# Statistics, normal distribution

• cdotter
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
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?

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?

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?

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?

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"

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.

Last edited:
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.

Last edited:
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

$$F(x) = \int_{-\infty}^{x} dt f(t)$$

You want the integral

$$I = \int_a^b dt f(t)$$

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

$$I = \int_a^{-\infty}dt f(t) + \int_{-\infty}^b dt f(t) = -F(a) + F(b)$$

Or, intuitively speaking: the cumulative function F(b) is just the area under f(t) from $-\infty$ up to the value b. You want from a to b, so you just need to subtract the area from $-\infty$ up to a, which is F(a).

Thank you Dick and Mute.

## 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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