# Calculating probability of a normal distribution, not getting correct answer

I'm doing a homework assignment and having some trouble matching the correct answers from my professor. As a reference, I'm calculating these answers using R.

The question is as follows:

Assuming $E(X_i) = \mu = 7$, $Var(X_i) = \sigma^2 = 0.12$, and $\bar{X_{50}} \sim N(7,0.12)$

What is $P(\mu - 2 \times \frac{\sigma}{\sqrt{50}} \leq \bar{X_{50}} \leq \mu + 2 \times \frac{\sigma}{\sqrt{50}})$?

The answer my professor gave is $0.9545$. He also stated that I don't need to know the values of $\mu$ and $\sigma$.

I don't know how he got this though? Doing the easy thing and plugging in 7 for $\mu$ and 0.12 for $\sigma^2$ I ended up getting:

$P(6.90202041 \leq \bar{X_{50}} \leq 7.09797959) = P(\bar{X_{50}} \leq 7.09797959) - P(\bar{X_{50}} \leq 6.90202041)$

Then in R I did pnorm(7.09797959,7,0.12) - pnorm(6.90202041,7,0.12) and got the result $0.5857838$

My question is what did I do wrong, and what should I be doing? This is the way he told us to do inequalities for probabilities like the one above. Also how could this be solved without having to use $\mu$ and $\sigma$?

Consider the more general probability $$\Pr\left[ \mu - z \frac{\sigma}{\sqrt{n}} < \bar X < \mu + z \frac{\sigma}{\sqrt{n}} \right],$$ for some $z > 0$, where $$\bar X = \frac{1}{n} \sum_{i=1}^n X_i$$ is the sample mean of $n$ independent and identically distributed observations from a normal distribution with mean $\mu$ and standard deviation $\sigma$. Then the sampling distribution of the sample mean is also normally distributed, with the same mean $$\mu_{\bar X} = \mu$$ but with standard deviation $$\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}.$$ Therefore, if we standardize $\bar X$, we find that the above probability is equivalent to $$\Pr\left[ -z < \frac{\bar X - \mu}{\sigma/\sqrt{n}} < z \right],$$ and the random variable $$Z = \frac{\bar X - \mu}{\sigma/\sqrt{n}}$$ is a standard normal random variable with mean $\mu_Z = 0$ and standard deviation $\sigma_Z = 1$. So the above probability is simply $$\Pr[|Z| < z] = 1 - 2 \Phi(-z) = 2\Phi(z) - 1.$$ Now you can see why your professor said that the values of $\mu$ and $\sigma$ do not matter: what matters is the constant $z = 2$. Looking up $\Phi(2) \approx 0.97725$ in a standard normal table, or using the R command pnorm(2,0,1), we then obtain the desired probability $$2(0.97725) - 1 = 0.9545,$$ as claimed.
Now, let's see why your calculation didn't work. First, we have to be clear about the difference between the normal distribution from which individual observations are drawn, and the normal distribution that represents the mean of the set of observations, which we called "the sampling distribution of the sample mean." Both distributions are normal, and both have the same mean, but the sampling distribution has a standard deviation that is a decreasing function of the sample size: $\sigma_{\bar X} = \sigma/\sqrt{n}$. Consequently, if individual observations follow a $\operatorname{Normal}(7,0.12)$ distribution, the sample mean of $n = 50$ such observations follows a $\operatorname{Normal}(7,0.0169706)$ distribution. This, in addition to an apparent arithmetic error, is why your pnorm() command is incorrect, because you are using the wrong standard deviation.
The correct R command should be
pnorm(7+2*0.12/sqrt(50),7,0.12/sqrt(50))-pnorm(7-2*0.12/sqrt(50),7,0.12/sqrt(50))