Question

Math

Posted 6 months ago

```
Zainab runs her own business where she leads guided day-long tours for small groups of tourists in her city. Each tour has a maximum of 3 guests. Let $X$ represent the number of customers Zainab gets on a randomly chosen day. Based on previous data, here is the probability distribution of $X$ along with summary statistics:
\begin{tabular}{lrrrr}
\hline$X=$ \# of customers & 0 & 1 & 2 & 3 \\
$P(X)$ & 0.20 & 0.10 & 0.40 & 0.30 \\
& & & Mean: & $\mu_{X}=1.8$ \\
& & & Standard deviation: & $\sigma_{X} \approx 1.08$
\end{tabular}
It costs Zainab $\$ 5$ each day to pay for her own transportation to her business, and then each customer she gets pays her $\$ 60$. Let $Y$ represent Zainab's net gain on a randomly chosen day.
What are the mean and standard deviation of $Y$ ?
$\begin{array}{ll}
\mu_{Y}=\square \text { dollars } \\
\sigma_{Y} \approx \square \text { dollars }
\end{array}$
```

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Answer from Sia

Posted 6 months ago

Solution by Steps

step 1

Calculate the mean of $Y$ using the formula $\mu_Y = \mu_X \cdot 60 - 5$

step 2

Substitute the given mean of $X$ ($\mu_X = 1.8$) into the formula: $\mu_Y = 1.8 \cdot 60 - 5$

step 3

Perform the calculation: $\mu_Y = 108 - 5$

step 4

Simplify to find the mean of $Y$: $\mu_Y = 103$

step 5

Calculate the standard deviation of $Y$ using the formula $\sigma_Y = \sigma_X \cdot 60$

step 6

Substitute the given standard deviation of $X$ ($\sigma_X \approx 1.08$) into the formula: $\sigma_Y = 1.08 \cdot 60$

step 7

Perform the calculation to find the standard deviation of $Y$: $\sigma_Y \approx 64.8$

[1] Answer

$\mu_Y = 103$ dollars

$\sigma_Y \approx 64.8$ dollars

Key Concept

Linear transformation of a random variable

Explanation

The mean of a linearly transformed random variable $Y = aX + b$ is given by $\mu_Y = a\mu_X + b$, and the standard deviation is given by $\sigma_Y = |a|\sigma_X$, where $a$ and $b$ are constants. In this case, $a = 60$ and $b = -5$.

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