a two - way frequency table shows grades for students in college and students in high school:\n| | high…

a two - way frequency table shows grades for students in college and students in high school:\n| | high school | college | total |\n|--|--|--|--|\n| gpa above 3.0 | 14 | 26 | 40 |\n| gpa below 3.0 | 46 | 14 | 60 |\n| | 60 | 40 | 100 |\nbased on this data, are \being in high school\ and \gpa above 3.0\ independent events?\nyes, p(high school ∩ gpa above 3.0) = p(high school) · p(gpa above 3.0)\nno, p(high school ∩ gpa above 3.0) = p(high school) · p(gpa above 3.0)\nyes, p(high school ∩ gpa above 3.0) ≠ p(high school) · p(gpa above 3.0)\nno, p(high school ∩ gpa above 3.0) ≠ p(high school) · p(gpa above 3.0)

a two - way frequency table shows grades for students in college and students in high school:\n| | high school | college | total |\n|--|--|--|--|\n| gpa above 3.0 | 14 | 26 | 40 |\n| gpa below 3.0 | 46 | 14 | 60 |\n| | 60 | 40 | 100 |\nbased on this data, are \being in high school\ and \gpa above 3.0\ independent events?\nyes, p(high school ∩ gpa above 3.0) = p(high school) · p(gpa above 3.0)\nno, p(high school ∩ gpa above 3.0) = p(high school) · p(gpa above 3.0)\nyes, p(high school ∩ gpa above 3.0) ≠ p(high school) · p(gpa above 3.0)\nno, p(high school ∩ gpa above 3.0) ≠ p(high school) · p(gpa above 3.0)

Answer

Explanation:

Step1: Calculate P(high school)

$P(\text{high school})=\frac{60}{100} = 0.6$

Step2: Calculate P(GPA above 3.0)

$P(\text{GPA above 3.0})=\frac{40}{100}=0.4$

Step3: Calculate P(high school ∩ GPA above 3.0)

$P(\text{high school}\cap\text{GPA above 3.0})=\frac{14}{100} = 0.14$

Step4: Calculate P(high school)×P(GPA above 3.0)

$P(\text{high school})\times P(\text{GPA above 3.0})=0.6\times0.4 = 0.24$

Step5: Compare the two - probabilities

Since $0.14\neq0.24$, i.e., $P(\text{high school}\cap\text{GPA above 3.0})\neq P(\text{high school})\times P(\text{GPA above 3.0})$, the events are not independent.

Answer:

No, P(high school ∩ GPA above 3.0) ≠ P(high school)·P(GPA above 3.0)