**Estimate and Confidence Interval for College Graduation Time**The data analysis of 42,000 college graduates indicates that the mean time required to graduate with a bachelor's degree is approximately **5.44 years**. This average is calculated with a standard deviation of **1.88 years**. To further explore this data point, a 95% confidence interval is sought for the mean time needed to graduate among all college graduates. This interval provides a range within which the true mean time is expected to lie with a high degree of certainty. The question remains to compute this 95% confidence interval:□ years < μ < □ years*Note: Please ensure to round your answers to two decimal places as necessary.*These statistical estimates allow educational institutions and policymakers to understand typical graduation timelines and better facilitate student success and resource allocation.

Given: sample size(n)=4200sample meanx¯=5.44standard deviation(s)=1.88 Estimate for the mean time…

Answered: estimate the mean time required to…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Similar questions

The mean number of carats for 313 diamonds in a sample data set is 0.506 carats, with a standard deviation of 0.145 carats. Use the mean and standard deviation to form an interval that will contain at least 75% of the carat values in the data set. (Round your answer to 3 decimal places)
to qualify for a police academy, applicants are given a test of physical fitness. The scores are normally distributed with a mean of 66 and a standard deviation of 10 . If only the top 13% of the applicants are selected, find the cutoff score. Round z-value calculations to 2 decimal places and the final answer up to the nearest whole number. Cutoff score
The mean SAT score for 13 male highschool students is 500 and the standard deviation is 100. the mean SAT score for 16 female highschool students is 490 and the standard deviation is 80. At a=.05 are male SAT scores higher than female SAT scores? (a=.05)
A standardized exam's scores are normally distributed. In a recent year, the mean test score was 1466 and the standard deviation was 315. The test scores of four students selected at random are 1890, 1210, 2200, and 1340. Find the z-scores that correspond to each value and determine whether any of the values are unusual. The Z-score for 1890 is. (Round to two decimal places as needed.).
A right skewed data set has a mean of 35 and a standard deviation of 25. What can be said about the amount of data contained within the interval (-15, 85)? 95% of the data is within this range 68% of the data is within this range Nothing At least 75% of data is within this range
How to find the standard deviation after we have found the range?
The combined SAT scores for the students at a local high school are normally distributed with a mean of 863 and a standard deviation of 156. The local college requires a minimum SAT score of 877 before students are considered for admision.What percentage of students from this school have SAT scores that do not satisfy the local college's admission requirement? Enter your answer as a percent accurate to 2 decimal places.
IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = IQ of an individual. The middle 30% fall between what two values? (round to 2 decimal places)
a sugar company packages sugar in 5kg bags. the amount of sugar per bag varies according to a normal distribution. a sample of 25 bags is selected from the days production and the mean for the day is 4.98 kg and the standard deviation is 0.05 kg. ( use 1-a= 0.95 ) the margin of error for the amount of sugar is?
he average amount of time boys and girls spend playing sports each week is believed to be the same. A sample of equal size n is taken from boys and girls , the mean of the boys is 29 hours with standard deviation 4 hours and the mean of the girls is 27 hours with standard deviation 3 hours Both populations have a normal distribution. The calculated value is 1.2 n =

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