inal Exami8Tire pressure monitoring systems (TPMS) warn the driver when the tire pressure of the vehicle is 24% below the target pressure.Suppose the target tire pressure of a certain car is 30 psi (pounds per square inch.)(a) At what psi will the TPMS trigger a warning for this car? (Round your answer to 2 decimal place.)ppedWhen the tire pressure isProbability(b) Suppose tire pressure is a normally distributed random variable with a standard deviation equal to 2 psi. If the car's average tirepressure is on target, what is the probability that the TPMS will trigger a warning? (Round your answer to 4 decimal places.)aw11- here to searchECpsi.1¹Saved< Prev18 of 19 HIINext >20H89°F Su Exam(b) Suppose tire pressure is a normally distributed random variable with a standard deviation equal to 2 psi. If the car's average tirepressure is on target, what is the probability that the TPMS will trigger a warning? (Round your answer to 4 decimal places.)Probability(c) The manufacturer's recommended correct inflation range is 28 psi to 32 psi. Assume the tires' average psi is on target. If a tire onthe car is inspected at random, what is the probability that the tire's inflation is within the recommended range? (Round yourintermediate calculations and final answer to 4 decimal places.)Probabilityto search10SavedC< Prev18 of 19Next >Help89°F SunnySa

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Answered: Lems (TPMS) warn the driver when the…

