Question

10.3 notetaking with vocabulary (continued)

1. in a survey, 214 ninth graders played video games every day of the week and 22 ninth graders did not play video games every day of the week. of those that played every day of the week, 36 had trouble sleeping at night. of those that did not play every day of the week, 7 had trouble sleeping at night. make a two - way table that shows the joint and marginal relative frequencies.
2. for financial reasons, a school district is debating about eliminating a computer programming class at the high school. the district surveyed parents, students, and teachers. the results as joint relative frequencies, are shown in the two - way table.

		population
response	yes	0.58	0.08	0.10
	no	0.06	0.15	0.03

a. what is the probability that a randomly selected parent voted to eliminate the class?
b. what is the probability that a randomly selected student did not want to eliminate the class?
c. determine whether voting to eliminate the class and being a teacher are independent?

Explanation:

Part 5a

Step1: Identify relevant data

We look at the two - way table. For parents, the joint relative frequency of voting "Yes" (to eliminate the class) is given as 0.58? Wait, no. Wait, the "Yes" in the response is for eliminating? Wait, the problem says "voted to eliminate the class". Wait, the table: Response "Yes" and "No" - we need to see what "Yes" means. The school district is debating about eliminating a Computer Pro class. So "Yes" would be voting to eliminate, "No" is not voting to eliminate. So for parents, the joint relative frequency for "Yes" (voting to eliminate) is 0.58? Wait, no, wait the table:

The table has Response (Yes/No) and Population (Parents/Students/Teachers). For Parents, Yes: 0.58, No: 0.06. Wait, but we need to find the probability that a randomly selected parent voted to eliminate the class. Wait, but first, we need to find the marginal relative frequency for parents? Wait, no, the joint relative frequencies are given, but to find the probability for a parent, we need to consider the parent's total. Wait, the marginal relative frequency for parents is the sum of their Yes and No: 0.58 + 0.06=0.64. But the joint relative frequency for parents voting Yes is 0.58. Wait, no, actually, in a two - way table with joint relative frequencies, the probability that a randomly selected parent voted Yes (to eliminate) is the joint relative frequency of (Parent, Yes) divided by the marginal relative frequency of Parent? Wait, no, no. Wait, joint relative frequencies are calculated as (frequency in cell)/(total number of observations). So the total of all joint relative frequencies should be 1. Let's check: 0.58 (Parent - Yes)+0.06 (Parent - No)+0.08 (Student - Yes)+0.15 (Student - No)+0.10 (Teacher - Yes)+0.03 (Teacher - No)=0.58 + 0.06=0.64; 0.08+0.15 = 0.23; 0.10+0.03=0.13; 0.64 + 0.23+0.13 = 1. So the marginal relative frequency for parents is 0.58+0.06 = 0.64. The joint relative frequency for parents who voted Yes is 0.58. But the probability that a randomly selected parent voted to eliminate the class is the conditional probability? Wait, no. Wait, the question is "What is the probability that a randomly selected parent voted to eliminate the class". So we are selecting a parent, so the sample space is all parents. The number of parents who voted Yes (to eliminate) divided by the total number of parents. But in terms of relative frequencies, the joint relative frequency of (Parent, Yes) is 0.58, and the marginal relative frequency of Parent is 0.58 + 0.06=0.64. Wait, no, that's not right. Wait, joint relative frequency is (count in cell)/total. So the probability that a randomly selected parent voted Yes is (joint relative frequency of Parent - Yes)/ (marginal relative frequency of Parent). Wait, no, marginal relative frequency of Parent is (number of parents)/total = 0.58+0.06 = 0.64. The joint relative frequency of Parent - Yes is 0.58. So the probability is 0.58/0.64? Wait, no, that's not correct. Wait, no, I think I made a mistake. Actually, in a two - way table with joint relative frequencies, the probability that a randomly selected parent voted Yes is the joint relative frequency of Parent - Yes divided by the marginal relative frequency of Parent? No, no. Wait, the joint relative frequency is (frequency of Parent - Yes)/N, where N is the total number of observations. The marginal relative frequency of Parent is (frequency of Parent)/N=(frequency of Parent - Yes + frequency of Parent - No)/N = 0.58+0.06 = 0.64. So the probability that a randomly selected parent (from all parents) voted Yes is (frequency of Pa…

Step1: Identify relevant data

We want the probability that a randomly selected student did not want to eliminate the class. So we look at the (Student, No) cell. The joint relative frequency for (Student, No) is 0.15, and the marginal relative frequency for students is the sum of (Student, Yes) and (Student, No), which is 0.08+0.15 = 0.23.

Step2: Calculate the probability

The probability \( P(\text{Student votes No})=\frac{\text{Joint relative frequency of (Student, No)}}{\text{Marginal relative frequency of Student}}=\frac{0.15}{0.08 + 0.15}=\frac{0.15}{0.23}=\frac{15}{23}\approx0.652 \)

Step1: Recall the independence formula

Two events \( A \) (voting to eliminate the class) and \( B \) (being a teacher) are independent if \( P(A\cap B)=P(A)\times P(B) \)

Step2: Calculate \( P(A) \), \( P(B) \), and \( P(A\cap B) \)

• Calculate \( P(A) \) (probability of voting to eliminate the class):

The total number of "Yes" votes (voting to eliminate) is the sum of (Parent - Yes), (Student - Yes), and (Teacher - Yes) joint relative frequencies. So \( P(A)=0.58 + 0.08+0.10 = 0.76 \)

• Calculate \( P(B) \) (probability of being a teacher):

The marginal relative frequency of teachers is the sum of (Teacher - Yes) and (Teacher - No) joint relative frequencies. So \( P(B)=0.10 + 0.03=0.13 \)

• Calculate \( P(A\cap B) \) (probability of being a teacher and voting to eliminate):

The joint relative frequency of (Teacher, Yes) is 0.10.

Step3: Check the independence condition

Calculate \( P(A)\times P(B)=0.76\times0.13 = 0.0988 \)
And \( P(A\cap B) = 0.10 \)
Since \( 0.0988
eq0.10 \), the events are not independent.

Answer:

\( \frac{29}{32} \) (or 0.90625)

Part 5b