I need help for a physics error analysis assignment and I have no background understanding whatsoever of statistics since I've never taken it and it all seems so foreign to me without any teaching. Can someone please help me with these questions?

1) One commonly used method is to find the "full width at half the maximum" (FWHM). To find this you determine where the number of data is one-half of the value of the maximum, i.e. where N(x) = A/2. There will be two such points for a bell shaped curve. Then the FWHM is the difference between the right hand side value and the left hand side value of x. For a Gaussian distribution what is the mathematical relationship between the FWHM and the standard deviation?

2) You have a large dataset that is normally distributed. If you choose one data point at random from the dataset, what is the probability that it will lie within one standard deviation of the mean?

Express the following quantities to the correct number of significant figures:

(a) 23.043 ± 4.321
(b) 67 ± 9.023
(c) 33.1492934 ± 2

Isn't (a) already the correct number of significant digits?
Would (b) be 67.000 ± 9.023?
Would (c) be 33.1492934 ± 2.0000000

or would
(a) 23.04 ± 4.321
(b) 67 ± 9.0
(c) 30 ± 2

A student measures three lengths a, b and c in cm and a time t in seconds:

a = 40 ± 5
b = 30 ± 3
c = 20 ± 1
t = 1.2 ± 0.1

What is a+b
a+b+c
a/t
(a+c)/t

Calculate (1.23 ± 0.03) + pi

Question 9.4. Calculate (1.23 ± 0.03) × pi

Can you do a sample calculation of each of those please?

You are determining the period of oscillation of a pendulum. One procedure would be to measure the time for 20 oscillations, t20, and repeat the measurement 5 times. Another procedure would be to measure the time for 5 oscillations, t5, and repeat the measurement 20 times. Assume, reasonably, that the error in the determination of the time for 20 oscillations is the same as the error in the determination of the time for 5 oscillations. Calculate the error in the period for both procedures to determine which will give the smallest error in the value of the period?

dont really know

Sure, I can help you with your physics error analysis assignment. Let's break down each question step by step.

1) For a Gaussian distribution (also known as a normal distribution), the standard deviation (σ) is related to the full width at half the maximum (FWHM) by the equation: FWHM = 2√(2ln2)σ.

2) In a normal distribution, approximately 68% of the data lies within one standard deviation of the mean. Therefore, the probability of choosing a data point within one standard deviation of the mean is approximately 0.68.

Now let's move on to the next part:

(a) The number of significant figures in a measurement is determined by the uncertainty. In this case, 4.321 has four significant figures, so the value 23.043 ± 4.321 already has the correct number of significant figures.

(b) In this case, the uncertainty is given as 9.023. If you want to express the value to the correct number of significant figures, the uncertainty should also have three significant figures. Therefore, the value 67 ± 9.023 can be expressed as 67.0 ± 9.0.

(c) The value 2 has only one significant figure. Therefore, the uncertainty should also have one significant figure. The value 33.1492934 ± 2 can be expressed as 33 ± 2.

Moving on to the next part:

To calculate a+b, simply add the values: (40 ± 5) + (30 ± 3) = 70 ± 8.

To calculate a+b+c, add all three values: (40 ± 5) + (30 ± 3) + (20 ± 1) = 90 ± 9.

To calculate a/t, divide a by t: (40 ± 5) / (1.2 ± 0.1) = 33.33 ± 4.17.

To calculate (a+c)/t, first add a and c, then divide by t: (40 ± 5 + 20 ± 1) / (1.2 ± 0.1) = 50 ± 6.67.

Next:

To calculate (1.23 ± 0.03) + π, simply add the two values: 1.23 + π = 4.37.

To calculate (1.23 ± 0.03) × π, multiply the two values: 1.23 × π = 3.86.

Finally:

For the first procedure, the error in the determination of the time for 20 oscillations is the same as the error in the determination of the time for 5 oscillations. Therefore, the error in the period for the first procedure will be the same as the error in the period for the second procedure. To determine which procedure gives the smallest error in the value of the period, you would need to compare the actual values of the errors.

I hope this step-by-step explanation helps you with your physics error analysis assignment! Let me know if you have any further questions.

Sure, I can help you with your physics error analysis assignment. Let's go through each question one by one.

1) The relationship between the FWHM and the standard deviation for a Gaussian distribution is given by the equation FWHM = 2√(2ln2) * σ, where FWHM is the full width at half maximum and σ is the standard deviation. To derive this relationship, we start with the fact that the FWHM represents the width of the distribution at half the maximum value, which corresponds to ±σ for a Gaussian distribution. By solving the equation N(x) = A/2 for x, we find the right and left-hand side values for x. The difference between these values gives us the FWHM.

2) If a dataset is normally distributed and you choose one data point at random, the probability that it will lie within one standard deviation of the mean is approximately 68.27%. This is because in a normal distribution, about 68.27% of the data falls within one standard deviation of the mean.

Now let's move on to the next set of questions about expressing quantities to the correct number of significant figures:

(a) The correct way to express the quantity 23.043 ± 4.321 to the correct number of significant figures is 23.0 ± 4.3. This is because when expressing the uncertainty, you should round it to have the same number of decimal places as the value itself.

(b) The correct way to express the quantity 67 ± 9.023 to the correct number of significant figures is 67 ± 9.0. In this case, the value 67 has two significant figures, so the uncertainty should also be rounded to have two decimal places.

(c) The correct way to express the quantity 33.1492934 ± 2 to the correct number of significant figures is 33.1 ± 2. This is because the value 33.1492934 has four significant figures, so the uncertainty should also be rounded to have one decimal place.

Now let's move on to the next set of questions involving calculations:

To find a+b, simply add the values together:

a = 40 ± 5
b = 30 ± 3

a+b = (40+30) ± (√(5^2 + 3^2)) = 70 ± √34

To find a+b+c, add all three values together:

a = 40 ± 5
b = 30 ± 3
c = 20 ± 1

a+b+c = (40+30+20) ± (√(5^2 + 3^2 + 1^2)) = 90 ± √35

To find a/t, divide the value of a by the value of t:

a = 40 ± 5
t = 1.2 ± 0.1

a/t = (40/1.2) ± (√((5/40)^2 + (0.1/1.2)^2)) = 33.3 ± 0.25

To find (a+c)/t, add the values of a and c together, then divide by t:

a = 40 ± 5
c = 20 ± 1
t = 1.2 ± 0.1

(a+c)/t = ((40+20)/1.2) ± (√((5/40)^2 + (1/20)^2 + (0.1/1.2)^2)) = 50 ± 0.43

To calculate (1.23 ± 0.03) + pi, simply add the values together:

(1.23 ± 0.03) + pi = (1.23 + 3.14159) ± (0.03 + 0) = 4.37 ± 0.03

To calculate (1.23 ± 0.03) × pi, multiply the values together:

(1.23 ± 0.03) × pi = (1.23 × 3.14159) ± (0.03 × 3.14159) = 3.86 ± 0.09

Finally, let's tackle the last question regarding the error in the period of a pendulum:

Since the error in the determination of the time for 20 oscillations is assumed to be the same as the error in the determination of the time for 5 oscillations, we can conclude that the error in the period will be smaller for the procedure measuring the time for 20 oscillations. This is because the error in the determination of the period can be calculated as the error in the time measurement divided by the number of oscillations. As the number of oscillations increases, the error in the period decreases. Therefore, the procedure measuring the time for 20 oscillations will give the smallest error in the value of the period.

I hope this helps you with your physics error analysis assignment! If you have any further questions, feel free to ask.