How Do You Calculate Uncertainties and Probabilities in Various Scenarios?

[jnimagine]

How do yo do these calculations??
help please!

for ex would a + b be, 90.043 ± 14.344?

Calculate:
(a) 23.043 ± 4.321
(b) 67 ± 9.023
(c) 33.1492934 ± 2
a = 40 ± 5
b = 30 ± 3
c = 20 ± 1
t = 1.2 ± 0.1
Calculate a + b, a + b + c, a / t, and (a + c) / t.

Calculate (1.23 ± 0.03) +pi . (pi is the irrational number 3.14159265…)

Calculate (1.23 ± 0.03) × pi


Totally lost for these ones... :|

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?

What is the probability of rolling a seven 10 times in a row?

Amazingly, you have rolled a seven 9 times in a row. What is the probability that you will get a seven on the next roll?

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?

 

[lippyka]

The answers to these questions are as follows: (a) 23.043 ± 4.321(b) 67 ± 9.023(c) 33.1492934 ± 2a = 40 ± 5b = 30 ± 3c = 20 ± 1t = 1.2 ± 0.1Calculate a + b: 70 ± 6.021Calculate a + b + c: 90 ± 8.021Calculate a / t: 33.333 ± 4.167Calculate (a + c) / t: 53.333 ± 4.167Calculate (1.23 ± 0.03) + pi: 4.37059265 ± 0.03Calculate (1.23 ± 0.03) × pi: 3.8462684 ± 0.09For the first two questions, the procedure that measures the time for 5 oscillations and repeats the measurement 20 times will give the smallest error in the value of the period.The probability of rolling a seven 10 times in a row is very small and is equal to 1 in 10^13.The probability that you will get a seven on the next roll is 1/6 since the probability of rolling a seven on any single roll is 1/6.The probability that a data point chosen at random from a normally distributed dataset lies within one standard deviation of the mean is 68.27%.