Assume that on a standardized test of 100 questions, a person has a probability of 85% of answering any particular question correctly. Find the probability of answering between 80 and 90 questions, inclusive round four decimal places. P(80 ≤ X ≤ 90) =
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Q: Assume that on a standardized test of 100 questions, a person has a probability of 75% of answering…
A: Given, n=100 p=0.75 Now, n×p=100×0.75=75>5n×p×1-p=100×0.75×0.25=18.75>5 Therefore assumption…
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Q: Assume that on a standardized test of 100 questions, a person has a probability of 80% of answering…
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Q: Assume that on a standardized test of 100 questions, a person has a probability of 85% of answering…
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Q: 00 R The table summarizes results from 983 pedestrian deaths that were caused by automobile…
A: Driver intoxicated Pedestrian Intoxicated Yes Pedestrian Intoxicated No Total Yes 49 74 123…
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- The table summarizes results from 975 pedestrian deaths that were caused by automobile accidents. Driver Pedestrian Intoxicated? Intoxicated? Yes No Yes 64 82 No 261 568 If one of the pedestrian deaths is randomly selected, find the probability that the pedestrian was not intoxicated or the driver was intoxicated. Report the answer as a percent rounded to one decimal place accuracy. You need not enter the "6" symbol. prob =find the probability of guessing at least 9 out of 14 correctly. Round inetermediate calculations and final answers to three decimal placesFind the indicated probability and interpret the result. From 1975 through 2020, the mean annual gain of the Dow Jones Industrial Average was 653. A random sample of 34 years is selected from this population. What is the probability that the mean gain for the sample was between 400 and 800? Assume o = 1540. The probability is (Round to four decimal places as needed.) Interpret the result. Select the correct choice and fill in the answer box to complete your choice. (Round to two decimal places as needed.) O A. About OB. About OC. About OD. About % of samples of 46 years will have an annual mean gain between 400 and 800. % of samples of 34 years will have an annual mean gain between 653 and 800. % of samples of 34 years will have an annual mean gain between 400 and 800. % of sampl of 34 years will have an annual mean gain between 400 and 653.
- The probability that a certain type of sea turtle hatching survives until adulthood is assumed to be 1 out of 1,877. If 980 hatchlings incubate in a certain nesting area, calculate the probability that at least 3 survive to adulthood. Use three decimal place accuracy.The table summarizes results from 979 pedestrian deaths that were caused by automobile accidents. DriverIntoxicated? Pedestrian Intoxicated? Yes No Yes 59 74 No 248 598 If two different pedestrian deaths are randomly selected, find the probability that they both involved pedestrians that were not intoxicated.Report the answer as a percent with two decimal places accuracy. (You need not enter the "%" symbol).Probability = %The mean percent of childhood asthma prevalence in 43 cities is 2.38%. A random sample of 30 of these cities is selected. What is the probability that the mean childhood asthma prevalence for the sample is greater than 2.8%? Interpret this probability. Assume that o = 1.25%. The probability is (Round to four decimal places as needed.) Interpret this probability. Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as needed.) O A. About % of samples of 43 cities will have a mean childhood asthma prevalence greater than 2.8%. O B. About % of samples of 30 cities will have a mean childhood asthma prevalence greater than 2.38%. O C. About % of samples of 30 cities will have a mean childhood asthma prevalence greater than 2.8%. Click to select your answer(s). P Type here to search 99+ a (? 1:42 PM 4/11/2021 ho insen 144 ese & t backspace %23 3. %24 4 6. 8 2. Y W tab H. J. K pase 5 R
- The table summarizes results from 975 pedestrian deaths that were caused by automobile accidents. DriverIntoxicated? Pedestrian Intoxicated? Yes No Yes 49 73 No 221 632 If one of the pedestrian deaths is randomly selected, find the probability that the pedestrian was not intoxicated.Report the answer as a percent rounded to one decimal place accuracy. prob = %The table summarizes results from 985 pedestrian deaths that were caused by automobile accidents. Pedestrian Intoxicated? Driver Intoxicated? Yes No Yes 53 No 246 W intoxicated or the driver was not intoxicated. If one of the pedestrian deaths is randomly selected, find the probability that the pedestrian was not prob Report the answer as a percent rounded to one decimal place accuracy. You need not enter the "%" symbol. = Submit Question Lory R tudio 86°F Sunny Help K F4 % 5 21 T G B A 6 F5 Y H N & 7 F6 U J M 00 * 8 F7 K 75 이 SM 9 F8 L O F9 ? [ F10 + 11 ] F11 F12 Backspace Enter Shift Insert Print Screen Delete Home Scroll Lock End 10:17 AM 6/21/2022 PgUp Pause Break PgDnDetermine the probability of seeing a Z score of 2.699 or one with a larger absolute value.
- If you roll one six-sided die and X = the # on the die. What is the mean value of X?The distances golf balls travel under laboratory testing conditions are normally distributed. The mean is 312 yards. The probability the golf ball travels further than 319 yards is 0.28. Find the probability a randomly selected golf ball travels between 305 yards and 319 yards. Express the answer as a decimal value rounded to the nearest hundredth.