Mixing
Finding the probability of the sample standard deviation using Normal Distribution
$begingroup$
The maximum wind speed on any given day in Dublin is modelled as
Normally distributed with mean 10 knots and standard deviation 2 knots.
(a) Explain potential deficiencies of using the Normal distribution to model
wind speeds.
(b) In a 5 day working week what is the probability that at least 2 of the
days have maximum wind speed of more than 9 knots, using the
Normal distribution?
(c) In a 7 day week, what is the probability that the sample standard
deviation is less than 2.5 knots, using the Normal distribution?
I have solved the a) and b) part of the questions however I am a bit confused for the c) part. If anyone can help me over how to solve this.
probability normal-distribution standard-deviation
asked Jan 16 at 14:11
surbhi groversurbhi grover
61
$endgroup$
add a comment |
$begingroup$
The maximum wind speed on any given day in Dublin is modelled as
Normally distributed with mean 10 knots and standard deviation 2 knots.
(a) Explain potential deficiencies of using the Normal distribution to model
wind speeds.
(b) In a 5 day working week what is the probability that at least 2 of the
days have maximum wind speed of more than 9 knots, using the
Normal distribution?
(c) In a 7 day week, what is the probability that the sample standard
deviation is less than 2.5 knots, using the Normal distribution?
I have solved the a) and b) part of the questions however I am a bit confused for the c) part. If anyone can help me over how to solve this.
probability normal-distribution standard-deviation
asked Jan 16 at 14:11
surbhi groversurbhi grover
61
$endgroup$
$begingroup$
The statistic $left(n - 1right)S^{2}/sigma^{2}$ will be Chi-squared distributed with $n - 1$ degrees of freedom, where $S^{2}$ is the sample variance, $n$ is the sample size and $sigma$ is the population standard deviation.
$endgroup$
– rzch
Jan 16 at 15:35
$begingroup$
But I have to find it using the normal distribution.
$endgroup$
– surbhi grover
Jan 16 at 16:39
1
$begingroup$
I took it to mean to use the normal distribution in the sense of 'applying the normal assumption on wind speeds'. Either way, the Chi-squared distribution can be derived from the normal distribution I so don't see why it can't be used.
$endgroup$
– rzch
Jan 16 at 17:19
add a comment |
$begingroup$
The maximum wind speed on any given day in Dublin is modelled as
Normally distributed with mean 10 knots and standard deviation 2 knots.
(a) Explain potential deficiencies of using the Normal distribution to model
wind speeds.
(b) In a 5 day working week what is the probability that at least 2 of the
days have maximum wind speed of more than 9 knots, using the
Normal distribution?
(c) In a 7 day week, what is the probability that the sample standard
deviation is less than 2.5 knots, using the Normal distribution?
I have solved the a) and b) part of the questions however I am a bit confused for the c) part. If anyone can help me over how to solve this.
probability normal-distribution standard-deviation
asked Jan 16 at 14:11
surbhi groversurbhi grover
61
$endgroup$
The maximum wind speed on any given day in Dublin is modelled as
Normally distributed with mean 10 knots and standard deviation 2 knots.
(a) Explain potential deficiencies of using the Normal distribution to model
wind speeds.
(b) In a 5 day working week what is the probability that at least 2 of the
days have maximum wind speed of more than 9 knots, using the
Normal distribution?
(c) In a 7 day week, what is the probability that the sample standard
deviation is less than 2.5 knots, using the Normal distribution?
I have solved the a) and b) part of the questions however I am a bit confused for the c) part. If anyone can help me over how to solve this.
probability normal-distribution standard-deviation
probability normal-distribution standard-deviation
asked Jan 16 at 14:11
surbhi groversurbhi grover
61
asked Jan 16 at 14:11
surbhi groversurbhi grover
61
asked Jan 16 at 14:11
surbhi groversurbhi grover
61
asked Jan 16 at 14:11
surbhi groversurbhi grover
61
asked Jan 16 at 14:11
surbhi groversurbhi grover
61
61
$begingroup$
The statistic $left(n - 1right)S^{2}/sigma^{2}$ will be Chi-squared distributed with $n - 1$ degrees of freedom, where $S^{2}$ is the sample variance, $n$ is the sample size and $sigma$ is the population standard deviation.
$endgroup$
– rzch
Jan 16 at 15:35
$begingroup$
But I have to find it using the normal distribution.
$endgroup$
– surbhi grover
Jan 16 at 16:39
1
$begingroup$
I took it to mean to use the normal distribution in the sense of 'applying the normal assumption on wind speeds'. Either way, the Chi-squared distribution can be derived from the normal distribution I so don't see why it can't be used.
$endgroup$
– rzch
Jan 16 at 17:19
add a comment |
$begingroup$
The statistic $left(n - 1right)S^{2}/sigma^{2}$ will be Chi-squared distributed with $n - 1$ degrees of freedom, where $S^{2}$ is the sample variance, $n$ is the sample size and $sigma$ is the population standard deviation.
$endgroup$
– rzch
Jan 16 at 15:35
$begingroup$
But I have to find it using the normal distribution.
$endgroup$
– surbhi grover
Jan 16 at 16:39
1
$begingroup$
I took it to mean to use the normal distribution in the sense of 'applying the normal assumption on wind speeds'. Either way, the Chi-squared distribution can be derived from the normal distribution I so don't see why it can't be used.
$endgroup$
– rzch
Jan 16 at 17:19
$begingroup$
The statistic $left(n - 1right)S^{2}/sigma^{2}$ will be Chi-squared distributed with $n - 1$ degrees of freedom, where $S^{2}$ is the sample variance, $n$ is the sample size and $sigma$ is the population standard deviation.
$endgroup$
– rzch
Jan 16 at 15:35
$begingroup$
The statistic $left(n - 1right)S^{2}/sigma^{2}$ will be Chi-squared distributed with $n - 1$ degrees of freedom, where $S^{2}$ is the sample variance, $n$ is the sample size and $sigma$ is the population standard deviation.
$endgroup$
– rzch
Jan 16 at 15:35
$begingroup$
But I have to find it using the normal distribution.
$endgroup$
– surbhi grover
Jan 16 at 16:39
$begingroup$
But I have to find it using the normal distribution.
$endgroup$
– surbhi grover
Jan 16 at 16:39
1
1
$begingroup$
I took it to mean to use the normal distribution in the sense of 'applying the normal assumption on wind speeds'. Either way, the Chi-squared distribution can be derived from the normal distribution I so don't see why it can't be used.
$endgroup$
– rzch
Jan 16 at 17:19
$begingroup$
I took it to mean to use the normal distribution in the sense of 'applying the normal assumption on wind speeds'. Either way, the Chi-squared distribution can be derived from the normal distribution I so don't see why it can't be used.
$endgroup$
– rzch
Jan 16 at 17:19
add a comment |
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$begingroup$
The statistic $left(n - 1right)S^{2}/sigma^{2}$ will be Chi-squared distributed with $n - 1$ degrees of freedom, where $S^{2}$ is the sample variance, $n$ is the sample size and $sigma$ is the population standard deviation.
$endgroup$
– rzch
Jan 16 at 15:35
$begingroup$
But I have to find it using the normal distribution.
$endgroup$
– surbhi grover
Jan 16 at 16:39
1
$begingroup$
I took it to mean to use the normal distribution in the sense of 'applying the normal assumption on wind speeds'. Either way, the Chi-squared distribution can be derived from the normal distribution I so don't see why it can't be used.
$endgroup$
– rzch
Jan 16 at 17:19