I'm trying to help my brother with his statistic question to study for his exam. I had a statistic class 5 years ago back in my country, amazingly I attained nothing of it, and this question is college grad. If anyone could give a pointer as to how to solve the problem, I'd really appreciate it. The simpler, the better since I also find it difficult to understand the english math terms.
Question: Survey found that the height for occupants of building A is 1.7m with standard deviation of 0.15m.
a. Probability that a randomly selected occupant measures less than 1.8m?
b. In a random sample of 22 people, how many are expected to be less than 1.5m?
c. An occupant is at the 20th percentile of the height in target population. What's the height of this occupant?
d Another occupant is at 25th percentile in a sample of 34 occupants. What's his height?
Thank you!
Question: Survey found that the height for occupants of building A is 1.7m with standard deviation of 0.15m.
a. Probability that a randomly selected occupant measures less than 1.8m?
b. In a random sample of 22 people, how many are expected to be less than 1.5m?
c. An occupant is at the 20th percentile of the height in target population. What's the height of this occupant?
d Another occupant is at 25th percentile in a sample of 34 occupants. What's his height?
Thank you!
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This question is related normal distribution.
a) z = (X-Mean)/Standard deviation
z = (1.8-1.7)/0.15 = + 0.67
The area under the standard normal curve left to z = + 0.67 indicates the required probability.
Required probability = 0.5000 (area on left of mean) + 0.2486 (area on right of mean) = 0.7486
b) z = (1.5-1.7)/0.15 = - 1.33
The area under the standard normal curve left to z = - 1.33 indicates the probability.
This area lies in the extreme left tail of the normal curve.
Probability = 0.5000 - 0.4082 = 0.0918
Expected number of people to be < 1.5m = 0.0918*22 = 2
c) for finding X value z score corresponding to 0.2000 (20th percentile) area in the extreme left tail is to be located from the table.
0.5000 - 0.2000 (area left to z) = 0.3000 (area right to z)
The z value representing 0.3000 area is - 0.84 (approximately)
z value is negative because it lies on the left side of the normal curve.
- 0.84 = (X-1.7)/0.15
- 0.126 = X-1.7
X = 1.7- 0.126 = 1.574m
The occupant's height = 1.574m
d) The area in the extreme left tail = 0.2500(25th percentile)
0.500 - 0.2500 (area left to z) = 0.2500 (area right to z)
The z score representing 0.2500 area is - 0.675 (approximately)
- 0.675 = (X-1.7)/0.15
- 0.10125 = X-1.7
X = 1.7 - 0.10125 = 1.59875 or 1.6m
The height of the occupant = 1.6m
a) z = (X-Mean)/Standard deviation
z = (1.8-1.7)/0.15 = + 0.67
The area under the standard normal curve left to z = + 0.67 indicates the required probability.
Required probability = 0.5000 (area on left of mean) + 0.2486 (area on right of mean) = 0.7486
b) z = (1.5-1.7)/0.15 = - 1.33
The area under the standard normal curve left to z = - 1.33 indicates the probability.
This area lies in the extreme left tail of the normal curve.
Probability = 0.5000 - 0.4082 = 0.0918
Expected number of people to be < 1.5m = 0.0918*22 = 2
c) for finding X value z score corresponding to 0.2000 (20th percentile) area in the extreme left tail is to be located from the table.
0.5000 - 0.2000 (area left to z) = 0.3000 (area right to z)
The z value representing 0.3000 area is - 0.84 (approximately)
z value is negative because it lies on the left side of the normal curve.
- 0.84 = (X-1.7)/0.15
- 0.126 = X-1.7
X = 1.7- 0.126 = 1.574m
The occupant's height = 1.574m
d) The area in the extreme left tail = 0.2500(25th percentile)
0.500 - 0.2500 (area left to z) = 0.2500 (area right to z)
The z score representing 0.2500 area is - 0.675 (approximately)
- 0.675 = (X-1.7)/0.15
- 0.10125 = X-1.7
X = 1.7 - 0.10125 = 1.59875 or 1.6m
The height of the occupant = 1.6m
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Ans:
a. Need to find P(x < 1.8) =?
z = (1.8 - 1.7)/0.15 = 0.67
P(x < 1.8) = P(z < 0.67)=0.7485(Ans.)
From TutorTeddy
a. Need to find P(x < 1.8) =?
z = (1.8 - 1.7)/0.15 = 0.67
P(x < 1.8) = P(z < 0.67)=0.7485(Ans.)
From TutorTeddy