Grear Tire Company has produced a new tire with an estimated mean lifetime mileage of 36,500 miles. Management also believes that the standard deviation is 5,000 miles and that tire mileage is normally distributed. To promote the new tire, Grear has offered to refund some money if the tire fails to reach 30,000 miles before the tire needs to be replaced. Specifically, for tires with a lifetime below 30,000 miles, Grear will refund a customer $1 per 100 miles short of 30,000. a) For each tire sold, what is the expected cost of the promotion?
b) Please perform the simulation for 1000 times, what is the probability that Grear will refund more than $50 for a tire?

Step-by-step explanation:

standard deviation σ =5000

Mean μ = 36500

X = 30000

Probability that tire fails to reach 30000 miles

P(< 30000) = P( z < (30000-36500) / 5000 )

= P ( z < - 1.3 )

= .0968 ( from z table )

refund per mile = .01

refund for 30000 mile = .01 x 30000 = 300

expected cost of promotion

= 300 x .0968

= 29.04

b )

refund of $50 is equivalent to shortage of milage by 50 / .01 = 5000 miles

Probability of miles less than 30000 - 5000 = 25000

P ( X < 25000 ) = P ( z < (25000 - 36500) / 5000 )

P ( z < - 2.3 )

= 0.0107 Answer .

answer: 59

step-by-step explanation:

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2