Answer
P(four to six (inclusive) rooms) $=0.183+0.230+0.204=0.617$
If we randomly select 1000 U.S. housing units, we would expect about 617 to be a housing unit that has four to six (inclusive) rooms.
Work Step by Step
The event "from four to six (inclusive) rooms" = "four or five or six rooms".
The events "four rooms", "five rooms" and "six rooms" are mutually exclusives. So:
P(four to six (inclusive) rooms) = P(four or five or six rooms) =
=P(four)+P(five)+P(six) $=0.183+0.230+0.204=0.617.$
