A data scientist is testing his predictor (based on historical data) that would make predictions about Philadelphia’s temperatures. A prediction is judged "correct" if it falls within three standard deviations from the target value. Assume that the expected value of each prediction equals the target value. What is the accuracy of the predictor if the distribution of measurements is uniform?
Uniform Distribution Accuracy
probabilityprobability distributions
Related Solutions
Statistical meanings of 'accuracy' and 'precision' are as follows:
Precision refers to the variability of the measurements. For example, if repeated measurements using the pipet are normally distributed with mean $\mu$ and standard deviation $\sigma,$ then small $\sigma$ means high precision. How reproducible are the repeated results?
Accuracy refers to whether the mean $\mu$ of the repeated measurements is correct. Are repeated measurements systematically too small or systematically too large.
In terms of target practice at a rifle range: Precision has to do with whether I can hit approximately the same spot on the target every time. Accuracy has to do with whether that spot is the bull's eye.
To find out precisely how the manufacturer determines accuracy and precision, you'd have to ask--and then not be terribly surprised if no one knows the exact answer. For example the precision of $\pm 0.02\, mL$ might mean that $\sigma = .01,$ so that an interval $\pm 0.02$ would contain 95% of repeated determinations. Or it might mean that $\sigma = .02.$ (I'm a statistician and it has been about 50 years since I've had anything to do with pipettes, but I suspect @JohnBenton is correct that there is no uniform standard. Also, I suspect information provided is for ideal circumstances, and assumes an experienced persion is using the pipette.)
Here is one method that is sometimes used (especially in theories of estimation) to combine accuracy and precision. Suppose the standard deviation is $\sigma$ so that the variance is $\sigma^2.$ Suppose that the bias is $b$ (roughly, the average difference between true and experimental mean). Then the mean squared error is $MSE = \sigma^2 + b^2,$ which is expressed in squared units. To return to original units (e.g, mL, in your case), you can take the square root to get $RMSE = \sqrt{\sigma^2 + b^2}.$
$P(11.1\le X\le28.9)=P(12\le X\le28)$ since $X$ is discrete. So what you calculated is not correct.
Also, they might use normal approximation of a Poisson random variable.
$$\frac{X-\lambda}{\sqrt\lambda}\sim N(0,1)$$
Best Answer
For those who were curious, here is the answer provided by my professor: