Suppose 80% of people who buy a new car say they are satisfied with the car when surveyed one year after purchase. Let X be the number of people in a group of 60 randomly chosen new car owners who report satisfaction with their car. Let Y be the number of satisfied owners in a second (independent) survey of 62 randomly chosen new car buyers. Using a suitable approximation, find $ P(|X-Y| \geq 3) $. A continuity correction is expected.
We have $X - Bi(60,0.8) \Rightarrow X - N(48,9.6)$ and $Y - Bi(62,0.8) \Rightarrow Y - N(49.6,9.92) $ so $(X-Y) - N(-1.6,19.52)$. Therefore, we have $$ P(|X-Y| \geq 3) = 1 - P(|X-Y| < 3) = 1 - P(-3 < X - Y < 3) = 1 - P(-3.5 < X - Y < 3.5) = 1 - P(-0.43 < Z < 1.15) = 0.45867$$ But the answer given is 0.5969. Any help would be appreciated.
The event $[-3 < X - Y < 3]$ is $[-2.5 < X - Y < 2.5]$, not $[-3.5 < X - Y < 3.5]$. Using $[-3.5 < X - Y < 3.5]$ yields $P(|X-Y|\gt3)=P(|X-Y|\geqslant4)$, not $P(|X-Y|\geqslant3)$.