Resistors to be used in a circuit have average resistance 200 ohms and standard deviation 10 ohmsâ¦
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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:
$$P((199-200)/10/(sqrt25) < (X-200)/10/(sqrt25) < (202-200)/10/(sqrt25)$$
$$P(-0.5 < Z < 1)$$
Wouldn't this be $$Phi(1)-[1-Phi(0.5)]?$$
probability statistics
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up vote
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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:
$$P((199-200)/10/(sqrt25) < (X-200)/10/(sqrt25) < (202-200)/10/(sqrt25)$$
$$P(-0.5 < Z < 1)$$
Wouldn't this be $$Phi(1)-[1-Phi(0.5)]?$$
probability statistics
$0.8413447 - (1-0.6914625)= 0.8413447-0.3085375=0.5328072gt 0$
â 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).
â stretch
Jul 8 '17 at 13:37
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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:
$$P((199-200)/10/(sqrt25) < (X-200)/10/(sqrt25) < (202-200)/10/(sqrt25)$$
$$P(-0.5 < Z < 1)$$
Wouldn't this be $$Phi(1)-[1-Phi(0.5)]?$$
probability statistics
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:
$$P((199-200)/10/(sqrt25) < (X-200)/10/(sqrt25) < (202-200)/10/(sqrt25)$$
$$P(-0.5 < Z < 1)$$
Wouldn't this be $$Phi(1)-[1-Phi(0.5)]?$$
probability statistics
edited Nov 3 '15 at 16:03
Empty
7,90642254
7,90642254
asked Dec 9 '13 at 19:12
user108626
497
497
$0.8413447 - (1-0.6914625)= 0.8413447-0.3085375=0.5328072gt 0$
â 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).
â stretch
Jul 8 '17 at 13:37
add a comment |Â
$0.8413447 - (1-0.6914625)= 0.8413447-0.3085375=0.5328072gt 0$
â 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).
â stretch
Jul 8 '17 at 13:37
$0.8413447 - (1-0.6914625)= 0.8413447-0.3085375=0.5328072gt 0$
â Henry
Jul 7 '16 at 7:32
$0.8413447 - (1-0.6914625)= 0.8413447-0.3085375=0.5328072gt 0$
â 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).
â 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).
â stretch
Jul 8 '17 at 13:37
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
This is not negative.
$$Phi(1) + Phi(0.5)-1=0.5318$$
What values are you using for your phi's?
â user108626
Dec 9 '13 at 19:21
@user108626 The values from a (standard) Normal Distribution Table.
â 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!
â user108626
Dec 9 '13 at 19:30
1
@user108626 And now you can do part(b) as well without problems.
â JohnK
Dec 9 '13 at 19:41
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
This is not negative.
$$Phi(1) + Phi(0.5)-1=0.5318$$
What values are you using for your phi's?
â user108626
Dec 9 '13 at 19:21
@user108626 The values from a (standard) Normal Distribution Table.
â 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!
â user108626
Dec 9 '13 at 19:30
1
@user108626 And now you can do part(b) as well without problems.
â JohnK
Dec 9 '13 at 19:41
add a comment |Â
up vote
0
down vote
This is not negative.
$$Phi(1) + Phi(0.5)-1=0.5318$$
What values are you using for your phi's?
â user108626
Dec 9 '13 at 19:21
@user108626 The values from a (standard) Normal Distribution Table.
â 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!
â user108626
Dec 9 '13 at 19:30
1
@user108626 And now you can do part(b) as well without problems.
â JohnK
Dec 9 '13 at 19:41
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This is not negative.
$$Phi(1) + Phi(0.5)-1=0.5318$$
This is not negative.
$$Phi(1) + Phi(0.5)-1=0.5318$$
answered Dec 9 '13 at 19:19
JohnK
2,73011532
2,73011532
What values are you using for your phi's?
â user108626
Dec 9 '13 at 19:21
@user108626 The values from a (standard) Normal Distribution Table.
â 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!
â user108626
Dec 9 '13 at 19:30
1
@user108626 And now you can do part(b) as well without problems.
â JohnK
Dec 9 '13 at 19:41
add a comment |Â
What values are you using for your phi's?
â user108626
Dec 9 '13 at 19:21
@user108626 The values from a (standard) Normal Distribution Table.
â 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!
â user108626
Dec 9 '13 at 19:30
1
@user108626 And now you can do part(b) as well without problems.
â JohnK
Dec 9 '13 at 19:41
What values are you using for your phi's?
â user108626
Dec 9 '13 at 19:21
What values are you using for your phi's?
â user108626
Dec 9 '13 at 19:21
@user108626 The values from a (standard) Normal Distribution Table.
â JohnK
Dec 9 '13 at 19:22
@user108626 The values from a (standard) Normal Distribution Table.
â 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!
â 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!
â user108626
Dec 9 '13 at 19:30
1
1
@user108626 And now you can do part(b) as well without problems.
â JohnK
Dec 9 '13 at 19:41
@user108626 And now you can do part(b) as well without problems.
â JohnK
Dec 9 '13 at 19:41
add a comment |Â
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$0.8413447 - (1-0.6914625)= 0.8413447-0.3085375=0.5328072gt 0$
â 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).
â stretch
Jul 8 '17 at 13:37