Statistics, normal distribution

[cdotter]

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?

 

[Dick]

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?

 

[cdotter]

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?

 

[Dick]

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?

 

[cdotter]

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"

 

[Dick]

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.

 

[cdotter]

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.

 

[Mute]

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).

 

[cdotter]

Thank you Dick and Mute.