The tip $ X $ (in Euros) that a waiter in a cafe / restaurant receives from a guest is distributed as follows:

  • If the guest orders a hot meal, $X$ has expectation $2$ and variance $2$

  • If the guest does not order a hot meal (for example, just a drink), $X$ has expectation $0, 5$ and variance $0, 3$

The probability $p$ that a guest orders a hot meal is $ 0, 3$.

a) Calculate the expected value $\mu_X$ and the variance $\sigma^2$ of $X$.

b) Determine the (approximative) propability that the waiter of $n=100$ guesrs gets less that $80$ Euro tips.


At the question a) do we have to define a new random variable that is a linear combination of the random variables $X_1, X_2$, where these two describe the tips if the guest orders a meal or a drink respectively?

  • $\begingroup$ We usually call $X$ has a mixture distribution, which is different from a linear combination of $X_1, X_2$. But the expectation of $X$ indeed is a linear combination of the expectation of $X_1, X_2$. $\endgroup$ – BGM Apr 9 '18 at 4:00

For the expected value we have


By BGM's comment, this is a mixture distribution. Let $p_A$ denote the probability of ordering a hot meal and $p_B$ denote the probability of not ordering a hot meal. Then by this we get

$$\begin{align*} \sigma_X^2 &=p_A\sigma_A^2+p_B\sigma_B^2+p_Ap_B(\mu_A−\mu_B)^2\\\\ &=0.3\cdot2+0.7\cdot0.3+0.3\cdot0.7(2-0.5)^2\\\\ &=1.2825 \end{align*}$$

Thus, $X\sim N(0.95,1.2825)$.

Let $$Z=\sum_{i=1}^{100}X_i$$


$$Z\sim N(100\cdot0.95, 100\cdot1.2825)$$


$$P(Z\lt80)=\Phi\left(\frac{80-95}{\sqrt{100\cdot1.2825}}\right)\approx 0.0927$$

In R statistical software

> pnorm((80-95)/(sqrt(100*1.2825)))
[1] 0.09266315

Note: Must be checked as this is the first I've heard of a mixture distribution.

  • $\begingroup$ Let $X_1, X_2$ be the random variables as I described in my original post. Does it hold that $X=p_A\cdot X_1+p_B\cdot X_2$ ? If this is true then we have that $E(X)=E(p_A\cdot X_1+p_B\cdot X_2)=p_A\cdot E(X_1)+p_B\cdot E(X_2)$. Is this correct? $\endgroup$ – Mary Star Apr 9 '18 at 7:22
  • $\begingroup$ If it is like that, can we calculate the variance as follows? \begin{align*}\sigma_X^2&=V(X) \\ & =V(0.3\cdot X_1+0.7\cdot X_2) \\ & =0.3^2\cdot V(X_1)+0.7^2\cdot V(X_2)+2\cdot 0.3\cdot 0.7\cdot Cov(X_1, X_2) \\ & =0.09\cdot V(X_1)+0.49\cdot V(X_2)+0.42\cdot Cov(X_1, X_2) \\ & =0.09\cdot 2+0.49\cdot 0.3+0.42\cdot Cov(X_1, X_2) \\ & =0.18+0.147+0.42\cdot Cov(X_1, X_2) \\ & =0.327+0.42\cdot Cov(X_1, X_2)\end{align*} But how could we calculate the covariance? $\endgroup$ – Mary Star Apr 9 '18 at 7:50
  • $\begingroup$ Is the Random variable normally distributed? Or how do we know that $Z$ is normally distributed? $\endgroup$ – Mary Star Apr 9 '18 at 9:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.