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Math
Posted 3 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 XX represent the number of customers Zainab gets on a randomly chosen day. Based on previous data, here is the probability distribution of XX along with summary statistics:
\begin{tabular}{lrrrr}
\hlineX=X= \# of customers & 0 & 1 & 2 & 3 \\
P(X)P(X) & 0.20 & 0.10 & 0.40 & 0.30 \\
& & & Mean: & μX=1.8\mu_{X}=1.8 \\
& & & Standard deviation: & σX1.08\sigma_{X} \approx 1.08
\end{tabular}

It costs Zainab $5\$ 5 each day to pay for her own transportation to her business, and then each customer she gets pays her $60\$ 60. Let YY represent Zainab's net gain on a randomly chosen day.

What are the mean and standard deviation of YY ?
μY= dollars σY dollars 
\begin{array}{ll}
\mu_{Y}=\square \text { dollars } \\
\sigma_{Y} \approx \square \text { dollars }
\end{array}
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Answer from Sia
Posted 3 months ago
Solution by Steps
step 1
Calculate the mean of YY using the formula μY=μX605\mu_Y = \mu_X \cdot 60 - 5
step 2
Substitute the given mean of XX (μX=1.8\mu_X = 1.8) into the formula: μY=1.8605\mu_Y = 1.8 \cdot 60 - 5
step 3
Perform the calculation: μY=1085\mu_Y = 108 - 5
step 4
Simplify to find the mean of YY: μY=103\mu_Y = 103
step 5
Calculate the standard deviation of YY using the formula σY=σX60\sigma_Y = \sigma_X \cdot 60
step 6
Substitute the given standard deviation of XX (σX1.08\sigma_X \approx 1.08) into the formula: σY=1.0860\sigma_Y = 1.08 \cdot 60
step 7
Perform the calculation to find the standard deviation of YY: σY64.8\sigma_Y \approx 64.8
[1] Answer
μY=103\mu_Y = 103 dollars
σY64.8\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+bY = aX + b is given by μY=aμX+b\mu_Y = a\mu_X + b, and the standard deviation is given by σY=aσX\sigma_Y = |a|\sigma_X, where aa and bb are constants. In this case, a=60a = 60 and b=5b = -5.

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