Help with probability and standard deviation

Question

Suppose you have measured a quantity x for 6 times, which results in: $$M=\{51, 53, 54, 55, 52, 53\}$$ Assuming these measurements are normally distributed:

1. What should be your best estimates for the true value $X$ and the standard deviation $\sigma$?
2. Based on these estimates, what is the probability that a seventh measurement would fall outside the range of the first six(given that your results are rounded to the nearest integer, such range would be $x\leq 50.5 \: x \geq 55.5$?

Answer 1

I suppose the last observation should be 53, not 53!. Assuming normal data the best estimate of the population mean $\mu$ (not $X$) is the sample mean $\bar X = 53$, and the best estimate of of the population SD $\sigma$ is the sample SD $S = 1.4142,$ as you have said.

The answer to the second question depends on the exact interpretation of "based on these estimates." One way to model the approximate distribution of a predicted seventh observation is $X_7 \stackrel{aprx}{\sim} \mathsf{Norm}(\bar X, S\sqrt{1 + 1/6}),$ where the second argument is the population SD. In that case the requested probability is about 0.1. [A similar method is used to predict an additional observation $Y$ after finding a regression model of $Y$ on $x.$]

Roughly, the rationale for this model of $X_7$ is that one would model $\bar X \stackrel{aprx}{\sim} \mathsf{Norm}(\bar X, S\sqrt{1/6}),$ but that the additional observation was not used to find $\bar X$, hence the estimated SD for $X_7$ is $S\sqrt{1 + 1/6}$.

There are other ways to model an additional observation so you should verify this method by looking in your text or notes.

Computations in R statistical software follow. If you use standardization and printed normal tables, you may get a slightly different answer for the second part.

x = c(51,53,54,55,52,53) 
mean(x)
## 53
sd(x)
## 1.414214
1-diff(pnorm(c(50.5,55.5), mean(x), sd(x)*sqrt(1+1/6)))
## 0.1017069

Note: If we ignore the estimates and model the new observation as $X_7 \sim \mathsf{Norm}(\mu, \sigma)$ (whatever $\mu$ and $\sigma$ may be), then by a simple combinatorial argument there are 2 chances in 7 that $X_7$ will be either the minimum or the maximum of the sample of $n+1 = 7$ observations. It does not seem to me that this is the intended approach.