The lifetime of a certain electronic component is a random variable with the expectation of 5000 hours and a standard deviation of $200$ hours. Using Central Limit Theorem, find the probability that the average lifetime of $100$ components is less than $4650$ hours?
Let $X$ be a random variable which represents the lifetime if certain electronic components with mean of $500$ hours & standard deviation $200$ hours. $\mu=5000 $ hours $, \sigma= 200$ hours , n = $100$.
My main question is for the Standard Normal Distribution do I use the formula
.A Standard Normal variable, usually denoted by Z, can be obtained from a non-standard Normal($μ, σ)$ random variable X by standardizing, that is, subtracting the mean and dividing by the standard deviation,Using these transformations, any Normal random variable can be obtained from a Standard Normal variable Z; therefore, we need a table of Standard Normal Distribution only Table. $$Z= \frac{X-\mu}{\sigma}\\ X = μ + σZ.$$ or do I use $$Z = \frac{x̄ -\mu}{\frac{\sigma}{\sqrt{n}}}\\ $$.
Depending on the selected formula the answer changes accordingly..... Any advise?
This is a common source of confusion among students of statistics. The first formula, $$Z = \frac{X - \mu}{\sigma}, \tag{1}$$ follows from the fact that if $X$ is normally distributed with mean $\mu$ and standard deviation $\sigma$, then $Z$ will be normally distributed with mean $0$ and standard deviation $1$--i.e., $Z = (X - \mu)/\sigma$ is standard normal. As such, we call the transformation described in equation $(1)$ the standardization of $X$ or standardizing $X$.
The second formula, $$Z = \frac{\bar X - \mu}{\sigma/\sqrt{n}}, \tag{2}$$ is related to the first formula, except $X$ is replaced by $\bar X$ and $\sigma$ is replaced by $\sigma/\sqrt{n}$. This formula refers to what we call the standardization of the sampling distribution of the sample mean of $X$. Specifically, if $X_1, X_2, \ldots, X_n$ are independent and identically distributed random variables, each of which are normally distributed with mean $\mu$ and standard deviation $\sigma$, then the sample mean $$\bar X = \frac{1}{n} \sum_{i=1}^n X_i = \frac{X_1 + X_2 + \cdots + X_n}{n} \tag{3}$$ will be normally distributed with the same mean $\mu$ as $X$ itself, but the standard deviation will be smaller: it will be $\sigma/\sqrt{n}$ instead of $\sigma$. The reason should be intuitively obvious: the more data we observe in our sample, the more the sample mean will tend to converge toward the true mean $\mu$--in other words, the variance will decrease.
We call the distribution of $\bar X$ the sampling distribution of the sample mean. And so, when we standardize it, the appropriate transformation is shown in equation $(2)$.
Now, which formula to use depends on what you mean by $X$ and $\bar X$. If by $X$ you are referring to the lifetime of a single electronic component, then $$X \sim \operatorname{Normal}(\mu = 5000, \sigma = 200).$$ Then the sample mean $\bar X$ of $n = 100$ components will be distributed as follows: $$\bar X \sim \operatorname{Normal}(\mu = 5000, \sigma_{\bar X} = \sigma/\sqrt{n} = 200/\sqrt{100} = 20),$$ and so when we standardize the sample mean, it looks like this: $$Z = \frac{\bar X - \mu}{\sigma/\sqrt{n}} = \frac{\bar X - 5000}{20}. \tag{4}$$
But you could also have taken $X$ to represent the sample mean itself and use equation $(1)$ directly, which would give you the same result as equation $(4)$. You just have to remember that if $X$ represents the sample mean, then its standard deviation is $\sigma = 200/\sqrt{100} = 20$, rather than $200$. However, in practice, it is clearer in your specific case to define $X$ as the lifetime of a single component, and $\bar X$ as the sample mean of some number of components.