A factory produces links for heavy metal chains. The research lab of the factory models the length (in cm) of a link by the random variable ${X}$, with expected value ${E(X) = 5}$ and variance ${Var(X) = 0.04}$. The length of a link is defined in such a way that the length of a chain is equal to the sum of the lengths of its links. The factory sells chains of 50 meters; to be on the safe side 1002 links are used for such chains. The factory guarantees that the chain is not shorter than 50 meters. If by chance a chain is too short, the customer is reimbursed, and a new chain is given for free.  Give an estimate of the probability that for a chain of at least 50 meters more than 1002 links are needed. For what percentage of the chains does the factory have to reimburse clients and provide free chains?    Number of links in a chain: ${n = 1002}$  Mean lengh of 1002 links: ${ \mu = 1002 \times 0.05 = 50.1 }$  We know variance, so we can find the standard deviation: ${ \sigma = \sqrt{0.04} = 0.2 }$  We are interested how often chain is shorter then 50 meters: ${ \bar{x} = 50 }$  Thus let define an appropriate test statistic: ${ \displaystyle z = \frac{ \mu - \bar{x} }{ \sigma } = \frac{50.1 - 50}{0.2} = \frac{0.1}{0.2} = 0.5 }$  Probability of getting value in a lower tail is: ${ \displaystyle P(z &lt; 0.5) = \frac{1}{2} + \frac{1}{\sqrt{2\pi}} \int\limits_{0}^{0.5} e^{-\frac{z^2}{2}} dz }$  By making the substitution ${ z = \sqrt{2} x }$, we get:  ${ \displaystyle \frac{1}{\sqrt{2\pi}} \int\limits_{0}^{0.5} e^{-\frac{z^2}{2}} dz = \frac{1}{\sqrt{\pi}} \int\limits_{0}^{\frac{0.5}{\sqrt{2}}} e^{-x^2} dx }$  As we can not solve integral above, let's consider ${ e^{-x^2} }$ as a Taylor series: ${ \displaystyle \sum\limits_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!} }$  Then the approximate probability is: ${ \displaystyle P(z &lt; 0.5) = \frac{1}{2} + \frac{1}{\sqrt{\pi}} \sum\limits_{n=0}^{\infty} \frac{(-1)^n (\frac{c}{\sqrt{2}})^{2n+1} }{ n! (2n+1) } \simeq 0.691462 }$    But this probability looks too large. What I am doing wrong?
121463
A factory produces links for heavy metal chains. The research lab of the factory models the length (in cm) of a link by the random variable ${X}$, with expected value ${E(X) = 5}$ and variance ${Var(X) = 0.04}$. The length of a link is defined in such a way that the length of a chain is equal to the sum of the lengths of its links. The factory sells chains of 50 meters; to be on the safe side 1002 links are used for such chains. The factory guarantees that the chain is not shorter than 50 meters. If by chance a chain is too short, the customer is reimbursed, and a new chain is given for free.  Give an estimate of the probability that for a chain of at least 50 meters more than 1002 links are needed. For what percentage of the chains does the factory have to reimburse clients and provide free chains?    Number of links in a chain: ${n = 1002}$  Mean lengh of 1002 links: ${ \mu = 1002 \times 0.05 = 50.1 }$  We know variance, so we can find the standard deviation: ${ \sigma = \sqrt{0.04} = 0.2 }$  We are interested how often chain is shorter then 50 meters: ${ \bar{x} = 50 }$  Thus let define an appropriate test statistic: ${ \displaystyle z = \frac{ \mu - \bar{x} }{ \sigma } = \frac{50.1 - 50}{0.2} = \frac{0.1}{0.2} = 0.5 }$  Probability of getting value in a lower tail is: ${ \displaystyle P(z &lt; 0.5) = \frac{1}{2} + \frac{1}{\sqrt{2\pi}} \int\limits_{0}^{0.5} e^{-\frac{z^2}{2}} dz }$  By making the substitution ${ z = \sqrt{2} x }$, we get:  ${ \displaystyle \frac{1}{\sqrt{2\pi}} \int\limits_{0}^{0.5} e^{-\frac{z^2}{2}} dz = \frac{1}{\sqrt{\pi}} \int\limits_{0}^{\frac{0.5}{\sqrt{2}}} e^{-x^2} dx }$  As we can not solve integral above, let's consider ${ e^{-x^2} }$ as a Taylor series: ${ \displaystyle \sum\limits_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!} }$  Then the approximate probability is: ${ \displaystyle P(z &lt; 0.5) = \frac{1}{2} + \frac{1}{\sqrt{\pi}} \sum\limits_{n=0}^{\infty} \frac{(-1)^n (\frac{c}{\sqrt{2}})^{2n+1} }{ n! (2n+1) } \simeq 0.691462 }$    But this probability looks too large. What I am doing wrong?