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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Assume that a randomly selected subject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1. Draw a graph and find P15, the 15th percentile. This is the bone density score separating the bottom 15% from the top 85%. Which graph represents P15? Choose the correct graph below. O A. 1 P15 Q The bone density score corresponding to P15 is (Round to two decimal places as needed.) B. P15 Q Q -C*** O C. A P15 O D. A P15 Q
For a certain group of individuals, the average heart rate is 74 beats per minute. Assume the variable is normally distributed and the standard deviation is 4 beats per minute. If a subject is selected at random, find the probability that the person has the following heart rate. Use a graphing calculator. Round the answers to four decimal places. Less than 78 beats per minute =
Terri Vogel, an amateur motorcycle racer, averages 129.45 seconds per 2.5 mile lap (in a 7 lap race) with a standard deviation of 2.29 seconds . The distribution of her race times is normally distributed. We are interested in one of her randomly selected laps. (Source: log book of Terri Vogel) Let X be the number of seconds for a randomly selected lap. Round all answers to 4 decimal places where possible.a. What is the distribution of X? X ~ N()b. Find the proportion of her laps that are completed between 130.57 and 132.63 seconds. c. The fastest 3% of laps are under seconds.d. The middle 50% of her laps are from seconds to seconds.
Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 240 feet and a standard deviation of 48 feet. Let X = distance in feet for a fly ball. If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled fewer than 186 feet? (Round your answer to four decimal places.) ch6#4
The data represent the results for a test for a certain disease. Assume one individual from the group is randomly selected. Find the probability of getting someone who tested positive, given that he or she had the disease. The probability is approximately (Round to three decimal places as needed.) View an example @ 2 s W S X 3 дв command # Get more help 26 80 F3 E D A GA C a R F tv % 5 V T G The individual actually had the disease O Yes No 135 Positive Negative 13 VAG 810 MacBook Air 6 B Y & 7 H F7 U 25 127 N * 8 DII 1 M ( 9 K O 0 I H qqqqqqqqı L A F10 Clear all P > л. : ; { + [ = Check answer I command option 11 18 ? F12 1 } 1 delete
Students in a class took a test whose results were normally distributed. The mean score was an 81 and the standard deviation was 4.5. What score is at the 83rd percentile (round to the nearest whole number.) a.84 b.83 c.85 d.86
Pl Assume that thermometer readings are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A thermometer is randomly selected and tested. For the case below, draw a sketch, and find the probability of the reading. (The given values are in Celsius degrees.) Between 0.50 and 2.00 Click to view page 1 of the table. Click to view page 2 of the table. Draw a sketch. Choose the correct graph below. OA. OB. О в. C. z-0.50 z-2.00 z-0.50 z-2.00 z-0.50 z-2.00 The probability of getting a reading between 0.50°C and 2.00°C is. Click to select your answer(s). atv 4 4. MacBook Air F10 F11 F9 F6 F7 F8 임미 F3 F5 F4 F2 F1 & %$4 4. #3 9- 6.
Assume that adults have IQ scores that are normally distributed with a mean of u = 105 and a standard deviation a = 15. Find the probability that a randomly selected adult has an IQ between 91 and 119. Click to view page 1 of the table. Click to view page 2 of the table. The probability that a randomly selected adult has an IQ between 91 and 119 is. (Type an integer or decimal rounded to four decimal places as needed.) Enter your answer in the answer box. MacBook Air esc 888 %23 & 2 3 4 5 6 7 8 Q E R T Y tab P A D G H caps lock K V в N M > ? control option command command option
Assume that thermometer readings are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A thermometer is randomly selected and tested. For the case below, draw a sketch, and find the probability of the reading. (The given values are in Celsius degrees.) Between 1.50 and 2.25 Click to view page 1 of the table. Click to view page 2 of the table. Draw a sketch. Choose the correct graph below. O A. The probability of getting a reading between 1.50°C and 2.25°C is (Round to four decimal places as needed.) d Normal Table (Page 2) .00 Standard Normal (z) Distribution: Cumulative Area from the LEFT .5000 5398 5793 .6179 .6554 .6915 7257 .7580 7881 .8159 .8413 8643 .8849 .9032 9192 9332 9452 .9554 .9641 9713 9772 .9821 9861 0007 .01 5040 5438 5832 6217 .6591 z=1.50 z=2.25 6950 7291 .7611 7910 .8186 .8438 8665 8869 9049 9207 9345 9463 .9564 9649 .9719 9778 9826 9864 .02 5080 5478 5871 6255 6628 .6985 7324 7642 7939 .8212 8461 8686 8888 .9066 9222 9357 9474 .9573 9656…
Terri Vogel, an amateur motorcycle racer, averages 129.52 seconds per 2.5 mile lap (in a 7 lap race) with a standard deviation of 2.28 seconds . The distribution of her race times is normally distributed. We are interested in one of her randomly selected laps. (Source: log book of Terri Vogel) Let X be the number of seconds for a randomly selected lap. Round all answers to 4 decimal places where possible.a. What is the distribution of X? X ~ N(,)b. Find the proportion of her laps that are completed between 128.63 and 129.99 seconds.c. The fastest 3% of laps are under seconds.d. The middle 80% of her laps are from seconds to seconds.
Assume that a randomly selected subject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1. Draw a graph and find P15, the 15th percentile. This is the bone density score separating the bottom 15% from the top 85%. Which graph represents P15? Choose the correct graph below. O A. AS P15 The bone density score corresponding to P15 is (Round to two decimal places as needed.) O B. A P15 Q C O C. A P15 Q Q O D. A P15
Systolic Blood Pressure Assume that the mean systolic blood pressure of normal adults is 120 millimeters of mercury (mmHg) and the standard deviation is 5.6. Assume the variable is normally distributed. Round the answers to at least 4 decimal places and intermediate z-value calculations to 2 decimal places. Part 1 of 3 If an individual is selected, find the probability that the individual's pressure will be between 117.3 and 120 mmHg. P(117.3 < X < 120) = 111 Part: 1/3 - Part 2 of 3 If a sample of 28 adults is randomly selected, find the probability that the sample mean will be between 117.3 and 120 mmHg. Assume that the sample is taken from a large population and the correction factor can be ignored. P(117.3 < X < 120) = Next Part olo Submit Assignment 2020 meoraw Thill LLC. All Rights Reserved. Terms of Use | Privacy Center

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