Question

a) \

$$\begin{tabular}{c|ccc}x & 0 & 1 & 2 \\\\ \\hline p(x = x) & 0.2 & 0.4 & 0.4 \\end{tabular}$$

b) \

$$\begin{tabular}{c|cccc}x & 100 & 200 & 300 & 400 \\\\ \\hline p(x = x) & 0.1 & 0.2 & 0.5 & 0.2 \\end{tabular}$$

1. swing club leslie the park manager wants to determine if there are enough swings to go around in one of the local parks. she goes to the park and develops the following probability model for the usage level of the six swings. how many swings are in use, on average?

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$$\begin{tabular}{l|ccccccc}number of swings in use & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\\\ \\hline probability & 0.12 & 0.19 & 0.26 & 0.28 & 0.10 & 0.03 & 0.02 \\end{tabular}$$

1. caffeinated a coffee shop tracks sales and has observed the distribution in the following table. what is the average daily sales that it can expect?

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$$\begin{tabular}{l|ccccc}\\# of sales & 145 & 150 & 155 & 160 & 170 \\\\ \\hline probability & 0.15 & 0.22 & 0.37 & 0.19 & 0.07 \\end{tabular}$$

1. swing some more what is the standard deviation of exercise 3?
2. caffeinated again what is the standard deviation for exercise 4?
3. make a bet? your crafty teacher makes every student in the class a simple offer. your teacher has an octahedron-of-chance (an eight - sided die). every student rolls once. if you roll an 8, you get five extra credit points on the next test. if you roll anything else you roll again. if on the second roll, you roll a number 3 or higher, you get a two - point bonus on the test, but if not, you lose one point! (this is not legal in most states.)

a) create a probability model for the number of points a student in the class will receive/lose.
b) find the expected number of points the teacher will award to students in this class on average.
c) would you play? why or why not?

1. you bet! you roll a die. if it comes up a 6, you win $100. if not, you get to roll again. if you get a 6 the second time, you win $50. if not, you lose.

Explanation:

Step1: Recall the formula for the expected value of a discrete random variable

The expected value \( E(X) \) of a discrete random variable \( X \) is calculated as \( E(X)=\sum_{i}x_{i}P(X = x_{i}) \), where \( x_{i} \) are the possible values of \( X \) and \( P(X = x_{i}) \) are their corresponding probabilities.

Step2: Identify the values of \( x_i \) and \( P(X = x_i) \) from the table

For the "Caffeinated" problem, the number of sales (\( x_i \)) are 145, 150, 155, 160, 170 and their probabilities (\( P(X = x_i) \)) are 0.15, 0.22, 0.37, 0.19, 0.07 respectively.

Step3: Calculate each term \( x_iP(X = x_i) \)

• For \( x = 145 \): \( 145\times0.15 = 21.75 \)
• For \( x = 150 \): \( 150\times0.22 = 33 \)
• For \( x = 155 \): \( 155\times0.37 = 57.35 \)
• For \( x = 160 \): \( 160\times0.19 = 30.4 \)
• For \( x = 170 \): \( 170\times0.07 = 11.9 \)

Step4: Sum up all the terms

\( E(X)=21.75 + 33+57.35 + 30.4+11.9 \)
\( E(X)=21.75+33 = 54.75 \); \( 54.75 + 57.35=112.1 \); \( 112.1+30.4 = 142.5 \); \( 142.5+11.9 = 154.4 \)

Answer:

The average daily sales the coffee shop can expect is \( \boldsymbol{154.4} \)