It could. If you turn it into an experiment. Let's say you take 100 students from each grade, and you ask them if they exercise. And then you wanna create a graph. And find out out of those students how many of them exercise and how many of them do not. Then turn it into a percent.
I hope this helps. If you have any further questions on this or any other questions just ask. I am here to help you, as needed.
As A Simplified Fraction She Only Has 1/4 Of Her Allowance Left
Answer:
a) the probability that the minimum of the three is between 75 and 90 is 0.00072
b) the probability that the second smallest of the three is between 75 and 90 is 0.396
Step-by-step explanation:
Given that;
fx(x) = { 1/5 ; 50 < x < 100
0, otherwise}
Fx(x) = { x-50 / 50 ; 50 < x < 100
1 ; x > 100
a)
n = 3
F(1) (x) = nf(x) ( 1-F(x)^n-1
= 3 × 1/50 ( 1 - ((x-50)/50)²
= 3/50 (( 100 - x)/50)²
=3/50³ ( 100 - x)²
Therefore P ( 75 < (x) < 90) = ⁹⁰∫₇₅ 3/50³ ( 100 - x)² dx
= 3/50³ [ -2 (100 - x ]₇₅⁹⁰
= (3 ( -20 + 50)) / 50₃
= 9 / 12500 = 0.00072
b)
f(k) (x) = nf(x) ( ⁿ⁻¹_k₋ ₁) ( F(x) )^k-1 ; ( 1 - F(x) )^n-k
Now for n = 3, k = 2
f(2) (x) = 3f(x) × 2 × (x-50 / 50) ( 1 - (x-50 / 50))
= 6 × 1/50 × ( x-50 / 50) ( 100-x / 50)
= 6/50³ ( 150x - x² - 5000 )
therefore
P( 75 < x2 < 90 ) = 6/50³ ⁹⁰∫₇₅ ( 150x - x² - 5000 ) dx
= 99 / 250 = 0.396
This cannot be expressed as an answer but the expression for that is 3L-5