90068
I wish to better understand how the  to the binomial distribution for the normal approximation was derived.  What method was used to decide we should add 1/2 (why not another number?). Any explanation (or a link to suggested reading, other than , would be appreciated).
110990
I was asked this question with $(n, k) = (400, 220)$ in an interview.  Is there a "correct" answer?  Assume the tosses are i.i.d. and the probability of heads is $p=0.5$.  The distribution of the number of heads in 400 tosses should then be close to Normal(200, 10^2), so that 220 heads is 2 standard deviations away from the mean.  The probability of observing such an outcome (i.e. more 2 SDs away from the mean in either direction) is slightly less than 5%.  The interviewer told me, essentially, "if I observe something >= 2 SDs from the mean, I conclude that something else is going on. I would bet against the coin being fair."  That's reasonable -- after all, that's what most hypothesis tests do. But is that the end of the story?  For the interviewer that seemed to be the "correct" answer.  What I'm asking here is whether some nuance is justified.  I couldn't help but point out that deciding that the coin is not fair is a bizarre conclusion in this coin-tossing context.  Am I right to say that?  I'll try and explain below.  First of all, I -- and I would assume most people as well -- have a strong prior about coins:  they're very likely to be fair.  Of course that depends on what we mean by fair -- one possibility would be to define "fair" as "having a probability of heads 'close' to 0.5, say between 0.49 and 0.51."  (You could also define 'fair' as meaning that the probability of heads is exactly 0.50, in which case having a perfectly fair coin now seems rather unlikely.)  Your prior might depend not only on your general beliefs about coins but also on the context. If you pulled the coin out of your own pocket, you might be virtually certain that it's fair; if your magician friend pulled it out of his, your prior might put more weight on double-headed coins.  In any case, it's easy to come up with reasonable priors that (i) put a large probability on the coin being fair and (ii) lead your posterior to be quite similar, even after observing 220 heads. You'd then conclude that the coin was very likely to be fair, despite observing an outcome 2 SDs from the mean.  In fact, you could also construct examples where observing 220 heads in 400 tosses makes your posterior put more weight on the coin being fair, for example if all unfair coins have a probability of heads in $\{0, 1\}$.  Can anyone shed some light on this for me?    After writing this question I remembered that I had heard about this general situation before -- isn't it ?  Whuber put a very interesting link in the comments: .  From page 3:     It does not make sense to say that the coin has a probability p of   heads, because it can be completely determined by the manner in which   it is tossed— unless it is tossed high in the air with a rapid spin   and caught in the air with no bouncing, in which case p = 1/2.   Pretty cool!  This ties into my question in an interesting way:  suppose we know that the coin is being "tossed high in the air with a rapid spin and caught in the air with no bouncing."  Then we definitely shouldn't reject the hypothesis that the coin is fair (where "fair" now means "having p=1/2 when tossed in the way described above"), because we effectively have a prior that puts all probability on the coin being fair. Maybe that justifies to some degree why I am uncomfortable rejecting the null after 220 heads are observed.
51308
There are $n$ coupons in a collection. A collector has the ability to purchase a coupon, but can't choose the coupon he purchases. Instead, the coupon is revealed to be coupon $i$ with probability $p_i=\frac 1 n$. Let $N$ be the number of coupons he'll need to collect before he has at least one coupon of each type. Find the expected value and variance of $N$. Bonus: generalize to the case where the probability of collecting the $j$th coupon is $p_j$ with $\sum\limits_{j=1}^n p_j=1$  I recently came across this problem and came up with/ unearthed various methods to solve it. I'm intending this page as a wiki with various solutions. I'll be posting all the solutions I'm aware of (4 so far) over time.    EDIT: As mentioned in the comments, this question is different from the one people are saying it is a duplicate of since (for one thing) it includes an expression for the variance and it covers the general case where all coupons have unequal probabilities. The case of calculating the variance for the general case of coupons having unequal probabilities has not been covered anywhere on the site apart from , which this one intends to consolidate along with other approaches to solve this problem.    EDIT: Paper on the solutions on this page submitted to ArXiv:
56137
There is a probabilistic method to solve it. But I am not familiar with probability. I am trying to compute it by analytic method, such as using L Hospital's rule or Stolz formula, but they are not working.
