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OleMash [197]
3 years ago
12

De un saco de frutas que contiene 3 naranjas, 2 manzanas, y 3 plátanos, una muestra aleatoria de 2 piezas de se selecciona la fr

uta. Si X es el número de naranjas e Y es el número de manzanas en la muestra, encuentre
(a) la distribución de probabilidad conjunta de X y Y, muestre la tabla de distribución;
(b) Pruebe que la suma de columnas y renglones nos entregan las distribuciones marginales de X , Y
(c) Calcule P(Y = 0|X = 2).
(d) Demuestre que las variables aleatorias X ,Y no son estadísticamente independientes
Mathematics
1 answer:
rodikova [14]3 years ago
3 0

Answer:

speak english plz

Step-by-step explanation:

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Factor the expression completely. 10x^5 -7×^3?​
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What 75% of two hundred
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To find 75% of 200, we can multiply two fractions.

75/100 * 200/1

Now, just multiply the numerators and the denominators separately.

75 * 200 = 15000

100 * 1 = 100

Now, divide.

15000/100 = 150

<h3>75% of 200 is 150.</h3>
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3 years ago
I need help please help m
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there are 10 numbers you want either a 4 or 7 ( 2 numbers)

 so probability is 2/10 which reduces to 1/5

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2 years ago
99.7 of all newborn babies in the United States weigh between
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Answer:

2000-5000 g

Step-by-step explanation:

Averagely, the weight of a new born baby is 7.5 lb, which is equal to 3.5 kg.

Converting this weight in kg to grams,

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4 0
3 years ago
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A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute
nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

P(k\geq8)=1-P(k

P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\

P(k

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002

5 0
3 years ago
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