Respuesta :
Answer:
a) X is a discrete random variable.
b) 2% probability of choosing a unit with 10 rooms
c) 98% probability that a unit chosen at random is not a 10-room unit
d) 30% probability that a unit chosen at random has less than five rooms
e) 7% probability that a unit chosen at random has less than five rooms
Step-by-step explanation:
The distribution means that:
7% probability that a randomly chosen owner-occupied housing unit in a certain city has 3 rooms.
23% probability that a randomly chosen owner-occupied housing unit in a certain city has 4 rooms.
36% probability that a randomly chosen owner-occupied housing unit in a certain city has 5 rooms.
19% probability that a randomly chosen owner-occupied housing unit in a certain city has 6 rooms.
5% probability that a randomly chosen owner-occupied housing unit in a certain city has 7 rooms.
5% probability that a randomly chosen owner-occupied housing unit in a certain city has 8 rooms.
3% probability that a randomly chosen owner-occupied housing unit in a certain city has 9 rooms.
?% probability that a randomly chosen owner-occupied housing unit in a certain city has 10 rooms.
(a) Is X a discrete or continuous random variable?
X is the number of rooms in a randomly selected house. The number of rooms is a countable variable, that is, only assumes values like 1,2,3,4,... You cannot have 3.5 rooms, for example.
So X is a discrete random variable.
(b) What must be the probability of choosing a unit with 10 rooms?
The sum of all probabilities must be 100%(1 decimal). So
0.07 + 0.23 + 0.36 + 0.19 + 0.05 + 0.05 + 0.03 + ? = 1
0.98 + ? = 1
? = 0.02
2% probability of choosing a unit with 10 rooms
(c) What is the probability that a unit chosen at random is not a 10-room unit?
0.07 + 0.23 + 0.36 + 0.19 + 0.05 + 0.05 + 0.03 = 0.98
98% probability that a unit chosen at random is not a 10-room unit
(d) What is the probability that a unit chosen at random has less than five rooms?
7% probability it has 3 rooms
23% probability it has 4 rooms. So
7+23 = 30% probability that a unit chosen at random has less than five rooms
(e) What is the probability that a unit chosen at random has three rooms?
7% probability that a randomly chosen owner-occupied housing unit in a certain city has 3 rooms. So
7% probability that a unit chosen at random has less than five rooms
Probabilities are used to determine the chances of an event.
- X is a discrete random variable
- The probability of choosing a unit with 10 rooms must be 0.08
- The probability of choosing a unit that is not 10 rooms is 0.92
- The probability of choosing a unit that has less than 5 rooms is 0.30
- The probability of choosing a unit that has 3 rooms is 0.07
(a) Is X a discrete or a continuous random variable
From the table, the values of  X are natural numbers.
Only discrete random variables take natural numbers for their dataset.
Hence, X is a discrete random variable
(b) P(x = 10)
From the table, we have:
[tex]\mathbf{P(3\le x \le 9) = 0.07 + 0.23 + 0.30 + 0.19 + 0.05 + 0.05 + 0.03}[/tex]
[tex]\mathbf{P(3 \le x \le 9) = 0.92}[/tex]
Using the complement rule, we have:
[tex]\mathbf{P(x = 10) = 1 - P(3\e x \le 9)}[/tex]
So, we have:
[tex]\mathbf{P(x = 10) = 1 - 0.92}[/tex]
[tex]\mathbf{P(x = 10) = 0.08}[/tex]
Hence, the probability of choosing a unit with 10 rooms must be 0.08
(c) P (x ! = 10)
In (b), we have [tex]\mathbf{P(3 \le x \le 9) = 0.92}[/tex]
This means that:
The probability of choosing a unit that is not 10 rooms is 0.92
(d) P(x < 5)
This is calculated as:
[tex]\mathbf{P(x < 5) = P(x =3) + P(x = 4)}[/tex]
So, we have:
[tex]\mathbf{P(x < 5) =0.07 + 0.23}[/tex]
[tex]\mathbf{P(x < 5) =0.30}[/tex]
Hence, the probability of choosing a unit that has less than 5 rooms is 0.30
(e) P(x = 3)
From the table, we have:
[tex]\mathbf{P(x = 5) =0.07}[/tex]
Hence, the probability of choosing a unit that has 3 rooms is 0.07
Read more about probabilities at:
https://brainly.com/question/11234923