3733
What is the distribution of the square of a non-standard normal random variable (i.e., the mean is not equal to 0 and the variance is not equal to 1)?
95747
Suppose we have $\mu = 19.1$ and $\sigma = 1.7$.  I need to find $a \in \mathbb{R}$ such that: $P(|X-19.1|\leq a) = 0.98$ Now in the correction it is given that:  $$P(|X-19.1|\leq a) = 0.98 \iff P(\frac{-a}{1.7} \leq \frac{X - 19.1}{1.7} \leq \frac{a}{1.7})=0.98 \iff P(\frac{X-19.1}{1.7} \leq \frac{a}{1.7})=0.99$$  Can someone explain to me why is the last equality correct?
52420
Suppose $X_1, X_2, X_3, \ldots, X_n$ be independent and identically distributed random variables having an exponential distribution with mean $\frac{1}{\lambda}$.  If $S_n = X_1 + X_2 + \ldots + X_n$ and $N = \inf\{n \geq 1: S_n &gt; 1\}$, then what would be the value of $\mathrm{Var}(N)$ ?
80037
To construct the half-normal probability plot, plot the absolute values in a certain statistical diagnostic (residual, leverage, Cook distance and others) versus $z_i$ where:  $\displaystyle z_{i}  = \Phi^{-1} \left(\frac{k + n - \frac{1}{8}}{2n +\frac{1}{2}}\right)$   $\therefore$ Applied Linear Statistical Models, (Kutner,Nachtsheim,Neter &amp; Li, 2005) Fifth Edition, Pag 596  or  $\displaystyle z_{i} = \Phi^{-1} \left(\frac{k + n + \frac{1}{2}}{2n +\frac{9}{8}}\right)$  $\therefore$ Generalized Linear Models, (P. McCullagh &amp; J.A. Nelder, 1989) Second Edition,  Pag 407  where  $\Phi$ is the cumulative distribution function of the standard normal distribution.  $k$ is the  $k$-th ordered absolute residual   $n$ is the sample size  I could not locate the mathematical justification for the difference of the $z_i$, and more importantly I don't have any intuition about where it comes from.      Any help? Could you please guide me?
771
Let $X_1, X_2, \ldots X_n$ be random variables with mean $\mu$ and variance $\sigma^2$. We define the sample mean $\bar{X} = \sum_{i = 1}^n X_i$.  I am having trouble understanding why the variance of $\bar{X}$ is $\sigma^2 / n$.  Here is what I worked out thus far:  $$\begin{gather} \sigma_\bar{X}^2 = E((\bar{X} - \mu)^2) = E(\bar{X}^2 - 2\bar{X}\mu + \mu^2) = E(\bar{X}^2) - 2\mu E(\bar{X}) + E(\mu^2) = E(\bar{X}^2) - \mu. \end{gather}$$  However, I'm not sure what to do with the $E(\bar{X}^2)$.
86454
I am working on a paper, and have a problem regarding ellipsoid and Trivariate Normal Distribution. Suprisingly I can't find much in literature but I found your in one of your answers:     Because this construction has nothing to do with "confidence" per se,   the objective is to establish some convention for describing the shape   and relative size of the points. Using 1.96 sort of works (for three   variables): it contains about 72% of the probability of the trivariate   normal distribution. But as the number of variables increases this   method produces ellipses that are far too small. For instance, with 10   variables it will contain only 4.6% of the probability; using 4.28   instead of 1.96 in this case will contain 95% of the probability.   How did you get this number 72%? Or do you have some literature to recommend to me in which I can find this. I would appreciate it very much!
