Assume that the per capita income of residents in a country is normally distributed with mean μ = $1000 and variance σ2 = 10,000 ($ squared).
a. What is the probability that the per capita income lies between $800 and $1200?
b. What is the probability that it exceeds $1200?
c. What is the probability that it is less than $800?
d. Is it true that the probability of per capita income exceeding $5000 is practically zero?
Answers
Given : capita income of residents in a country is normally distributed with mean μ = $1000 and variance σ2 = 10,000 ($ squared).
To find : a. What is the probability that the per capita income lies between $800 and $1200?
b. What is the probability that it exceeds $1200?
c. What is the probability that it is less than $800?
Solution:
Mean μ = 1000 $
variance σ2 = 10,000
=> Standard deviation SD σ = 100
Z score = ( Value - Mean)/SD
Z score = (Value - μ )/ σ
$800 and $1200?
Z score for 800 = ( 800 - 1000)/100 = -2 => 0.0228
Z score for 1200 = ( 1200 - 1000)/00 = 2 => 0.9772
Probability that the per capita income lies between $800 and $1200 = 0.9772 - 0.0228 = 0.9544
probability that it exceeds $1200 = 1 - 0.9772 = 0.0228
probability that it is less than $800 = 0.0228
Probability for μ ± 3σ is considered Zero
1000 ± 3(100)
700 to 1300
so its true that the probability of per capita income exceeding $5000 is practically zero
Learn more:
Assume that adults have iq scores that are normally distributed with ...
https://brainly.in/question/11133397
The mean length of mature whiting is 35 cm and the standard ...
https://brainly.in/question/13450142
The value of the cumulative standardized normal distribution at z is ...
https://brainly.in/question/11376268
