The lifetimes of the tread of a certain automobile tire are normally distributed with mean 45,000 miles and standard deviation 3,000miles.For each of the questions below make sure to show your work. Round any z-score calculated to two decimal places to find your result.1. What percent of tires have a tread life between 39,000 and 51,000 miles?2. The manufacturer wants to know what percent of times tread life is less than 37,500 miles.3. Find the tread life for the 90th percentile. Round to nearest integer (if needed)- No text entered -Hide Feedback

Answered: The lifetimes of the tread of a certain…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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A recent Nielsen analysis found that the typical U.S. smartphone user is spending an average of 220 minutes per day using apps. We speculate that the average time spent using apps for Generation X smartphone users (aged 41-56 years) is less than the national average. The hypotheses are Ho: H = 220 minutes versus Ha: µ < 220 minutes. Summary results from R are provided. Summary Statistics Std. Dev (s) Mean Sample Size (n) 205 minutes 63.25 minutes 31
Answer number 3
Please help by providing detailed workings. Answers to a minimum of 6 d.p. Thanks
The blood pressure in millimeters was measured for a large sample of people. The average pressure is 140 mm, and the SD of the measurements is 20 mm. The histogram looks reasonably like a normal curv Use the normal curve to estimate the following percentages. Choose the answer that is closest to being correct. A 19.1% B 30.9% C 38.2% D 69.1% E 68.27% A The percentage of people with blood pressure between 130 and 150 mm. B The percentage of people with blood pressure between 140 and 150 mm. D The percentage of people with blood pressure over 150 mm.
Using the given data, find the following. 28 3 15 14 11 22 Identify the five-number summary. Round your answers to one decimal place as needed. Lo=, Q1=, MD=, Q3=, High= Find the interquartile range. Round your answer to one decimal place as needed. IQR=
The average height of 1,000 children is 150cm with a standard deviation of 10cm. How many children stand between 155cm and 150cm ? A. Find the raw score for z= -1,-2,-3,+1,+2,+3 B. Sketch the graph C. What z score corresponds to 50? D. What z score corresponds to 60? E. Find the total area . F. Multiply the total area by 1000. G. Make a conclusion.
Use the following set of values for #10 - 13. Round answers to nearest hundredth, if necessary. .3, .6, 9, 1.3, .4, .6, 1.2, 1.4, 1.1, .2, .2 10. Find the mean, median, and mode(s) 11. Find the range 12. Find the interquartile range 13. Find the standard deviation
Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 52.5 degrees. Low Temperature (o F) 40 - 44 45 - 49 50 - 54 55 - 59 60 – 64 D Frequency 2 9. ... The mean of the frequency distribution is degrees. (Round to the nearest tenth as needed.) Which of the following best describes the relationship between the computed mean and the actual mean? O A. The computed mean is close to the actual mean because the difference between the means is more than 5% of the actual mean. B. The computed mean is not close to the actual mean because the difference between the means is less than 5% of the actual mean. C. The computed mean is not close to the actual mean because the difference between the means is more than 5% of the actual mean. D. The computed mean is close to the actual mean because the difference between the means is less than 5% of the actual mean.
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Number of Students 30 25 2 15 10 5 0 S 10 1 15 3 20 SAMSUNG 2.5 24 3 30 Grade 28 35 SUBMIT QUESTION 19 40 25 45 SAVE PROGRESS 50 55 SU
Using the given data, find the following. 6.8, 15.2, 16.2, 7.4, 18.3, 5.6, 13.5, 11.3, 4.6, 9.7 Send data to Excel Part: 0 / 2 Part 1 of 2 Identify the five-number summary. Round your answers to one decimal place as needed. 0.e,-0, MD =D, e,=D, High = Q3 Low =

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