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- Assume a recent sociological report states that university students drink 5.10 alcoholic drinks per week on average, with a standard deviation of 1.3401. Suppose Jason, a policy manager at a local university, decides to take a random sample of 150 university students to survey them about their drinking habits. Determine the mean and standard deviation of the sampling distribution of the sample mean alcohol consumption. Provide your answer with precision to two decimal places. mean of the sampling distribution = standard deviation of the sampling distribution =1. Identify and explain the three defining characteristics of the normal probability distribution 2. What other famous name is the normal probability distribution know by? 3. Describe the role the following play in the normal probability distribution: Mean (mu) а. b. Standard deviation (sigma) Z-score c. d. Write the formula for the Z-scoreThe standard deviation for length of stay in a hotel is 7.2 days. For the variable “length of hospital stay” the sampling distribution when there are 67 guests is approximately normal and has a standard deviation of 0.8796 days. What is the probability that the sampling error made in estimating the population mean length of stay by the mean length of stay of a sample of 67 patients will be at most 2 days
- Solve for the:1. Mean2. Median3. Mode 4. Standard DeviationHelp!!For a certain type of computers, the length of time between charges of the battery is normally distributed with a mean of 62 hours and a standard deviation of 11 hours. John owns one of these computers and wants to know the number of hours when the computer would have 12% of its battery life left. Sketch the region in question. Write a sentence describing what your answer represents about the problem's specefics.
- Cotinine levels were measured in a group of non-smokers exposed to tobacco smoke (n =40), the mean of the sample was 60.58 ng/Ml and the standard deviation was 138.08 ng/Ml. Also, cotinine levels were measured in a group of non-smokers not exposed to tobacco smoke(n =40), the mean of the sample was 16.35 ng/mL and the standard deviation was 62.53 ng/mL. Cotinine is a metabolite of nicotine, meaning that when nicotine is absorbed by the body, cotinine is produced. Use a 0.05 significance level to test the claim that non-smokers exposed to tobacco smoke have a higher mean nicotine level than nonsmokers not exposed to tobacco smoke. a)Define the parameter and state null and alternative hypothesis b)Test Statistics c) Critical Value Approach d) ConclusionSuppose that heights of 10-year-old boys in the US vary according to a normal distribution with mean of 138 cm and standard deviation of 7. Compute the Z-score for a boy with height of 150 cm and find the correct interpretations. a. Boy's height is 0.71 standard deviation lower the national average b. Boy's height is 1.71 standard deviation above the national average c. Boy's height is 1.71 standard deviation lower than the national average d. Boy's height is 2.71 standard deviation above the national averageThe distribution of the student heights at a large college is approximately bell shaped. If the mean height is 66 inches,and approximately 95% of the heights fall between 32 and 100 inches, then, the standard deviation of the heightdistribution is approximately equal to
- Suppose that a category of world class runners are known to run a marathon (26 miles) in an average of 146 minutes with a standard deviation of 16 minutesConsider 49 of the races Find the probabillity that the average of the sample will be between 144 and 146 min in the 49 marathons B find the 80th percentile for the average of these 49 marathons C find the medianSuppose that a sample of size 52 is drawn from a population with mean 23 and standard deviation 39 . Find the value of , the standard deviation of the distribution of sample means.The ages of a group of 138 randomly selected adult females have a standard deviation of 17.5 years. Assume that the ages of female statistics students have less variation than ages of females in the general population, so leto= 17.5 years for the sample size calculation. How many female statistics student ages must be obtained in order to estimate the mean age of all female statistics students? Assume that we want 95% confidence that the sample mean is within one-half year of the population mean. Does it seem reasonable to assume that the ages of female statistics students have less variation than ages of females in the general population? The required sample size is (Round up to the nearest whole number as needed.)