A different random sample of 1,000 people yielded the results recorded in table 2. Use this data to answer the questions that fo
llow.
Part A
What percentage of Americans would you predict wear corrective lenses?
NOTE: this is table 2 just in case you wanted to know or if u got confused/just a double check :)
Part A
What percentage of Americans would you predict wear corrective lenses?
NOTE: this is table 2 just in case you wanted to know or if u got confused/just a double check :)
2 answers:
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Answer:
a) 0.96
b) 0.016
c) 0.018
d) 0.982
e) x = 2
Step-by-step explanation:
We are given with the Probability density function f(x)= 2/x^3 where x > 1.
<em>Firstly we will calculate the general probability that of P(a < X < b) </em>
P(a < X < b) = =
= { Because
}
= =
= =
a) Now P(X < 5) = P(1 < X < 5) {because x > 1 }
Comparing with general probability we get,
P(1 < X < 5) = =
= 0.96 .
b) P(X > 8) = P(8 < X < ∞) = 1/ - 1/∞ = 1/64 - 0 = 0.016
c) P(6 < X < 10) = =
= 0.018 .
d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)
= + (1/
- 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982
e) We have to find x such that P(X < x) = 0.75 ;
⇒ P(1 < X < x) = 0.75
⇒ = 0.75
⇒ = 1 - 0.75 = 0.25
⇒ =
⇒
= 4 ⇒ x =
Therefore, value of x such that P(X < x) = 0.75 is 2.