A quantity in a standard time or period maybe lol
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
Answer:
C would be the correct answer. :D
Answer is 75% hopefully I mean I don't really know that much
