
497
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
probability statistics
edited Nov 3 '15 at 16:03
Empty
8,07252559
asked Dec 9 '13 at 19:12
user108626
497
edited Nov 3 '15 at 16:03
Empty
8,07252559
asked Dec 9 '13 at 19:12
user108626
497
edited Nov 3 '15 at 16:03
Empty
8,07252559
edited Nov 3 '15 at 16:03
Empty
8,07252559
edited Nov 3 '15 at 16:03
Empty
8,07252559
8,07252559
asked Dec 9 '13 at 19:12
user108626
497
asked Dec 9 '13 at 19:12
user108626
497
asked Dec 9 '13 at 19:12
user108626
497
497
$0.8413447 - (1-0.6914625)= 0.8413447-0.3085375=0.5328072gt 0$
– Henry
Jul 7 '16 at 7:32
@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).
– stretch
Jul 8 '17 at 13:37
add a comment |
$0.8413447 - (1-0.6914625)= 0.8413447-0.3085375=0.5328072gt 0$
– Henry
Jul 7 '16 at 7:32
@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).
– stretch
Jul 8 '17 at 13:37
$0.8413447 - (1-0.6914625)= 0.8413447-0.3085375=0.5328072gt 0$
– Henry
Jul 7 '16 at 7:32
$0.8413447 - (1-0.6914625)= 0.8413447-0.3085375=0.5328072gt 0$
– Henry
Jul 7 '16 at 7:32
@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).
– stretch
Jul 8 '17 at 13:37
@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).
– stretch
Jul 8 '17 at 13:37
add a comment |
1 Answer
1
active
oldest
votes
This is not negative.
$$Phi(1) + Phi(0.5)-1=0.5318$$
answered Dec 9 '13 at 19:19
JohnK
2,79811637
What values are you using for your phi's?
– user108626
Dec 9 '13 at 19:21
@user108626 The values from a (standard) Normal Distribution Table.
– JohnK
Dec 9 '13 at 19:22
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!
– user108626
Dec 9 '13 at 19:30
1
@user108626 And now you can do part(b) as well without problems.
– JohnK
Dec 9 '13 at 19:41
add a comment |
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1 Answer
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active
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1 Answer
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This is not negative.
$$Phi(1) + Phi(0.5)-1=0.5318$$
answered Dec 9 '13 at 19:19
JohnK
2,79811637
What values are you using for your phi's?
– user108626
Dec 9 '13 at 19:21
@user108626 The values from a (standard) Normal Distribution Table.
– JohnK
Dec 9 '13 at 19:22
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!
– user108626
Dec 9 '13 at 19:30
1
@user108626 And now you can do part(b) as well without problems.
– JohnK
Dec 9 '13 at 19:41
add a comment |
This is not negative.
$$Phi(1) + Phi(0.5)-1=0.5318$$
answered Dec 9 '13 at 19:19
JohnK
2,79811637
What values are you using for your phi's?
– user108626
Dec 9 '13 at 19:21
@user108626 The values from a (standard) Normal Distribution Table.
– JohnK
Dec 9 '13 at 19:22
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!
– user108626
Dec 9 '13 at 19:30
1
@user108626 And now you can do part(b) as well without problems.
– JohnK
Dec 9 '13 at 19:41
add a comment |
This is not negative.
$$Phi(1) + Phi(0.5)-1=0.5318$$
answered Dec 9 '13 at 19:19
JohnK
2,79811637
This is not negative.
$$Phi(1) + Phi(0.5)-1=0.5318$$
answered Dec 9 '13 at 19:19
JohnK
2,79811637
answered Dec 9 '13 at 19:19
JohnK
2,79811637
answered Dec 9 '13 at 19:19
JohnK
2,79811637
answered Dec 9 '13 at 19:19
JohnK
2,79811637
2,79811637
What values are you using for your phi's?
– user108626
Dec 9 '13 at 19:21
@user108626 The values from a (standard) Normal Distribution Table.
– JohnK
Dec 9 '13 at 19:22
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!
– user108626
Dec 9 '13 at 19:30
1
@user108626 And now you can do part(b) as well without problems.
– JohnK
Dec 9 '13 at 19:41
add a comment |
What values are you using for your phi's?
– user108626
Dec 9 '13 at 19:21
@user108626 The values from a (standard) Normal Distribution Table.
– JohnK
Dec 9 '13 at 19:22
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!
– user108626
Dec 9 '13 at 19:30
1
@user108626 And now you can do part(b) as well without problems.
– JohnK
Dec 9 '13 at 19:41
What values are you using for your phi's?
– user108626
Dec 9 '13 at 19:21
What values are you using for your phi's?
– user108626
Dec 9 '13 at 19:21
@user108626 The values from a (standard) Normal Distribution Table.
– JohnK
Dec 9 '13 at 19:22
@user108626 The values from a (standard) Normal Distribution Table.
– JohnK
Dec 9 '13 at 19:22
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!
– user108626
Dec 9 '13 at 19:30
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!
– user108626
Dec 9 '13 at 19:30
1
1
@user108626 And now you can do part(b) as well without problems.
– JohnK
Dec 9 '13 at 19:41
@user108626 And now you can do part(b) as well without problems.
– JohnK
Dec 9 '13 at 19:41
add a comment |
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$0.8413447 - (1-0.6914625)= 0.8413447-0.3085375=0.5328072gt 0$
– Henry
Jul 7 '16 at 7:32
@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).
– stretch
Jul 8 '17 at 13:37