Probability and Statistics

[lucyle992]

Can someone check my answer, I'm not sure I did it right.

The Pew Research Center asked a random sample of 2024 adult cell-phone owners from the United States their age and which type of cell phone they own: Phone, Android, or other (including non-smartphones). The two-way table summarizes the data.

A) The person is not age 18-34 and does not own an iPhone?

My answer is: P(person is not age 18 and does not own an iPhone) = 1 - P(person is age 18-34 and owns an iPhone) = 1 - 169/2024 = 0.916

 

[Romsek]

No this isn't correct.

[MATH]\text{Let $A$ be the event the person is 18-34 and $B$ be the event that they use an Iphone}\\ P[\neg A \cap \neg B] = P[\neg(A \cup B)] = 1 - P[A \cup B] = 1 - (P[A]+P[ B]-P[A\cap B])[/MATH]
See if you can complete it. Alternatively (and just as easy) just add up the the folks that aren't 18-34 and don't use an Iphone
and divide it by the total number of people in the survey.

 

[lucyle992]

189+100+277+643=1209/2024=0.597