Question

There is a​ 10% chance of the mean​ oil-change time being at or below the sample...

There is a​ 10% chance of the mean​ oil-change time being at or below the sample mean for which there is an area of 0.10 to the left under the normal curve with mu Subscript x overbar Baseline equals 16.2 and sigma Subscript x overbar Baseline equals 0.213 . Use technology to find this value for x overbar.

Homework Answers

Answer #1

= 16.2, = 0.213

area under normal curve to the left of x is = 0.10

Here we want to find the 10th percentile to get the value of x

using TI-83 PLUS calculator we get value of as follows

we get

= 15.927

2)

Manually you can do it as

find the z-score related to area 0.10 using z table we get

z-score = -1.281

formula for z-score is

convert it for x as follows

X= ( z * ) +

X = ( -1.281 * 0.213 ) + 16.2

= 15.927

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
The shape of the distribution of the time is required to get an oil change at...
The shape of the distribution of the time is required to get an oil change at a 15-minute oil-change facility is unknown. However, records indicate that the mean time is 16.2 minutes, and the standard deviation is 4.2 minutes. Complete parts (a) through (c). (a) To compute probabilities regarding the sample mean using the normal model, what size sample would be required? A. The sample size needs to be less than or equal to 30. B. The sample size needs...
Suppose a simple random sample of size nequals36 is obtained from a population that is skewed...
Suppose a simple random sample of size nequals36 is obtained from a population that is skewed right with mu equals 81 and sigma equals 6. ​(a) Describe the sampling distribution of x overbar. ​(b) What is Upper P left parenthesis x overbar greater than 82.1 right parenthesis​? ​(c) What is Upper P left parenthesis x overbar less than or equals 78.5 right parenthesis​? ​(d) What is Upper P left parenthesis 79.9 less than x overbar less than 82.65 right parenthesis​?...
Construct a 99% confidence interval for mu 1 minus mu 2μ1−μ2 with the sample statistics for...
Construct a 99% confidence interval for mu 1 minus mu 2μ1−μ2 with the sample statistics for mean cholesterol content of a hamburger from two fast food chains and confidence interval construction formula below. Assume the populations are approximately normal with unequal variances. Stats x overbar 1 equals 134 mg comma s 1 equals 3.64 mg comma n 1 equals 20x1=134 mg, s1=3.64 mg, n1=20 x overbar 2 equals 88 mg comma s 2 equals 2.02 mg comma n 2 equals...
The shape of the distribution of the time required to get an oil change at a...
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 meantime is 11.8 minutes , and the standard deviation is 4.6 minutes.Complete parts (a) through (c) below. (a) To compute probabilities regarding the sample mean using the normal model, what size sample would be required? Choose the required sample size below. A. The sample size needs to be greater than 30. B. Any...
The shape of the distribution of the time required to get an oil change at a...
The shape of the distribution of the time required to get an oil change at a 15​-minute ​oil-change facility is unknown.​ However, records indicate that the mean time is 16.5 minutes​, and the standard deviation is 4.3 minutes. ​-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 35 oil changes between 10 A.M. and 12 P.M. Treating this as a random​ sample, there...
The shape of the distribution of the time required to get an oil change at a...
The shape of the distribution of the time required to get an oil change at a 20​-minute ​oil-change facility is unknown.​ However, records indicate that the mean time is 21.4 minutes​, and the standard deviation is 4.2 minutes. 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 45 oil changes between 10 A.M. and 12 P.M. Treating this as a random​ sample, there...
Use the central limit theorem to find the mean and standard error of the mean of...
Use the central limit theorem to find the mean and standard error of the mean of the indicated sampling distribution. Then sketch a graph of the sampling distribution. The per capita consumption of red meat by people in a country in a recent year was normally​ distributed, with a mean of 111 pounds and a standard deviation of 38.7 pounds. Random samples of size 17 are drawn from this population and the mean of each sample is determined. mu Subscript...
A random sample of n = 81 observations is drawn from a population with a mean...
A random sample of n = 81 observations is drawn from a population with a mean equal to 15 and a standard deviation equal to 9 . Complete parts a through g below. a. Give the mean and standard deviation of the? (repeated) sampling distribution x overbar . mu Subscript x overbar equalsnothingsigma Subscript x overbar equalsnothing ?(Type integers or? decimals.) b. Describe the shape of the sampling distribution of x overbar . Does this answer depend on the sample?...
The shape of the distribution of the time required to get an oil change at a...
The shape of the distribution of the time required to get an oil change at a 20​-minute ​oil-change facility is unknown.​ However, records indicate that the mean time is 21.4 minutes​, and the standard deviation is 3.7 minutes. Complete parts ​(a) through ​(c). ​(a) To compute probabilities regarding the sample mean using the normal​ model, what size sample would be​ required? ​(b) What is the probability that a random sample of nequals35 oil changes results in a sample mean time...
According to an automobile​ association, the average cost of a gallon of regular unleaded fuel at...
According to an automobile​ association, the average cost of a gallon of regular unleaded fuel at gas stations in a certain month was ​$2.8282.828. Assume that the standard deviation of such costs is ​$0.150.15. Suppose that a random sample of nequals=100 gas stations is selected from the population and the​ month's cost per gallon of regular unleaded fuel is determined for each. Consider x overbarx​, the sample mean cost per gallon. Complete part a through d. LOADING... Click the icon...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT