Question

In: Economics

Bird’s data analyst wants to model the sample mean distance of scooter rides. They plan to...

Bird’s data analyst wants to model the sample mean distance of scooter rides. They plan to take a random sample of 41 scooter rides and to record the distance traveled on each ride (distance before the rider “releases” the scooter and the charge is processed). They decide it is reasonable to assume that the distribution of distances is approximately normal, but they don’t know the variance of the distribution. In the sample, the mean is 0.54 mile and the standard deviation is 0.1 mile.

If the population mean of the distribution of distances is 0.5, what is the probability that the sample mean will be 0.54 or larger? Four decimals

Solutions

Expert Solution


Related Solutions

Bird’s data analyst wants to model the sample mean and sample variance of net revenue of...
Bird’s data analyst wants to model the sample mean and sample variance of net revenue of scooters. They plan to take a random sample of 25 of its scooters in Austin. They will record weekly net revenue brought in by each scooter. They decide it is reasonable to assume that weekly revenue of scooters is approximately normally distributed with a standard deviation of $50 (variance of 2,500 squared dollars). Find values a and b such that (i) the probability is...
A quality analyst wants to construct a sample mean chart for controlling a packaging process. Each...
A quality analyst wants to construct a sample mean chart for controlling a packaging process. Each day last week, he randomly selected six packages and weighed each. The data from that activity appears below.                   Weight (in grams) Day Package 1 Package 2 Package 3 Package 4 Package 5 Package 6 Monday 23 22 23 24 23 21 Tuesday 23 21 19 21 24 20 Wednesday 20 19 20 21 19 20 Thursday 18 19 20 19 21 22 Friday...
A quality analyst wants to construct a sample mean chart for controlling a packaging process. He...
A quality analyst wants to construct a sample mean chart for controlling a packaging process. He knows from past experience that whenever this process is in control, package weight is normally distributed with a mean of 20 ounces and a standard deviation of two ounces. Each day last week, he randomly selected four packages and weighed each: Day Weight (ounces) Monday 23 22 23 24 Tuesday 23 21 19 21 Wednesday 20 19 20 21 Thursday 18 19 20 19...
An analyst wants to determine whether the cost of a flight depends on the distance. She...
An analyst wants to determine whether the cost of a flight depends on the distance. She collects a set of data of the round-trip fare for flights between Boston and some other cities on major airlines and the distance between cities. The following table gives the descriptive statistics of the distance (in miles) and the cost (in dollars). Mean Standard Deviation Miles (x) 1650.4 1074.4 Costs (y) 291 111.71 r = 0.9352 Compute the coefficient of determination and interpret your...
A sales analyst wants to determine whether there is difference in the mean monthly sales (in...
A sales analyst wants to determine whether there is difference in the mean monthly sales (in RM) of cosmetic company of six sales region in Johor. Several salespersons from each region are chosen and their monthly sales were recorded in Table Q2. Monthly sale (RM) Company A Company B Company C Company D Company E Company F 4250 3930 3960 4350 4150 3970 3980 4010 4020 4450 3890 4160 4020 4050 4130 3880 4568 4370 4130 4220 4350 4574 4480...
For a sample of 25 monthly observations, a financial analyst wants to regress the percentage rate...
For a sample of 25 monthly observations, a financial analyst wants to regress the percentage rate of return (Y) of the common stock of a corporation on the percentage rate of return ( X) of standard and Poor’s 500 Index. The following information is available: Σ??=??.????=? , Σ??=??.????=? , Σ???=???.????=? ,Σ???=???.????=? , Σ????=???.????=? a. Compute the sample correlation coefficient and interpret your answer. ( 3 marks) b. Test to determine the significance of the linear relationship between X and Y....
For a sample of 25 monthly observations, a financial analyst wants to regress the percentage rate...
For a sample of 25 monthly observations, a financial analyst wants to regress the percentage rate of return (Y) of the common stock of a corporation on the percentage rate of return ( X) of standard and Poor’s 500 Index. The following information is available: ∑?? = ??. ? ?? ?=? , ∑?? = ??. ? ?? ?=? , ∑?? ? = ???. ? ?? ?=? ,∑?? ? = ???. ? ?? ?=? , ∑???? = ???. ? ?? ?=?...
. In a random sample of five people, the mean driving distance to work was 25.4...
. In a random sample of five people, the mean driving distance to work was 25.4 miles and the standard deviation was 5.2 miles. Assume the random variable is normally distributed and use a tdistribution to find the margin of error and construct confidence intervals for the population mean at the following confidence levels: (4 points each) a. 85% b. 90% c. 99%
One wants to study the daily mean travel distance of a delivery service. In trial runs...
One wants to study the daily mean travel distance of a delivery service. In trial runs of 18 randomly chosen delivery trucks, the mean and the standard deviation are found to be 310 km and 70km, respectively. Assume that the daily travel distance is normally distributed. At the 0.1 level of significance, test the claim that the daily mean travel distance is different from 350 km. What is the CONCLUSION? Select one: a. there is not enough assumptions to do...
A group of researchers wants to estimate the true mean skidding distance along a new road...
A group of researchers wants to estimate the true mean skidding distance along a new road in a certain forest. The skidding distances​ (in meters) were measured at 20 randomly selected road sites. These values are given in the accompanying table. Complete parts a through d. 488 347 454 196 286 408 571 438 546 385 300 432 184 263 272 402 311 315 140 426 a. Estimate the true mean skidding distance for the road with a 99% confidence...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT