Question

1. the recycling company has added a new feature to their trucks that weighs and records the total number of kilograms of each household recycling bin they collect on their daily routes around san diego. the recycling company analyzed the data it collected last week and found the distribution of weights of household recycling bins to be skewed right with a mean of 7.4 kg and standard deviation of 2.8 kg. let the random variable r represent the weight, in kilograms, of a randomly selected household recycling bin. the weight of this bin can be converted into pounds using the formula p = 2.2r. which of the following best describes the distribution of p?

a.) roughly symmetric with mean 16.28 pounds and standard deviation 6.16 pounds
b.) roughly symmetric with mean 16.28 pounds and standard deviation 2.8 pounds
c.) skewed to the right with mean 16.28 pounds and standard deviation 4.15 pounds
d.) skewed to the right with mean 16.28 pounds and standard deviation 2.8 pounds
e.) skewed to the right with mean 16.28 pounds and standard deviation 6.16 pounds

1. jerry owns a bike shop and has determined the mean and standard deviation for the number of youth and adult bikes he sells each day. both distributions can be modeled using a normal distribution, and the sale of the two types of bikes are independent of one another.

	mean	standard deviation
youth bikes	4.2	1.7
adult bikes	5.1	1.9

determine the likelihood that jerry sells more youth bikes than adult bikes on a randomly selected day.
a.) 0.031 b.) 0.284 c.) 0.362 d.) 0.391 e.) 0.415

1. you’re playing a game in which you win points by rolling a die and spinning a spinner. a fair six sided die is rolled and a fair spinner labeled with 1 - 10 is spun. if your points are determined by adding together the value of the die with the value of the spinner, what is the expected value of the sum?

a.) 8 points b.) 9 points c.) 10 points d.) 11 points e.) 12 points

Explanation:

Question 8 (Recycling Bin Weights)

Step1: Analyze Shape of Distribution

When a random variable is transformed by a linear transformation \( P = aR + b \) (here \( a = 2.2 \), \( b = 0 \)), the shape of the distribution remains the same. Since \( R \) is skewed right, \( P \) will also be skewed right.

Step2: Calculate Mean of \( P \)

The mean of a linear transformation \( P = aR + b \) is \( \mu_P = a\mu_R + b \). Given \( \mu_R = 7.4 \), \( a = 2.2 \), \( b = 0 \), so \( \mu_P = 2.2\times7.4 = 16.28 \) pounds.

Step3: Calculate Standard Deviation of \( P \)

The standard deviation of a linear transformation \( P = aR + b \) is \( \sigma_P = |a|\sigma_R \). Given \( \sigma_R = 2.8 \), \( a = 2.2 \), so \( \sigma_P = 2.2\times2.8 = 6.16 \) pounds.

Step1: Define the Difference Variable

Let \( Y \) be the number of youth bikes sold and \( A \) be the number of adult bikes sold. We want \( P(Y > A) \), which is equivalent to \( P(Y - A > 0) \).

Step2: Find Mean of \( Y - A \)

The mean of \( Y - A \) is \( \mu_{Y - A}=\mu_Y - \mu_A \). Given \( \mu_Y = 4.2 \), \( \mu_A = 5.1 \), so \( \mu_{Y - A}=4.2 - 5.1=- 0.9 \).

Step3: Find Standard Deviation of \( Y - A \)

Since \( Y \) and \( A \) are independent, the standard deviation of \( Y - A \) is \( \sigma_{Y - A}=\sqrt{\sigma_Y^2+\sigma_A^2} \). Given \( \sigma_Y = 1.7 \), \( \sigma_A = 1.9 \), so \( \sigma_{Y - A}=\sqrt{1.7^2 + 1.9^2}=\sqrt{2.89+3.61}=\sqrt{6.5}\approx2.55 \).

Step4: Standardize the Variable

We standardize \( Z=\frac{(Y - A)-\mu_{Y - A}}{\sigma_{Y - A}} \). We want \( P(Y - A>0) \), so \( Z=\frac{0 - (-0.9)}{2.55}=\frac{0.9}{2.55}\approx0.353 \).

Step5: Find Probability from Z - table

Looking up \( Z = 0.35 \) (close to 0.353) in the standard normal table, the area to the left of \( Z = 0.35 \) is approximately 0.6368, so the area to the right (which is \( P(Y - A>0) \)) is \( 1 - 0.6368 = 0.3632\approx0.362 \).

Step1: Find Expected Value of Die Roll

A fair six - sided die has values \( 1,2,3,4,5,6 \). The expected value of a die roll, \( E(D) \), is \( \frac{1 + 2+3+4+5+6}{6}=\frac{21}{6}=3.5 \).

Step2: Find Expected Value of Spinner

A fair spinner labeled 1 - 10 has values \( 1,2,\cdots,10 \). The expected value of the spinner, \( E(S) \), is \( \frac{1 + 2+\cdots+10}{10}=\frac{\frac{10\times(10 + 1)}{2}}{10}=\frac{55}{10}=5.5 \).

Step3: Find Expected Value of Sum

The expected value of the sum \( E(D + S)=E(D)+E(S) \). So \( E(D + S)=3.5 + 5.5 = 9 \)? Wait, no, wait: Wait, die is 1 - 6, expected value \( \frac{1 + 2+3+4+5+6}{6}=3.5 \), spinner 1 - 10, expected value \( \frac{1 + 2+\cdots+10}{10}=5.5 \). Then \( 3.5+5.5 = 9 \)? Wait, no, the options have 9 as option b? Wait, no, let's recalculate: Wait, die: \( 1 + 2+3+4+5+6 = 21 \), \( 21\div6 = 3.5 \). Spinner: \( 1+2+\cdots+10=\frac{10\times(10 + 1)}{2}=55 \), \( 55\div10 = 5.5 \). Sum: \( 3.5+5.5 = 9 \). Wait, but let's check again. Wait, the die is six - sided, spinner is 1 - 10. So expected value of die: \( E(D)=\frac{1 + 2+3+4+5+6}{6}=3.5 \), expected value of spinner: \( E(S)=\frac{1 + 2+\cdots+10}{10}=5.5 \). Then \( E(D + S)=E(D)+E(S)=3.5 + 5.5 = 9 \).

Answer:

e.) Skewed to the right with mean 16.28 pounds and standard deviation 6.16 pounds

Question 9 (Bike Shop Sales)