***************** Choose the correct answer and shows the computations for each

Question

***************** Choose the correct answer and shows the computations for each answer. Thank ********************************

9. Let X be a binomial random variable. Find the probability if x = 3 and n = 8.

10. If 4 of 20 housing projects violate the building code. What is the probability that a home inspector that randomly selects four of them find that:

to. violates any of the houses building code.
b. viola building code
c. two violate the building code
d. at least three violate the building code

11. The probability that a cell run for a year without fail is 0.95. Determine the probability that in a sample of 20

to. 12 last less than one year.
b. at most five less than last year.
c. at least 2 last less than one year.

12. A manager checks the balancing of random boxes of 12 employees. If 10% of tabulated are incorrect. Determine the probability that the next day:

to. none is descuadrado.
b. one out of square.
c. at least two out descuadrados.
d. more than three are with descuadrados.

13. In a bar examination found that 18% fail the test. If 25 people taking the test determine the probability that:
to. nobody pass the exam.

b. over 6 pass the exam.
c. of 4-8 pass the exam.
d. Determine the mean and standard deviation and those who pass the test.

14. A car dealer can close a business 10% of the time serving customers. If you expect the weekend to attend 25 clients. To finish the likelihood that sell:
  
to. 3 cars
b. more than 3 cars
c. within 3
d. 3 or less

***************** Choose the correct answer and shows the computations for each answer. Thank ********************************

Explanation / Answer

Ans: 9. In the given case n=8 and x=3. In such a scenario, probability that x=3 given n=8 at a probbaility of success being p and that of failure being q = 8C3 (p^3) (q^5)

10. The probability that any project violates the building code is = 4/20 = 0.2

a) probability that any of the 4 violate the codes= 1-p(X=0) = 1- 4C0 (0.2)^0 (0.8)^4 = 1- 0.4096 = 0.5904

b) probability taht all 4 of them violate = p(X=4) = 4C4 (0.2)^4 (0.8)^0 = 0.0016

c) Two violate the code= p(X=2) = 4C2 (0.2^2) (0.8^2) = 6* 0.04 *1.6 = 0.384

d) At least 3 violate = p(X>=3) = p(X=3) + p(X=4) = 4C3 (0.2^3) (0.8^1) + 0.0016 = 4*0.008*0.8

= 0.0256

(Ans)

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