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hammer [34]
3 years ago
5

En un hospital el 30 % de los pacientes que acuden de manera diaria presentan problemas cardíacos. Si un día acuden al hospital

70 pacientes, ¿Cuántos pacientes acudieron ese día con problemas cardíacos?
Mathematics
1 answer:
juin [17]3 years ago
6 0

Answer:

21 pacientes tienen problemas cardíacos de los 70 pacientes diagnosticados

Step-by-step explanation:

Paciente con problema cardíaco del total de pacientes diagnosticados = 30%

Total de pacientes que van al hospital en un día = 70

El total de pacientes con problemas cardíacos entre los 70 pacientes diagnosticados es

30% de 70 = 30/100 * 70 = 21

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The probability that a randomly selected box of a certain type of cereal has a particular prize is 0.20.
MatroZZZ [7]

Answer:

a) the probability is 0.0734 (7.34%)

b) the probability is 0.7382 (73.82%)

c) 10 boxes

Step-by-step explanation:

Since each box is independent of the others and defining the random variable X= number of boxes of cereal purchased without prices until having 2 prices , then X has a negative binomial distribution . Thus

P(X=k)= C(k+r-1,k)*p^k *(1-p)^r

where

p= probability to not obtain a price when a box is purchased = 1- 0.2 = 0.8

C ( ) = combinations

k = number of boxes without prices

r= number of boxes with prices=2

P(X=k) = probability of purchasing k boxes without prices until obtaining r boxes with prices

a) for k= 8 - r = 6

P(X=6)= C(7,6)*0.8^6*0.2^2 = 7*0.8^6*0.2^2 = 0.0734 (7.34%)

b) for k≥9-2=7

P(X≥7) = 1- P(X<7)= 1- Fn(X=7)

where Fn(X) is the cumulative negative binomial distribution. We can calculate it through its relationship with the cumulative binomial distribution Fb(X) that is easily found in tables :

Fn(k=7,r=2,p=0.8) = Fb(k=7,n=k+r=9,p=0.8) = 0.2618

thus

P(X≥7) = 1- P(X<7)= 1- Fn(X=7) = 1-  0.2618 = 0.7382 (73.82%)

c) the expected value E(X) for a negative binomial distribution is

E(X)= p*r/(1-p) = 0.8*2/0.2 = 8 boxes that do not contain prices

thus

n=k+r = 8 + 2 = 10 boxes

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