Question

The operating time of a machine manufactured by company A is a random variable (Unit: years) with a standard distribution of N (5; 3,25).
a. A person who has purchased this device has used it for 2 years, calculating the probability that he or she will use it for at least 4 more years
b. The company warranty for 3 years. Calculate the percentage of products that are warranted by the company
c. A person who buys 10 units of this company, calculated the probability of having no more than 2 machines must be warranted in these 10 machines.
d. Among 10,000 products of company A, the probability of having no more than 800 of these 10,000 products must be covered by a warranty

Answer 1

P(z<Z) table :

x = operating time

a.

required : P(x>=4+2 | x>=2)

= P(x>=6 | x>=2)

= P(x>=6) / P(x>=2)

P(x>=6) :

P(x>=2) :

P(x>=6) / P(x>=2) = 0.3792 / 0.822 = 0.4613

therefore,

P(x>=4+2 | x>=2) = 0.4613

b.

P(warranted) = P(x<=3)

P(x<=3) :

P(warranted) = 0.2692

percentage warranted = 26.92%

c.

let x = no. of machines warranted :

n=10

p = P(warranted) = 0.2692

P(x<=2) = P(0)+P(1)+P(2)

= 10C0*(0.2692^0)*(1-0.2692)^(10-0)+10C1*(0.2692^1)*(1-0.2692)^(10-1)+10C2*(0.2692^2)*(1-0.2692)^(10-2)

= 0.4688

d. Among 10,000 products of company A, the probability of having no more than 800 of these 10,000 products must be covered by a warranty

in binomial :

mean = n*p = 10000*0.2692 = 2692

SD = n*p*(1-p) = 10000*0.2692*(1-0.2692) = 1967.3136

estimate using normal distribution

as number of products is large enough

P(x<=800) :

probability of having no more than 800 of these 10,000 products must be covered by a warranty

= P(x<=800)

= 0.1681

(please UPVOTE)