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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.

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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?

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  • $\begingroup$ 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
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For the expected value we have

$$E(X)=0.3\cdot2+0.7\cdot0.5=0.95$$

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

Then

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

Finally

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

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.

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  • $\begingroup$ 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

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