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Sphinxa [80]
2 years ago
9

Fifty students in the fourth grade class listed their hair and eye colors in the table below: Brown hair Blonde hair Total Green

eyes 16 12 28 Brown eyes 14 8 22 30 20 50 Are the events "green eyes" and "brown hair" independent?
Mathematics
1 answer:
Flura [38]2 years ago
6 0

 

Two events are considered to be independent of each other when the probability that one event occurs in no way affects the probability of the other event occurring.  Since some students have green eyes and brown hair, some have green eyes and not brown hair, and not green eyes and brown hair, therefore the two events are not independent. To say, they are dependent.

We can prove this mathematically by:

<span>P(</span>A∩B<span>)=</span><span>P(</span>A<span><span>) * </span><span>P(</span></span>B<span>)             --> If equal, then independent</span>

Where A = green eyes

B = brown hair

 

From the table:

<span>P(</span>A∩B<span>)<span> = 16/50  (16 out of 50 has green eyes and brown hair)</span></span>

<span>P(A) = 28/50        (28 out of 50 has green eyes)</span>

<span>P(B) = 30/50        (30 out of 50 has brown hair)</span>

16/50 = (28/50) * (30/50)

<span>16/50 = 840 / 2500        (NOT TRUE)</span>


Answer:

Not independent

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The probabilities of the orphaned pets in six cities' animal shelters being different types of animals are given in the table. I
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Given the table below showing t<span>he probabilities of the orphaned pets in six cities' animal shelters being different types of animals.

</span>
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                         −Lhasa Apso −Mastiff −Chihuahua −Collie
Austin       24.5%      2.76%    2.86%     3.44%    2.65%
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The probability that a randomly selected orphaned pet in a St. Louis animal shelter is a mastiff given that it is a dog, is given by
\frac{P(St. \, Louis\, and\, dog-mastiff)}{P(dog)} = \frac{2.46}{11.26} \times100\%=21.85\%</span>
5 0
2 years ago
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let x = the amoun of raw sugar in tons a procesing plant is a sugar refinery process in one day . suppose x can be model as expo
anygoal [31]

Answer:

The answer is below

Step-by-step explanation:

A sugar refinery has three processing plants, all receiving raw sugar in bulk. The amount of raw sugar (in tons) that one plant can process in one day can be modelled using an exponential distribution with mean of 4 tons for each of three plants. If each plant operates independently,a.Find the probability that any given plant processes more than 5 tons of raw sugar on a given day.b.Find the probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day.c.How much raw sugar should be stocked for the plant each day so that the chance of running out of the raw sugar is only 0.05?

Answer: The mean (μ) of the plants is 4 tons. The probability density function of an exponential distribution is given by:

f(x)=\lambda e^{-\lambda x}\\But\ \lambda= 1/\mu=1/4 = 0.25\\Therefore:\\f(x)=0.25e^{-0.25x}\\

a) P(x > 5) = \int\limits^\infty_5 {f(x)} \, dx =\int\limits^\infty_5 {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_5=e^{-1.25}=0.2865

b) Probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day can be solved when considered as a binomial.

That is P(2 of the three plant use more than five tons) = C(3,2) × [P(x > 5)]² × (1-P(x > 5)) = 3(0.2865²)(1-0.2865) = 0.1757

c) Let b be the amount of raw sugar should be stocked for the plant each day.

P(x > a) = \int\limits^\infty_a {f(x)} \, dx =\int\limits^\infty_a {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_a=e^{-0.25a}

But P(x > a) = 0.05

Therefore:

e^{-0.25a}=0.05\\ln[e^{-0.25a}]=ln(0.05)\\-0.25a=-2.9957\\a=11.98

a  ≅ 12

6 0
2 years ago
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