How Do You Calculate Uncertainty in Ammeter Readings?

[songoku]

Homework Statement

The manufacturer of a digital ammeter quotes its uncertainty as ±1.5% ±2 digits.
a. Determine the uncertainty in a constant reading of 2.64 A
b. The meter is used to measure the current from a d.c. power supply. The current is found to fluctuate randomly between 1.58 A and 2.04 A. Determine the most likely value of the current, with its uncertainty

Homework Equations

Not sure

The Attempt at a Solution

The answer of (a) is ±0.06 A and (b) is (2.01 ± 0.09) A

I am not sure what ± 2 digits means. Is it ± 0.02 ?

My attempts:
(a) uncertainty = 2.64 x 1.5 % + 0.02 = 0.0596 ≈ 0.06 A

(b)
Average = (1.58 + 2.04) / 2 = 1.81 A
Uncertainty for 1.58 A = 1.58 x 1.5% + 0.02 = 0.0437 ≈ 0.04 A
Uncertainty for 2.04 A = 2.04 x 1.5% + 0.02 = 0.0506 ≈ 0.05 A
Total uncertainty = 0.04 + 0.05 = 0.09 A

Then stuck...

Thanks

 

[Simon Bridge]

I would have read the "digit" part as meaning that the uncertainty given has been rounded to 2 digits.
But $\small (2.64 \pm 0.06)\text{A}$ would be an uncertainty of 2.3%

Similarly, I'd have expected part b to be the mean of the measurements - with the uncertainty on the mean as the error.
I I guess a typo and that 2.04A should be 2.64A, then the mean becomes 2.11A - kinda the same sort of typo perhaps?

So you do need to check how you are expected to interpret that rating.
So I too await what others come up with.

Note: you've asked this question in lots of forums... if another forum beats us to it, please repost here.
Thanks.

 

[songoku]

I would have read the "digit" part as meaning that the uncertainty given has been rounded to 2 digits.
But $\small (2.64 \pm 0.06)\text{A}$ would be an uncertainty of 2.3%

Similarly, I'd have expected part b to be the mean of the measurements - with the uncertainty on the mean as the error.
I I guess a typo and that 2.04A should be 2.64A, then the mean becomes 2.11A - kinda the same sort of typo perhaps?

So you do need to check how you are expected to interpret that rating.
So I too await what others come up with.

Note: you've asked this question in lots of forums... if another forum beats us to it, please repost here.
Thanks.

Hm...yes.

0.06 / 2.64 x 100% ≈ 2.3 % not 1.5%. It is because of that ± 2 digits. If we subtract 0.02 from 0.06 then the uncertainty will be back to 1.5% but the measurement can't be written as 2.64 ± 0.06. This confuses me.

And actually it is not me posting the question in another forums. After reading your comment, I tried to google it and it surprised me this question has been asked by so many students, including here (although it had not been answered yet)

https://www.physicsforums.com/showthread.php?t=473262

Maybe I should google my question first. I depend too much on PF

There is another question on my book, like this:
The diameter of a ball is measured using a metre rule and two set squares. The readings on the rule are 16.8 cm and 20.4 cm. Each reading has an uncertainty of ±1 mm. Calculate, for the diameter of the ball, its actual uncertainty.

My answer is ± 2 mm, twice the uncertainty of each reading. But the answer on the book is ±0.2 mm. Is it a typo (maybe ±0.2 cm) or I made mistake?

Thank you for your reply