University Survey Results on Student Government Satisfaction

A university conducted a survey of 379 undergraduate students regarding satisfaction with student government. Results of the survey are shown in the table by class rank. Complete parts​ (a) through​ (d) below. Freshman Sophomore Junior Senior Satisfied 52 52 60 57 Neutral 26 20 20 10 Not satisfied 21 22 11 28 ​(a) If a survey participant is selected at​ random, what is the probability that he or she is satisfied with student​ government? ​P(satisfied)equals    enter your response here ​(Round to three decimal places as​ needed.) Part 2 ​(b) If a survey participant is selected at​ random, what is the probability that he or she is a​ junior? ​P(junior)equals    enter your response here ​(Round to three decimal places as​ needed.) Part 3 ​(d) If a survey participant is selected at​ random, what is the probability that he or she is satisfied or a​ junior? ​P(satisfied or ​junior)equals    enter your response here ​(Round to three decimal places as​ needed.)

To solve the problems, we first need to calculate the total number of students in each category and then use that information to find the probabilities.

Given Data:

• Freshman:
  • Satisfied: 52
  • Neutral: 26
  • Not satisfied: 21
• Sophomore:
  • Satisfied: 52
  • Neutral: 20
  • Not satisfied: 22
• Junior:
  • Satisfied: 60
  • Neutral: 20
  • Not satisfied: 11
• Senior:
  • Satisfied: 57
  • Neutral: 10
  • Not satisfied: 28

Total Students:

To find the total number of students surveyed, we sum all the students in each category:

[ \text{Total} = (52 + 26 + 21) + (52 + 20 + 22) + (60 + 20 + 11) + (57 + 10 + 28) ]

Calculating each class:

• Freshman: (52 + 26 + 21 = 99)
• Sophomore: (52 + 20 + 22 = 94)
• Junior: (60 + 20 + 11 = 91)
• Senior: (57 + 10 + 28 = 95)

Now summing these totals:

[ \text{Total} = 99 + 94 + 91 + 95 = 379 ]

Part (a): Probability that a participant is satisfied

To find the probability that a randomly selected participant is satisfied, we sum the number of satisfied students:

[ \text{Satisfied} = 52 + 52 + 60 + 57 = 221 ]

Now, we calculate the probability:

[ P(\text{satisfied}) = \frac{\text{Number of satisfied students}}{\text{Total number of students}} = \frac{221}{379} ]

Calculating this gives:

[ P(\text{satisfied}) \approx 0.583 ]

Rounded to three decimal places:

[ P(\text{satisfied}) \approx 0.583 ]

Part (b): Probability that a participant is a junior

To find the probability that a randomly selected participant is a junior, we need the total number of juniors:

[ \text{Juniors} = 60 + 20 + 11 = 91 ]

Now, we calculate the probability:

[ P(\text{junior}) = \frac{\text{Number of juniors}}{\text{Total number of students}} = \frac{91}{379} ]

Calculating this gives:

[ P(\text{junior}) \approx 0.240 ]

Rounded to three decimal places:

[ P(\text{junior}) \approx 0.240 ]

Part (d): Probability that a participant is satisfied or a junior

To find (P(\text{satisfied or junior})), we use the formula for the union of two events:

[ P(A \cup B) = P(A) + P(B) - P(A \cap B) ]

Where:

• (A) is the event of being satisfied.
• (B) is the event of being a junior.

We already have:

• (P(A) = P(\text{satisfied}) \approx 0.583)
• (P(B) = P(\text{junior}) \approx 0.240)

Now we need (P(A \cap B)), which is the probability of being both satisfied and a junior. The number of satisfied juniors is 60.

[ P(A \cap B) = \frac{60}{379} ]

Calculating this gives:

[ P(A \cap B) \approx 0.158 ]

Now we can calculate (P(A \cup B)):

[ P(\text{satisfied or junior}) = P(\text{satisfied}) + P(\text{junior}) - P(\text{satisfied and junior}) ]

Substituting the values:

[ P(\text{satisfied or junior}) \approx 0.583 + 0.240 - 0.158 ]

Calculating this gives:

[ P(\text{satisfied or junior}) \approx 0.665 ]

Rounded to three decimal places:

[ P(\text{satisfied or junior}) \approx 0.665 ]

Summary of Results:

• (a) (P(\text{satisfied}) \approx 0.583)
• (b) (P(\text{junior}) \approx 0.240)
• (d) (P(\text{satisfied or junior}) \approx 0.665)