We have a laser that measures length. Average measurement error is 0 cm and deviation of error is 1cm. We repeat measurements 20 times and calculate new average. What is the probability that error is below 0.5 cm.
The solution says it's 0.975, but i can't get the same result.
What I tried is this: P(X<=0.5) = fi(-0.5/sqrt(20)) + 0.5. Where fi is the error function. I would like to know what is wrong with my attempt.
Each observation is $X_i \sim \mathsf{Norm}(0, 1).$
Mean of $n=20$ observations is $\bar X \sim \mathsf{Norm}(\mu = 0, \sigma=1/\sqrt{20}).$
You seem to be trying to find $P(\bar X < 0.5),$ or perhaps $P(\bar X < -0.5).$
Focusing on the 'error' rather than the 'value' $\bar X$ itself, I think you are intended to compute $$P(|\bar X - 0| < .5) = P(-.5 < \bar X < .5) = P\left(\frac{-.5-0}{1/\sqrt{20}}<\frac{\bar X-0}{1/\sqrt{n}} <\frac{.5-0}{1/\sqrt{20}}\right)\\ = P(-2.236 < Z < 2.236) = 0.9746154.$$
Note: You mention the CLT in your title. Even if $X_i$ are not exactly normally distributed, the CLT "suggests" that $\bar X$ is more nearly normal than the data. Taking the limit as $n \rightarrow \infty,$ the CLT says that $\bar X$ is exactly normal (provided the population distribution of the $X_i$ has finite variance). However, for many physical measurements $n = 20$ is a large enough sample for $\bar X_{20}$ to be essentially normal. [For example, if $X_i \sim \mathsf{Unif}(-1, 1),$ then it is very difficult to distinguish the distribution of $\bar X_{20}$ from a normal distribution.]