Central limit theorem … need help
So far I have used Chebychev's inequality to calculate n = 157, and my initial thinking for applying the central limit theorem is -1.2815 = (5-0n)/(n* sqrt(391/n)) because P(Z > -1.2815) = 90%, however this isn't giving me the correct value for n so I'm not sure what to try now.
"You want to determine the melting point
c
of a new material. You have
n
specimens on each of which you make a measurement of the melting point in degrees Kelvin, giving you a dataset
m
1
,…,
m
n
. We model this with random variables
M
i
=c+
U
i
, where
U
i
is the random measurement error. It is known that
E[
U
i
]=0
and Var
(
U
i
)=391
for each
i
, and that we may consider the random variables
M
1
,
M
2
,…
as independent.
how many measurements do you need to perform to be
90%
sure that the average of the measurements is within
5
degrees of
c?
Use the normal approximation (central limit theorem rule of thumb) to find a value for
n
. You can use that the
0.95
quantile of the standard normal distribution is
q(0.95)≈1.645
."
probability probability-theory central-limit-theorem
asked Dec 28 '18 at 14:05
KiryneKiryne
135
add a comment |
So far I have used Chebychev's inequality to calculate n = 157, and my initial thinking for applying the central limit theorem is -1.2815 = (5-0n)/(n* sqrt(391/n)) because P(Z > -1.2815) = 90%, however this isn't giving me the correct value for n so I'm not sure what to try now.
"You want to determine the melting point
c
of a new material. You have
n
specimens on each of which you make a measurement of the melting point in degrees Kelvin, giving you a dataset
m
1
,…,
m
n
. We model this with random variables
M
i
=c+
U
i
, where
U
i
is the random measurement error. It is known that
E[
U
i
]=0
and Var
(
U
i
)=391
for each
i
, and that we may consider the random variables
M
1
,
M
2
,…
as independent.
how many measurements do you need to perform to be
90%
sure that the average of the measurements is within
5
degrees of
c?
Use the normal approximation (central limit theorem rule of thumb) to find a value for
n
. You can use that the
0.95
quantile of the standard normal distribution is
q(0.95)≈1.645
."
probability probability-theory central-limit-theorem
asked Dec 28 '18 at 14:05
KiryneKiryne
135
add a comment |
So far I have used Chebychev's inequality to calculate n = 157, and my initial thinking for applying the central limit theorem is -1.2815 = (5-0n)/(n* sqrt(391/n)) because P(Z > -1.2815) = 90%, however this isn't giving me the correct value for n so I'm not sure what to try now.
"You want to determine the melting point
c
of a new material. You have
n
specimens on each of which you make a measurement of the melting point in degrees Kelvin, giving you a dataset
m
1
,…,
m
n
. We model this with random variables
M
i
=c+
U
i
, where
U
i
is the random measurement error. It is known that
E[
U
i
]=0
and Var
(
U
i
)=391
for each
i
, and that we may consider the random variables
M
1
,
M
2
,…
as independent.
how many measurements do you need to perform to be
90%
sure that the average of the measurements is within
5
degrees of
c?
Use the normal approximation (central limit theorem rule of thumb) to find a value for
n
. You can use that the
0.95
quantile of the standard normal distribution is
q(0.95)≈1.645
."
probability probability-theory central-limit-theorem
asked Dec 28 '18 at 14:05
KiryneKiryne
135
So far I have used Chebychev's inequality to calculate n = 157, and my initial thinking for applying the central limit theorem is -1.2815 = (5-0n)/(n* sqrt(391/n)) because P(Z > -1.2815) = 90%, however this isn't giving me the correct value for n so I'm not sure what to try now.
"You want to determine the melting point
c
of a new material. You have
n
specimens on each of which you make a measurement of the melting point in degrees Kelvin, giving you a dataset
m
1
,…,
m
n
. We model this with random variables
M
i
=c+
U
i
, where
U
i
is the random measurement error. It is known that
E[
U
i
]=0
and Var
(
U
i
)=391
for each
i
, and that we may consider the random variables
M
1
,
M
2
,…
as independent.
how many measurements do you need to perform to be
90%
sure that the average of the measurements is within
5
degrees of
c?
