Calculating a normal distribution with a sample size?

Question

the sample of $n=25$ is what is throwing me off. I have no clue what to do with it.

Given a normal distribution with $\mu=101, \sigma=25$, and given you select of $n=25$

$A.)$ $P(\overline{X} <93) = \text{ ?} $.

Answer 1

hint: $P\left(\overline{X} < 93\right) = P\left(z < \dfrac{93-101}{\frac{25}{\sqrt{25}}}\right) = .....$. Can you work it out?

Answer 2

The key concept is that the distribution from which individual observations are drawn generally is not the same as the sampling distribution of the statistic.

If I give you a coin whose probability $p$ of landing heads is known only to me, you can try to make an inference about $p$ by tossing the coin many times. The outcome of a single toss, from your perspective, is a Bernoulli random variable where the probability of $1$ (heads) is $p$.

Your estimate of the true value of $p$ is the sample proportion of the number of heads you observed, divided by the number of times you tossed the coin; that is, $\hat p = X/n$, where $X \sim \operatorname{Binomial}(n,p)$ is a binomial random variable that counts the number of heads tossed. But $\hat p$ is not a binomial random variable and it isn't a Bernoulli variable either.

So, the sampling distribution of the statistic $\hat p$ is not the same as the distribution from which the observations were drawn (the individual coin flips).

The same notion applies to observations of a normal random variable. If $X_i \sim \operatorname{Normal}(\mu,\sigma)$ for $i = 1, 2, \ldots, n$ are independent and identically distributed observations from a normal distribution with mean $\mu$ and standard deviation $\sigma$, and $\bar X = (X_1 + X_2 + \cdots + X_n)/n$ is the sample mean, then $\bar X \sim \operatorname{Normal}(\mu, \sigma/\sqrt{n})$; that is to say, the sampling distribution is also normal with the same mean, but with a standard deviation that is smaller (and gets smaller as the sample size increases). This makes sense: if I assume that the heights of adult females in France is normally distributed with mean $\mu = 167$ cm and standard deviation $\sigma = 7.6$ cm, then the variability of a single adult female chosen at random from this population is modeled by $\sigma$. But if I sample $1000$ such females at random and average their heights, you would expect much less variability in their mean height because the individual variability from person to person tends to "cancel" each other. Some will be shorter than average, others will be taller than average. And if I sample even more people, the better I would expect the resulting sample mean to estimate the true population mean.