Normal Distribution - Statistics True or false statements


*

*The central limit theorem implies that any sample of any size can be approximated using
normal distribution.

*Let X be a normal random variable with mean 3 and variance 16. The value of k that
satisfies P(4 less than or equal to X less than or equal to k) = 0.1747, is k = 6.50.

*The random variable X is normally distributed with mean 2 and variance 4. Then the
probability that X is less or equal to 3 is 0.5199.

*A random variable, X, follows an exponential distribution with λ = 0.05. The random
variable Y = 2X . Thus, Y has mean equal to 0.1 and variance 0.01.
 A: *

*FALSE. The CLT will be a good approximation only if the sample size $n$ is big enough. Theoretically, $n\to\infty$.

*To answer the questions no. 2 and 3, let's brush up this thing first. If the random variable $X$ is normally distributed with mean $\mu$ and variance $\sigma^2$, in statistical term we call $X\sim\mathcal{N}(\mu,\sigma^2)$, then the random variable
$$
Z=\frac{X-\mu}{\sigma}
$$
will have a normal distribution with mean $0$ and variance $1$, in statistical term we call $Z$ has standard normal distribution, $Z\sim\mathcal{N}(0,1)$. The transformation is needed in this case to make it easy for us to check its probability by using standard normal or $Z$ table.$$$$$X\sim\mathcal{N}(3,16)$, then$$\begin{align}\text{Pr}[4\le X\le k]&=0.1747\\\text{Pr}\left[\frac{4-3}{4}\le\frac{X-\mu}{\sigma}\le\frac{k-3}{4}\right]&=0.1747\\\text{Pr}\left[0.25\le Z\le\frac{k-3}{4}\right]&=0.1747\\\text{Pr}\left[Z\le\frac{k-3}{4}\right]-\text{Pr}[Z\le0.25]&=0.1747\\\text{Pr}\left[Z\le\frac{k-3}{4}\right]-0.5987&=0.1747\\\text{Pr}\left[Z\le\frac{k-3}{4}\right]&=0.7734\\\frac{k-3}{4}&=0.7501\\k&=4(0.7501)+3\\&=6.004.\end{align}$$The answer is FALSE.

*$X\sim\mathcal{N}(2,4)$, then$$\begin{align}\text{Pr}[X\le 3]&=\text{Pr}\left[\frac{X-\mu}{\sigma}\le\frac{3-2}{2}\right]\\&=\text{Pr}[Z\le 0.5]\\&=0.6915.\end{align}$$The answer is FALSE.

*About the exponential distribution, I use Wikipedia as a reference. $X\sim\mathcal{E}(\lambda=0.05)$, then$$\begin{align}
\text{Pr}[X\le x]&=1-e^{-\lambda x}\\
\text{Pr}\left[\frac{Y}{2}\le x\right]&=1-e^{-\lambda x}\\
\text{Pr}[Y\le 2x]&=1-e^{-\lambda (2x)}\\
\text{Pr}[Y\le 2x]&=1-e^{-\lambda' x},\\
\end{align}$$ where $\lambda'=2\lambda=2(0.05)=0.1$. Therefore, $Y\sim\mathcal{E}(\lambda'=0.1)$. Thus, $$\mu=\text{E}[Y]=\dfrac{1}{\lambda'}=\dfrac{1}{0.1}=10$$and $$\sigma^2=\text{Var}[Y]=\dfrac{1}{\lambda'^2}=\dfrac{1}{0.1^2}=100.$$The answer is FALSE.


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$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