Use the normal approximation (central limit theorem rule of thumb) to find a value for
n
. You can use that the
0.95
quantile of the standard normal distribution is
q(0.95)≈1.645
."
probability probability-theory central-limit-theorem
probability probability-theory central-limit-theorem
asked Dec 28 '18 at 14:05
KiryneKiryne
135
asked Dec 28 '18 at 14:05
KiryneKiryne
135
asked Dec 28 '18 at 14:05
KiryneKiryne
135
asked Dec 28 '18 at 14:05
KiryneKiryne
135
asked Dec 28 '18 at 14:05
KiryneKiryne
135
135
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
begin{align}
P left(left| frac{sum_{i=1}^nM_i}{n}-cright|<5 right)&=P left(sqrt{n}left| frac{sum_{i=1}^nM_i}{n}-cright|<5sqrt{n} right)\
&approx Pleft(|Z|<frac{5sqrt{n}}{sqrt{Var(U_i)}}right)
end{align}
That is we want $$frac{5sqrt{n}}{sqrt{Var(U_i)}}ge 1.645$$
answered Dec 28 '18 at 14:25
Siong Thye GohSiong Thye Goh
99.6k1464117
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
begin{align}
P left(left| frac{sum_{i=1}^nM_i}{n}-cright|<5 right)&=P left(sqrt{n}left| frac{sum_{i=1}^nM_i}{n}-cright|<5sqrt{n} right)\
&approx Pleft(|Z|<frac{5sqrt{n}}{sqrt{Var(U_i)}}right)
end{align}
That is we want $$frac{5sqrt{n}}{sqrt{Var(U_i)}}ge 1.645$$
answered Dec 28 '18 at 14:25
Siong Thye GohSiong Thye Goh
99.6k1464117
add a comment |
begin{align}
P left(left| frac{sum_{i=1}^nM_i}{n}-cright|<5 right)&=P left(sqrt{n}left| frac{sum_{i=1}^nM_i}{n}-cright|<5sqrt{n} right)\
&approx Pleft(|Z|<frac{5sqrt{n}}{sqrt{Var(U_i)}}right)
end{align}
That is we want $$frac{5sqrt{n}}{sqrt{Var(U_i)}}ge 1.645$$
answered Dec 28 '18 at 14:25
Siong Thye GohSiong Thye Goh
99.6k1464117
add a comment |
begin{align}
P left(left| frac{sum_{i=1}^nM_i}{n}-cright|<5 right)&=P left(sqrt{n}left| frac{sum_{i=1}^nM_i}{n}-cright|<5sqrt{n} right)\
&approx Pleft(|Z|<frac{5sqrt{n}}{sqrt{Var(U_i)}}right)
end{align}
That is we want $$frac{5sqrt{n}}{sqrt{Var(U_i)}}ge 1.645$$
answered Dec 28 '18 at 14:25
Siong Thye GohSiong Thye Goh
99.6k1464117
begin{align}
P left(left| frac{sum_{i=1}^nM_i}{n}-cright|<5 right)&=P left(sqrt{n}left| frac{sum_{i=1}^nM_i}{n}-cright|<5sqrt{n} right)\
&approx Pleft(|Z|<frac{5sqrt{n}}{sqrt{Var(U_i)}}right)
end{align}
That is we want $$frac{5sqrt{n}}{sqrt{Var(U_i)}}ge 1.645$$
answered Dec 28 '18 at 14:25
Siong Thye GohSiong Thye Goh
99.6k1464117
answered Dec 28 '18 at 14:25
Siong Thye GohSiong Thye Goh
99.6k1464117
answered Dec 28 '18 at 14:25
Siong Thye GohSiong Thye Goh
99.6k1464117
answered Dec 28 '18 at 14:25
Siong Thye GohSiong Thye Goh
99.6k1464117
99.6k1464117
add a comment |
add a comment |
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