Let the random variable X be the number of rooms in a randomly chosen owner-occupied housing unit in a certain city. The distribution for the units is given below. X 3 4 5 6 7 8 9 10P(X) 0.1 0.24 0.33 0.17 0.05 0.05 0.04 ?(a) Is X a discrete or continuous random variable? (Type: DISCRETE or CONTINUOUS)ANSWER:(b) What must be the probability of choosing a unit with 10 rooms? P(X = 10) =(c) What is the probability that a unit chosen at random has more than 5 rooms? P(X > 5) =(d) What is the probability that a unit chosen at random has three rooms? P(X=3) =(e) What is the probability that a unit chosen at random has between four and six rooms? P(4 \leq X \leq 6)

a) DISCRETE

b) P(X=10)=0.04

c) P(X>5)=0.64

d) P(X=3)=0.1

e) P(4≤X≤6)=0.74

Step-by-step explanation:

X       3      4       5      6       7      8     9      10

P(X) 0.1 0.24 0.33 0.17 0.05 0.05 0.04  0.00

(NOTE: the probability for X=10 is not shown)

a) X is a discrete random variable, as is possible value are only integer numbers (number of rooms: 3, 4, 5, ... , 10).

b) The probability of choosing a unit with ten rooms is: P(X=10)=0.04, according to the table.

c) The probability of choosing a unit that has more than 5 rooms is equal to the sum of probablities of choosing a unit with 6, 7, 8, 9 or 10 rooms.

P(X > 5) = P(6)+P(7)+P(8)+P(9)+P(10)

P(X > 5) = 0.17+0.05+0.05+0.04+0.00 = 0.31

d) The probability that a unit chosen at random has three rooms is P(X=3)=0.1.

e) The probability that a unit chosen at random has between four and six rooms is

P(4≤X≤6)=P(4)+P(5)+P(6) = 0.24+0.33+0.17 = 0.74

36/60

36/2=18

60/2=30

18/30

18/3=6

30/3=10

6/10

6/2=3

10/2=5

your final answer would be 3/5 of an

hope this !