Guys, does anyone know the answer?
get the distribution of time needed to complete a certain programming task is approximately normal, with mean 47 minutes and standard deviation 6 minutes. which of the following is closest to the probability that a randomly chosen task will take less than 34 minutes or more than 60 minutes to complete? from EN Bilgi.
Solved The distribution of time needed to complete a certain
Answer to Solved The distribution of time needed to complete a certain
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Using the Normal Distribution – Introductory Statistics
What is the probability of spending more than two days in recovery?
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Based upon the given information and numerically justified, would you be surprised if it took less than one minute to find a parking space?
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Find the probability that it takes at least eight minutes to find a parking space.
Seventy percent of the time, it takes more than how many minutes to find a parking space?
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<!– <solution id=”eip-idp25493760″> N(100, 15) The probability that a person has an IQ greater than 120 is 0.0918. A person has to have an IQ over 130 to qualify for MENSA. The middle 50% of IQ scores falls between 89.95 and 110.05. –>
Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet.
<!– <solution id=”eip-idp105794048″> X ~ N(250, 50) The probability that a fly ball travels less than 220 feet is 0.2743. Eighty percent of the fly balls will travel less than 292 feet. –>
In China, four-year-olds average three hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.5 hours and the amount of time spent alone is normally distributed. We randomly select one Chinese four-year-old living in a rural area. We are interested in the amount of time the child spends alone per day.
<!– <solution id=”eip-idp28319056″> X ~ N(1956.8, 572.3) This is a population mean, because all election districts are included. The probability that a district had less than 1,600 votes for President Clinton is 0.2676. 0.3798 Seventy-five percent of the districts had fewer than 2,340 votes for President Clinton. –>
Suppose that the duration of a particular type of criminal trial is known to be normally distributed with a mean of 21 days and a standard deviation of seven days.
Terri Vogel, an amateur motorcycle racer, averages 129.71 seconds per 2.5 mile lap (in a seven-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.
<!– <solution id=”eip-idp63928720″> X = the distribution of race times that Terry Vogel produces X ~ N(129.71, 2.28) Terri completes 55.17% of her laps in less than 130 seconds. Terri completes 55.17% of her laps in less than 130 seconds. 124.4 and 135.02 –>
Thuy Dau, Ngoc Bui, Sam Su, and Lan Voung conducted a survey as to how long customers at Lucky claimed to wait in the checkout line until their turn. Let X = time in line. (Figure) displays the ordered real data (in minutes):
Suppose that Ricardo and Anita attend different colleges. Ricardo’s GPA is the same as the average GPA at his school. Anita’s GPA is 0.70 standard deviations above her school average. In complete sentences, explain why each of the following statements may be false.
An expert witness for a paternity lawsuit testifies that the length of a pregnancy is normally distributed with a mean of 280 days and a standard deviation of 13 days. An alleged father was out of the country from 240 to 306 days before the birth of the child, so the pregnancy would have been less than 240 days or more than 306 days long if he was the father. The birth was uncomplicated, and the child needed no medical intervention. What is the probability that he was NOT the father? What is the probability that he could be the father? Calculate the z-scores first, and then use those to calculate the probability.
<!– <solution id=”eip-102″> For x = 240, X−μ σ = 240−280 13 =−3.0769 For x = 306, 306−280 13 =2 P(240 < x < 306) = P(–3.0769 < z < 2) = normalcdf(–3.0769,2,0,1) = 0.9762. According to the scenario given, this means that there is a 97.62% chance that he is not the father. To answer the second part of the question, there is a 1 – 0.9762 = 0.0238 = 2.38% chance that he is the father. –>
A NUMMI assembly line, which has been operating since 1984, has built an average of 6,000 cars and trucks a week. Generally, 10% of the cars were defective coming off the assembly line. Suppose we draw a random sample of n = 100 cars. Let X represent the number of defective cars in the sample. What can we say about X in regard to the 68-95-99.7 empirical rule (one standard deviation, two standard deviations and three standard deviations from the mean are being referred to)? Assume a normal distribution for the defective cars in the sample.
We flip a coin 100 times (n = 100) and note that it only comes up heads 20% (p = 0.20) of the time. The mean and standard deviation for the number of times the coin lands on heads is µ = 20 and σ = 4 (verify the mean and standard deviation). Solve the following:
<!– <solution id=”eip-171″> There is about a 68% chance that the number of heads will be somewhere between 16 and 24. z = ±1: x1 = µ + zσ = 20 + 1(4) = 24 and x2 = µ-zσ = 20 – 1(4) = 16. There is about a 95% chance that the number of heads will be somewhere between 12 and 28. For this problem: normalcdf(12,28,20,4) = 0.9545 = 95.45% There is about a 99.73% chance that the number of heads will be somewhere between eight and 32. For this problem: normalcdf(8,32,20,4) = 0.9973 = 99.73%. –>
Facebook provides a variety of statistics on its Web site that detail the growth and popularity of the site.
On average, 28 percent of 18 to 34 year olds check their Facebook profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent.
<!– <solution id=”eip-11″> X = the percent of 18 to 34-year-olds who check Facebook before getting out of bed in the morning. X ~ N(28, 5) P(x ≥ 30) = 0.3446; normalcdf(30,1EE99,28,5) = 0.3446 invNorm(0.95,0.28,0.05) = 0.3622.95% of the percent of 18 to 34 year olds who check Facebook before getting out of bed in the morning is at most 36.22%. P(25 < x < 55). P(25 < x < 55) = normalcdf(25,55,28,5) = 0.7257(0.7257)(400) = 290.28 –>