88708
Let there be a random variable $X$ and $N(0, \lambda ^2)$ given. I am to find $E[X^n]$ knowing that $\Gamma \big(\frac{1}{2} \big) = \frac{\pi}{2}$.     I started with the definition: $$E[X^n] = \frac{1}{\sqrt{2 \pi}\lambda}\int \limits_{- \infty}^{\infty}x^ne^{\frac{-x^2}{2 \lambda^2}} \mbox{d}x$$ I figured out that: $$E[X^n] = \begin{cases} 0 &amp;\text{for } n = 2k + 1; k \in \mathbb{N} \cup\{0\}\\\text{something}&amp;\text{for } n = 2k; k \in \mathbb{N} \end{cases}$$    So now: $$ \frac{1}{\sqrt{2 \pi}\lambda}\int \limits_{- \infty}^{\infty}x^ne^{\frac{-x^2}{2 \lambda^2}} \mbox{d}x =  \frac{2}{\sqrt{2 \pi}\lambda}\int \limits_{0}^{\infty}x^ne^{\frac{-x^2}{2 \lambda^2}} \mbox{d}x$$ I substitute: $$t = \frac{x}{\lambda}$$ $$\mbox{d}x = \lambda \mbox{d}t$$  Thus: $$\frac{2}{\sqrt{2 \pi}\lambda}\int \limits_{0}^{\infty}x^ne^{\frac{-x^2}{2 \lambda^2}} \mbox{d}x = \frac{2 \lambda^{n}}{\sqrt{2 \pi}}\int \limits_{0}^{\infty}t^ne^{\frac{-t^2}{2}} \mbox{d}t.$$ By integrating the formula above by parts I do get:  $$\frac{2 \lambda^{n}}{\sqrt{2 \pi}}(n-1) \int \limits_{0}^{\infty}t^{n-2}e^{\frac{-t^2}{2}} \mbox{d}t$$ I don't really know where to go from here. And unfortunately I don't see how can I use the Gamma Function, which is given by this formula: $$\Gamma (z) = \int \limits_{0}^{\infty} x^{z-1}e^{-x} \mbox{d}x,$$ isn't it?
76285
I am referring to this  page 198-9:    I have troubles understanding the case with unequal variance.  In general it :   or on :    Therefore if we assume unequal variance, I would think like this:  $s_{\bar{X}_T-\bar{X}_C}=\sqrt{s^2_T/n_1+s^2_C/n_2}$  Howerver, they give this for the standard error:    I have troubles with deriving these two different formulas.
43100
I'm having a little trouble with the proof that the expected value of $x_i$ is $ \bar{X} $.  What I have is   $E[x_i]=\sum_{j=1}^{N}X_j Pr(x_i=X_j) $  Then,  $Pr(x_i=X_j) = 1/N $  This is the bit I can't understand, how does that probability evaluate to that value.  I know the denominator is how many ways you can choose n out N. I think that the numerator should be how many ways you can choose (n-1) out of (N-1). But I seem to have an extra n.  Any advice ?
56651
I am trying to calculate the expected value of a Normal CDF, but I have gotten stuck. I want to find the expected value of $\Phi( \frac{a-bX}{c} )$ where $X$ is distributed as $\mathcal{N}(0,1)$ and $\Phi$ is the standard normal CDF.  I know I can transform $\frac{a-bX}{c}$ to be a normal random variable $\mathcal{N}\left(\frac{a}{c},\frac{b^2}{c^2}\right)$ where $\frac{b^2}{c^2}$ is the variance of the normal random variable. I'm not sure if this helps though.  I think that the expected value of a CDF is $0.5$ but since $\Phi$ is the CDF of a standard normal CDF and $\frac{a-bX}{c}$ is not standard normal I do not think the expected value should be 0.5. I tried integrating the CDF, but I do not believe I did it correctly.  When $a = -2.3338$, $b = 0.32164$, $c = 0.94686$, I believe the correct answer is approximately $0.009803$. I found this through simulation.  I would appreciate any help or suggestions.
19884
I was asked this question today in an interview and am still not sure of the answer. The question is as follows.   Part one:  Say I have flipped 100 coins already and they are hidden to you. Now I reveal these coins one by one and you are to guess if it is heads or tails before I reveal the coin. If you get it correct, you get $\$1$. If you get it incorrect, you get $\$0$. I will allow you to ask me one yes or no question about the sequence for free. What will it be to maximize your profit?  My approach for this part of the problem was to ask if there were more heads than tails. If they say yes, I will just guess all heads otherwise I just guess all tails. I know the expected value for this should be greater than 50 but is it possible to calculate the exact value for this? If so, how would you do it?   Part two: Same scenario as before but now I will charge for a question. I will allow you to ask me any amount of yes or no questions as I go through this process for $\$1$. What is your strategy to maximize your profit?  I was not sure about the answer to this part of the question. Would the best option be to guess randomly? I think the expected value of this should be 50. I am not sure about the expected value of part one but if it is greater than 51, I think I could also use that approach. Anyone have a good idea for this part?
115707
The exercise I'm doing describes the random variable $X$ as the following  | Number of cars |  0  |  1 |  2 |  3 |  4+ |  | % of families  |  15 | 45 | 25 | 13 |  2  |   Then it asks me to evaluate $E[X]$. But if there could be cases with 4+ cars, wouldn't that make it impossible to calculate the expectancy?