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arsen [322]
3 years ago
12

Let X1, X2 and X3 be three independent random variables that are uniformly distributed between 50 and 100.

Mathematics
1 answer:
sergejj [24]3 years ago
3 0

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

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Answer:

The equation representing how old Monique son is \mathbf{a = 2 + \dfrac{8}{21}(q-34)}

Step-by-step explanation:

From the given information:

A linear function can be used to represent the constant growth rate of Monique Son.

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q(t) = \hat q \times t + q_o

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q_o = initial height of Monique's son

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\hat q = 2 \dfrac{5}{8} \ in/yr

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The height of the son as a function of the age can now be expressed as:

q(t) = \dfrac{21}{8} \times t + 34

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Making t the subject;

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\mathbf{a = 2 + \dfrac{8}{21}(q-34)}

SO;

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a = 8.095 years

a ≅ 8 years

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Monique son will be 8 years at the time Monique is 50 inches tall.

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