A laboratory weighs filters from a coal mine

Asked by maham237 @ 10/01/2021 in Mathematics viewed by 245 People

A laboratory weighs filters from a coal mine to measure the amount of dust in the mine atmosphere. Repeated measurements of the weight of dust on the same filter vary normally with standard deviation  0.08 milligram (mg) because the weighing is not perfectly precise. The dust on a particular filter actually weighs 123 mg. Repeated weighings will then have the N(123, 0.08) distribution. a) The lab reports the mean of 3 weighings. What is the distribution of this mean? b) What is the probability that the lab reports a result of 124 mg or higher for this filter?

Answered by maham237 @ 10/01/2021

Answer:

a) $\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\bar X \sim N(\mu=123,\sigma_{\bar X} = \frac{0.08}{\sqrt{3}}= 0.046)$

b) $P(\bar X >124)=P(Z>\frac{124-123}{\frac{0.08}{\sqrt{3}}}=21.65)$

And using a calculator, excel or the normal standard table we have that:

$P(Z>21.65) \approx 0$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Part a

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(123,0.08)$

Where $\mu=123$ and $\sigma=0.08$

Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\bar X \sim N(\mu=123,\sigma_{\bar X} = \frac{0.08}{\sqrt{3}}= 0.046)$

Part b

We want to find this probability:

$P(\bar X >124)$

And we can use the z score given by:

$z = \frac{\bar X -\mu}{\sigma_{\bar X}}$

And using this formula we got:

$P(\bar X >124)=P(Z>\frac{124-123}{\frac{0.08}{\sqrt{3}}}=21.65)$

And using a calculator, excel or the normal standard table we have that:

$P(Z>21.65) \approx 0$

Answered by maham237 @ 10/01/2021

Answer:

a) $\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\bar X \sim N(\mu=123,\sigma_{\bar X} = \frac{0.08}{\sqrt{3}}= 0.046)$

b) $P(\bar X >124)=P(Z>\frac{124-123}{\frac{0.08}{\sqrt{3}}}=21.65)$

And using a calculator, excel or the normal standard table we have that:

$P(Z>21.65) \approx 0$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Part a

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(123,0.08)$

Where $\mu=123$ and $\sigma=0.08$

Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\bar X \sim N(\mu=123,\sigma_{\bar X} = \frac{0.08}{\sqrt{3}}= 0.046)$

Part b

We want to find this probability:

$P(\bar X >124)$

And we can use the z score given by:

$z = \frac{\bar X -\mu}{\sigma_{\bar X}}$

And using this formula we got:

$P(\bar X >124)=P(Z>\frac{124-123}{\frac{0.08}{\sqrt{3}}}=21.65)$

And using a calculator, excel or the normal standard table we have that:

$P(Z>21.65) \approx 0$

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A laboratory weighs filters from a coal mine

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