1. Find the z-scores that delineate the middle 95% of all values from the others
ID: 3369759 • Letter: 1
Question
1. Find the z-scores that delineate the middle 95% of all values from the others. Enter your answers to 2 or 3 decimal places. from_____to_____
2. Many educational institutions award three levels of Latin honors often based on GPA. These are cumlaude (with high praise), magna cumlaude (with great praise), and summa cumlaude (with highest praise). Requirements vary from school to school. Suppose the GPAs at State College are normally distributed with a mean of 2.95 and standard deviation of 0.43. (a) Suppose State College awards the top 2% of students (based on GPA) with the summa cumlaude honor. What GPA gets you this honor? Round your answer to 2 decimal places. GPA or higher (b) Suppose State College awards the top 10% of students (based on GPA) with the magna cumlaude honor. What GPA gets you this honor? Round your answer to 2 decimal places. GPA or higher (c) Suppose State College awards the top 20% of students (based on GPA) with the cumlaude honor. What GPA gets you this honor? Round your answer to 2 decimal places. GPA or higher
3. The bass in Clear Lake have weights that are normally distributed with a mean of 2.1 pounds and a standard deviation of 0.8 ____pounds. (a) Suppose you only want to keep fish that are in the top 15% as far as weight is concerned. What is the minimum weight of a keeper? Round your answer to 2 decimal places. ____pounds (b) Suppose you want to mount a fish if it is in the top 0.5% of those in the lake. What is the minimum weight of a bass to be mounted? Round your answer to 2 decimal places. ____pounds (c) Determine the weights that delineate the middle 99% of the bass in Clear Lake. Round your answers to 2 decimal places. from___ to pounds
4. On a stretch of Interstate-89, car speed is a normally distributed variable with a mean of 69.5 mph and a standard deviation of 3.9 mph. You are traveling at 76 mph. Approximately what percentage of cars are traveling faster than you? Enter your answer as a percentage with 1 decimal place. (%)
5. Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 9 ounces and a standard deviation of 0.9 ounces. Round your answers to 4 decimal places. (a) If one potato is randomly selected, find the probability that it weighs less than 7 ounces. (b) If one potato is randomly selected, find the probability that it weighs more than 12 ounces. (c) If one potato is randomly selected, find the probability that it weighs between 7 and 12 ounces.
6. The bass in Clear Lake have weights that are normally distributed with a mean of 1.9 pounds and a standard deviation of 0.8 pounds. (a) If you catch one random bass from Clear Lake, find the probability that it weighs less than 1 pound? Round your answer to 4 decimal places. (b) If you catch one random bass from Clear Lake, find the probability that it weighs more than 3 pounds? Round your answer to 4 decimal places. (c) If you catch one random bass from Clear Lake, find the probability that it weighs between 1 and 3 pounds? Round your answer to 4 decimal places.
7. Assume the average life-span of those born in the U.S. is 78.2 years with a standard deviation of 16 years. The distribution is not normal (it is skewed left). The good people at Live-Longer-USA (fictitious) claim that their regiment of acorns and exercise results in longer life. So far, 40 people on this program have died and the mean age-of-death was 83.7 years. (a) Calculate the probability that a random sample of 40 people from the general population would have a mean age-of-death greater than 83.7 years. Round your answer to 4 decimal places. (b) Which statement best describes the situation for those in the Live Longer program? This provides solid evidence that acorns and exercise have nothing to do with age-of-death. Since the probability of getting a sample of 40 people with a mean age-of-death greater than those in the Live Longer program is so small, this suggests that people enrolled in the program do actually live longer on average. This provides solid evidence that acorns and exercise cause people to live longer. (c) Why could we use the central limit theorem here despite the parent population being skewed? Because the sample size is greater than 30. Because the sample size is less than 100. Because skewed-left is almost normal. Because the sample size is greater than 20.
Explanation / Answer
1) Solution :
P(-z < Z < z) = 95%
P(Z < z) - P(Z < z) = 0.95
2P(Z < z) - 1 = 0.95
2P(Z < z) = 1 + 0.95
2P(Z < z) = 1.95
P(Z < z) = 1.95 / 2
P(Z < z) = 0.975
z =1.96
P(-1.96 < Z < 1.96) = 95%
Z-scores from : -1.96 to +1.96 .
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