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True or false statements

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

  2. 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.

  3. 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.

  4. 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.

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  • $\begingroup$ What is the question?. You seem to have posted the answeres $\endgroup$ – user130512 Apr 4 '14 at 14:13
  • $\begingroup$ Is this a jeopardy question, where we're to figure out the question given the answers :)? Also, the central limit theorem does not imply that any sample can be approximated by a normal distribution (which I guess is supposed to mean that the empirical distribution of a sample is approximately normal), it merely states that certain kinds of averaging operations yield (asymptotically) normally distributed things, when applied to sufficiently well behaved distributions. $\endgroup$ – Tilo Wiklund Apr 4 '14 at 14:17
  • $\begingroup$ Hi @TiloWiklund, I forgot to add, being inherently clumsy, that the above questions are of a true or false format. So I have to ascertain if each question is the former or latter. Although I'm sure the answer you have provided with regard to the central limit theorem is sound, I'm still struggling to understand it. I'm finding this particular module, Statistics, very difficult at the moment, mainly because my tutor is not very competent. If possible could you reply with a more simplified explanation of the reason why the first question is false. Thanks for the speedy and helpful reply, too! $\endgroup$ – Scott Goddard Apr 4 '14 at 15:28
  • $\begingroup$ @ScottGoddard Getting into details about CLT gets kind of technical, esp. once you get to its generalisations. For the common case replace "averaging operations" by "sample mean of independent variables" (say, the average of multiple independent measurements of some fixed quantity to reduce instrument noise). If you do this multiple times for different sets of independent observations, you get multiple such "averages", the CLT now states that as the number of observations in each sample gets large, the collections of averages look more and more like being taken from a normal distribution. $\endgroup$ – Tilo Wiklund Apr 4 '14 at 15:49
  • $\begingroup$ Note that one has to impose some requirements on the distribution you're taking averages of the observations of (which basically make sure that taking averages makes any sense to begin with, check wikipedia for details). $\endgroup$ – Tilo Wiklund Apr 4 '14 at 15:50
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  1. FALSE. The CLT will be a good approximation only if the sample size $n$ is big enough. Theoretically, $n\to\infty$.
  2. 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.

  3. $X\sim\mathcal{N}(2,4)$, 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.

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

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