Statistics, normal distribution

In summary: The cumulative distribution of a probability density function f(x) is defined asF(x) = \int_{-\infty}^{x} dt f(t)You want the integralI = \int_a^b dt f(t)You can relate this to the cumulative function using the property of integrals\int_a^b dt f(t) = \int_a^c dt f(t) + \int_c^b dt f(t)which holds for any c as long as f(t) is continuous on the integral regions.This gives
  • #1
cdotter
305
0

Homework Statement


The temperature reading from a thermocouple placed in a constant-temperature medium is normally distributed with a mean μ, the actual temperature of the medium, and standard deviation σ. What would the value of σ have to be to ensure that 95% of all readings are within 0.1 degree of σ?

The Attempt at a Solution



I looked up the solution and found the first step as P(-1.96 < x < 1.96) = 95%. This doesn't even make sense. Values of 1.96 on the normal distribution chart represent 0.975, or 97.5%. Can someone explain why I'm using these values?
 
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  • #2
cdotter said:

Homework Statement


The temperature reading from a thermocouple placed in a constant-temperature medium is normally distributed with a mean μ, the actual temperature of the medium, and standard deviation σ. What would the value of σ have to be to ensure that 95% of all readings are within 0.1 degree of σ?

The Attempt at a Solution



I looked up the solution and found the first step as P(-1.96 < x < 1.96) = 95%. This doesn't even make sense. Values of 1.96 on the normal distribution chart represent 0.975, or 97.5%. Can someone explain why I'm using these values?

The normal distribution chart is for sigma=1. How do you modify P to account for a sigma that's not equal to 1?
 
  • #3
Dick said:
The normal distribution chart is for sigma=1. How do you modify P to account for a sigma that's not equal to 1?

I know how to standardize for nonstandard normal distributions (Z=(X-μ)/σ). I don't see how it helps, or am I missing something really simple?
 
  • #4
cdotter said:
I know how to standardize for nonstandard normal distributions (Z=(X-μ)/σ). I don't see how it helps, or am I missing something really simple?

Z is the variable in the normal distribution chart. You don't want to make a different chart for every different value of mu and sigma, do you? So the chart is telling you P(-1.96 < Z < 1.96) = 95%. Put Z=(X-mu)/sigma into that. You want |X-mu|<0.1, yes?
 
  • #5
Dick said:
Z is the variable in the normal distribution chart. You don't want to make a different chart for every different value of mu and sigma, do you? So the chart is telling you P(-1.96 < Z < 1.96) = 95%. Put Z=(X-mu)/sigma into that. You want |X-mu|<0.1, yes?

Yes, I get that part. I'm confused about the 1.96 part. I only have a standard normal distribution chart to work from. I want to know how P(-1.96 < Z < 1.96) = 95% because on a standard normal distribution chart, those values give 97.5%. How did they get it to equal 95%? None of the solutions I've found state how, they just say "P(-1.96 < Z < 1.96) = .95"
 
  • #6
cdotter said:
Yes, I get that part. I'm confused about the 1.96 part. I only have a standard normal distribution chart to work from. I want to know how P(-1.96 < Z < 1.96) = 95% because on a standard normal distribution chart, those values give 97.5%. How did they get it to equal 95%? None of the solutions I've found state how, they just say "P(-1.96 < Z < 1.96) = .95"

Ohhh, ok, I get it now. I think you are looking at a cumulative distribution chart. It's giving you P(-infinity < Z < 1.96). Not P(-1.96 < Z < 1.96). Find a different chart or adjust the number you've got. What does the chart give for 0? If it's 0.5 then it's cumulative.
 
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  • #7
Dick said:
Ohhh, ok, I get it now. I think you are looking at a cumulative distribution chart. It's giving you P(-infinity < Z < 1.96). Not P(-1.96 < Z < 1.96). Find a different chart or adjust the number you've got. What does the chart give for 0? If it's 0.5 then it's cumulative.

It gives 0.5. How could I adjust the number? I glanced over the chapter again and it doesn't say anything about what to do in such a case, and it doesn't mention any other charts to use for the problem. Kind of ridiculous on the author's behalf.
 
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  • #8
cdotter said:
It gives 0.5. How could I adjust the number? I glanced over the chapter again and it doesn't say anything about what to do in such a case, and it doesn't mention any other charts to use for the problem. Kind of ridiculous and the author's behalf.

The cumulative distribution of a probability density function f(x) is defined as

[tex]F(x) = \int_{-\infty}^{x} dt f(t)[/tex]

You want the integral

[tex]I = \int_a^b dt f(t)[/tex]

You can relate this to the cumulative function using the property of integrals

[tex]\int_a^b dt f(t) = \int_a^c dt f(t) + \int_c^b dt f(t)[/tex]
which holds for any c as long as f(t) is continuous on the integral regions.

This gives

[tex]I = \int_a^{-\infty}dt f(t) + \int_{-\infty}^b dt f(t) = -F(a) + F(b)[/tex]

Or, intuitively speaking: the cumulative function F(b) is just the area under f(t) from [itex]-\infty[/itex] up to the value b. You want from a to b, so you just need to subtract the area from [itex]-\infty[/itex] up to a, which is F(a).
 
  • #9
Thank you Dick and Mute.
 

Related to Statistics, normal distribution

1. What is the normal distribution in statistics?

The normal distribution, also known as the Gaussian distribution, is a probability distribution that is commonly used in statistics to represent the distribution of a set of data. It is a bell-shaped curve that is symmetrical around the mean, with the majority of data points falling near the mean and fewer data points falling further away from the mean.

2. How is the normal distribution calculated?

The normal distribution is calculated using a mathematical formula called the Gaussian function. This formula takes into account the mean and standard deviation of the data set to determine the shape and location of the curve.

3. What is the significance of the normal distribution in statistics?

The normal distribution is significant in statistics because many real-world phenomena, such as human height and IQ scores, can be approximated by a normal distribution. This allows statisticians to make meaningful predictions and inferences about the data.

4. How is the normal distribution related to the central limit theorem?

The central limit theorem states that the sum of a large number of independent random variables will tend towards a normal distribution, regardless of the distribution of the individual variables. This means that the normal distribution is often used as a model for many real-world data sets, even if the data does not follow a normal distribution.

5. How can the normal distribution be used in hypothesis testing?

In hypothesis testing, the normal distribution is used to determine the probability of obtaining a certain result by chance. By calculating a z-score, which measures the number of standard deviations a data point falls above or below the mean, statisticians can determine the likelihood of obtaining a particular result under the assumption of a normal distribution. This can help determine the significance of a hypothesis and whether it should be accepted or rejected.

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