Concept explainers
To graph Problems 59-62, use a graphing calculator and refer to the normal
Graph equation(1) with
(A)
(B)
(C)
Graph all three in the same viewing window with
Want to see the full answer?
Check out a sample textbook solutionChapter 10 Solutions
Finite Mathematics for Business, Economics, Life Sciences and Social Sciences
Additional Math Textbook Solutions
Using and Understanding Mathematics: A Quantitative Reasoning Approach (6th Edition)
Calculus for Business, Economics, Life Sciences, and Social Sciences (13th Edition)
Calculus Volume 1
Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)
Using & Understanding Mathematics: A Quantitative Reasoning Approach (7th Edition)
Probability and Statistics for Engineers and Scientists
- What does the y -intercept on the graph of a logistic equation correspond to for a population modeled by that equation?arrow_forwardThe following data are annual 15-min peak rainfall intensities I (in./hr) for 9 years of record. Compute and plot the log,,-normal frequency curve and the data. Use the Weibull plotting position formula. Using both the curve and the mathematical equation estimate (a) the 25-yr, 15-min peak rainfall intensity; (b) the return period for an intensity of 7 in./hr; (c) the probability that the annual maximum 15-min rainfall intensity will be between 4 and 6 in./hr.arrow_forwardSuppose ₁ and ₂ are true mean stopping distances at 50 mph for cars of a certain type equipped with two different types of braking systems. The data follows: m = 9, x = 113.5, s₁=5.08, n = 9, y = 129.7, and s₂ = 5.34. Calculate a 95% CI for the difference between true average stopping distances for cars equipped with system 1 and cars equipped with system 2. (Round your answers to two decimal places.) USE SALT -22.73 X -10.12 1x )arrow_forward
- Suppose ₁ and ₂ are true mean stopping distances at 50 mph for cars of a certain type equipped with two different types of braking systems. The data follows: m = 9, x = 113.5, s₁ = 5.08, n = 9, y = 129.7, and s₂ = 5.34. Calculate a 95% CI for the difference between true average stopping distances for cars equipped with system 1 and cars equipped with system 2. (Round your answers to two decimal places.) USE SALTarrow_forwardFor Problem 19.27 , determine the probability (assuming normal distribution) that a car would need engine maintenance after 100,000 kilometers.arrow_forwardSuppose your manager indicates that for a normally distributed data set you are analyzing, your company wants data points between z=−0.5z=-0.5 and z=0.5z=0.5 standard deviations of the mean (or within 0.5 standard deviations of the mean). What percent of the data points will fall in that range?Answer: percent (Enter a number between 0 and 100, not 0 and 1 and round to 2 decimal places)arrow_forward
- For data that is not normally distributed we can't use z-scores. However, there is an equation that works on any distribution. It's called Chebyshev's formula. The formula is where P=1- 1/k2 p is the minimum percentage of scores that fall within kk standard deviations on both sides of the mean. Use this formula to answer the following questions. b) If you have scores that are normally distributed, find the percentage of scores that fall within 3.3 standard deviations on both sides of the mean? c) If you have scores and you don't know if they are normally distributed, how many standard deviations on both sides of the mean do we need to go to have 35 percent of the scores? Note: To answer part c you will need to solve the equation for k.k. A manufacturer knows that their items have a normally distributed length, with a mean of 5.3 inches, and standard deviation of 1.5 inches.If one item is chosen at random, what is the probability that it is less than 8.3 inches long?arrow_forwardSuppose that the return for a particular investment is normally distributed with a population mean of 10.1% and a population standard deviation of 5.4%. A person must score in the upper 5% of the population on an IQ test to qualify for a particular occupation If IQ scores are normally distributed with a mean of 100 and a standard deviation of 15, what score must a person have to qualify for this occupation?arrow_forwardIf X N(49, 16), then the standard deviation of X equalsarrow_forward
- A population has a mean of μ= 50 and a standard deviation of σ= 10. If 3 points were added to every score in the population, what would be the new values for the mean and standard deviation? If every score in the population were multiplied by 2, then what would be the new values for the mean and standard deviation?arrow_forwardWe analyze a data set with Y = stopping distance of a car and X = speed of the car when the brakes were applied, %3D and after running the data in STATISTICA, we obtain the following results. Std.Err. of b Std.Err. of b* t(61) p-value b* N=63 Intercept Speed -20.2734 3.1366 -6.26038 20.67978 0.000000 0.000000 3.238368 0.935504 0.045238 0.151674 Sums of df Mean p-value Squares Squares 59540.15 Effect 59540.15 427.6534 0.000000 Regress. Residual 1 8492.74 61 139.23 Total 68032.89 Speed X StopDist Y Speed squared StopDist squared Speed StopDist 65853 Total 1195 2471 28719 164951 One of the observations is (X = 39, Y = 138). The value of the internal studentized residual is . (final answer to 2 decimal places e.g. 2.12) Hence, the point (39, 138) an outlier. (choose from is or is not)arrow_forwardSuppose 4, and u, are true mean stopping distances at 50 mph for cars of a certain type equipped with two different types of braking systems. The data follows: m = 7, x = 113.8, s, = 5.09, n = 7, y = 129.5, and s, = 5.38. Calculate a 95% CI for the difference between true average stopping distances for cars equipped with system 1 and cars equipped with system 2. (Round your answers to two decimal places.) In USE SALT Does the interval suggest that precise information about the value of this difference is available? O Because the interval is so wide, it appears that precise information is not available. Because the interval is so narrow, it appears that precise information is available. O Because the interval is so wide, it appears that precise information is available. Because the interval is so narrow, it appears that precise information is not available.arrow_forward