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