Question

Suppose the amount of time needed to change the oil in a car is uniformly distributed between 14 minutes and 30 minutes, with a mean of 22 minutes and a standard deviation of 4.6 minutes.

Let X represent the amount of time, in minutes, needed to complete a randomly selected oil change. What is the probability that a randomly selected oil change takes at most 20 minutes to complete?

Assuming a randomly selected oil change has already taken 18 minutes and it is still not complete, what is the probability that the oil change takes no more than 5 additional minutes to complete?

Find the value that best completes the following statement: The slowest 20% of oil changes take at least __________ minutes to complete. Suppose 3 oil changes are selected at random.

Let Y be defined as the number of oil changes that take more than 20 minutes to complete in a random sample of 3. What is the probability that exactly 2 of the 3 randomly selected oil changes take more than 20 minutes to complete?

Suppose 500 oil changes are selected at random. What model would you use to approximate the probability that at least 190 of the randomly selected oil changes take at most 20 minutes to complete? Provide the approximate model with corresponding parameters.

Answer 1

1)

probability that a randomly selected oil change takes at most 20 minutes to complete

=P(X<20) =(20-14)/(30-14)=0.375

2)

The slowest 20% of oil changes take at least =14+(1-0.20)*(30-14)=26.8 minutes to complete

3)

probability that exactly 2 of the 3 randomly selected oil changes take more than 20 minutes to complete =(₃C₂)*(0.375)¹*(1-0.375)² =0.4395

4)

here since np =500*0.375 =187.5 and n(1-p) =312.5 both are greater than 5 , we can use normal approximation of binomial distribution:

parameter of normal distribution are:

here mean of distribution=μ=np=		187.50
and standard deviation σ=sqrt(np(1-p))=		10.83
for normal distribution z score =(X-μ)/σx

therefore from normal approximation of binomial distribution and continuity correction:

probability that at least 190 of the randomly selected oil changes take at most 20 minutes to complete :

probability =P(X>189.5)=P(Z>(189.5-187.5)/10.825)=P(Z>0.18)=1-P(Z<0.18)=1-0.5714=0.4286