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3 years ago

Three examples of discrete variables

Mathematics

1 answer:

CaHeK987 [17]3 years ago

7 0

Answer:

Step-by-step explanation:

Discrete Variables can meaningfully have only specific values:

Number of coin flips: 4.

Number of books published: 2.

Distance walked, rounded to the nearest kilometer: 3

(Must be rounded up or down)

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The can is 12 cm high and the diameter is 8 cm.

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3 years ago

A geologist has collected 5 specimens of basaltic rock and 7 specimens of granite. The geologist instructs a laboratory assistan

MaRussiya [10]

The rocks are chosen without replacement, which means that the hypergeometric distribution is used to solve this question. First we get the parameters, and then we answer the questions. From this, we get that:

$E(X) = 5.25, Var(X) = 0.5966$
P(X < 6) = 0.9545
P(all specimens of one of the two types of rock are selected for analysis) = 0.2046.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The mean and the variance are:

$\mu = \frac{nk}{N}$

$\sigma^2 = \frac{nk(N-k)(N-n)}{N^2(N-1)}$

We have that:

5 + 7 = 12 rocks, which means that $N = 12$

9 are chosen, which means that $n = 9$

7 are granite, which means that $k = 7$

Question a:

$E(X) = \mu = \frac{9\times7}{12} = 5.25$

$Var(X) = \sigma^2 = \frac{9\times7(12-7)(12-9)}{12^2(12-1)} = 0.5966$

Thus:

$E(X) = 5.25, Var(X) = 0.5966$

Question b:

Since there are only 5 specimens of basaltic rock, at least 9 - 5 = 4 specimens of granite are needed, which means that:

$P(X < 6) = P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 4) = h(4,12,9,7) = \frac{C_{7,4}*C_{5,5}}{C_{12,9}} = 0.1591$

$P(X = 5) = h(5,12,9,7) = \frac{C_{7,5}*C_{5,4}}{C_{12,9}} = 0.4773$

$P(X = 6) = h(6,12,9,7) = \frac{C_{7,6}*C_{5,3}}{C_{12,9}} = 0.3181$

Thus

$P(X < 6) = P(X = 4) + P(X = 5) + P(X = 6) = 0.1591 + 0.4773 + 0.3181 = 0.9545$

So P(X < 6) = 0.9545.

Question c:

5 of basaltic and 4 of granite: 0.1591 probability.

7 of granite is P(X = 7), in which

$P(X = 7) = h(7,12,9,7) = \frac{C_{7,7}*C_{5,2}}{C_{12,9}} = 0.0455$

0.1591 + 0.0455 = 0.2046, thus:

P(all specimens of one of the two types of rock are selected for analysis) = 0.2046.

A similar question is found at brainly.com/question/24008577

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3 years ago

Solve for t in terms of q, r, and s.<br> r = sqt<br> t=

mars1129 [50]

Answer:

t=r/(sq)

Step-by-step explanation:

r=sqt

=> r/(sq)=t

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Three examples of discrete variables ​

Three examples of discrete variables