The shape of the distribution of the time required to get an oil change at a 10-
ID: 3317316 • Letter: T
Question
The shape of the distribution of the time required to get an oil change at a 10-minute oil-change facility is unknown. However, records indicate that the mean tmeis 1.5 minutes, and the standard deviation is 4.8 minutes. Complete parts (a) (a) To compute probabilities regarding the sample mean using the normal model, what size sample would be required? O A. O B. ° C. The sample size needs to be less than or equal to 30. Any sample size could be used The normal model cannot be used if the shape of the dstribution is unknown. D. The sample size needs to be greater than or equal to 30. (b) what is the probability that a random sample of n=40 oil changes results in a sample mean time less than 10 minutes? The probability is approximately (Round to four decimal places as needed.) (c) Suppose the manager agrees to pay each employee a $50 bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 40 oil changes between 10 A.M. and 12 PM. Treatng this as a random sample, there would be a 10% chance of the mean oil-change time being at or below what value? This will be the goal established by the manager There is a 10% chance of being at or below a mean oil-change time of minutes. Round to one decimal place as needed.)Explanation / Answer
a. the sample size needs to be greater than equal to 30
b.
b.
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 11.5
standard Deviation ( sd )= 4.8/ Sqrt ( 40 ) =0.7589
sample size (n) = 40
P(X < 10) = (10-11.5)/4.8/ Sqrt ( 40 )
= -1.5/0.7589= -1.9764
= P ( Z <-1.9764) From Standard NOrmal Table
= 0.02405
c.
mean of the sampling distribution ( x ) = 11.5
standard Deviation ( sd )= 4.8/ Sqrt ( 40 ) =0.7589
sample size (n) = 40
P ( Z < x ) = 0.1
Value of z to the cumulative probability of 0.1 from normal table is -1.281552
P( x-u/s.d < x - 11.5/0.7589 ) = 0.1
That is, ( x - 11.5/0.7589 ) = -1.281552
--> x = -1.281552 * 0.7589 + 11.5 = 10.527431
it is less than 10.5
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.