Question

or a new type of​ tire, a racing car team found the average distance a set of tires would run during a race is 166 ​miles, with a standard deviation of 14 miles. Assume that tire mileage is independent and follows a Normal model. ​a) If the team plans to change tires twice during a​ 500-mile race, what is the expected value and standard deviation of miles remaining after two​ changes? ​b) What is the probability they​ won't have to change tires a third time​ (and use a fourth set of​ tires) before the end of a 500 mile​ race? ​a) The expected value for miles remaining is 168 miles. ​(Type an integer or decimal rounded to three decimal places as​ needed.) The standard deviation for miles remaining is 19.799 miles. ​(Type an integer or decimal rounded to three decimal places as​ needed.) ​b) The probability they​ won't have to change tires a third time is ​(Type an integer or decimal rounded to three decimal places as​ needed.)

Answer 1

Solution

Back-up Theory

Let X and Y be two random variables

E(X + Y) = E(X) + E(Y) ……………………………………………………………….....................…. (1)

V(X + Y) = V(X) + V(Y), when X and Y are Independent ……………………................………… (2)

Now to work out the solution,

Let

X = distance the first set of tires would run during a race

Y = distance the second set of tires would run during a race.

Then, the total distance two tire changes would run during a race, T = X + Y …………………….. (3)

Given ‘Assume that tire mileage is independent and follows a Normal model’   

T ~ N(µ, σ²) ……………………………………………………………………………................……….. (4)

X and Y are independent. …………………………………………………………….........…………….. (5)

Also given:

X ~ N(166, 14²) and Y ~ N(166, 14²) …………………………………………………..........………….. (6)

Part (a)

Vide (1) (3) and (6),

E(T) = 166 + 166 = 332.

Thus, the expected value of miles that will be covered by two tire changes = 332 miles. Hence,

expected value of miles remaining after two​ changes = 500 – 332 = 168 miles Answer 1

Number of miles remaining after two​ changes = 500 – T.

So, Variance of miles remaining after two​ changes = Var(500 – T)

= Var(T) [because variance of a constant is zero]

= Var(X) + Var(Y) [vide (2) and (5)]

= 14² + 14² [vide (6)]

= 392

Hence standard deviation = √392 = 19.799

Thus, standard deviation value of miles remaining after two​ changes = 19.799 Answer 2

Part (b)

Probability they​ won't have to change tires a third time

= Probability that the value of miles that will be covered by two tire changes is 500 or more

= P(T ≥ 500)

= P[Z ≥ {(500 - 332)/19.799}] [where Z ~ N(0, 1) and E(T) = 332 and SD(T) = 19.799]

= P(Z ≥ 8.486)

= 0 [Using Excel Function: Statistical NORMSDIST] Answer 3

DONE