Seventy million pounds of trout are grown in the U.S. every year. Farm-raised trout contain an average of 32 grams of fat per pound, with a standard deviationof 8 grams of fat per pound. A random sample of 40 farm-raised trout is selected. The mean fat content for the sample is 30.7 grams per pound. Find theprobability of observing a sample mean of 30.7 grams of fat per pound or less in a random sample of 40 farm-raised trout.Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.?

here random sample of 40 farm raised trout is selected here use central limit theorem for sampling…

Answered: probability of observing a sample mean…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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NORMAL DISTRIBUTION 1.) It is measured that the average time vehicles travel from San Marino to San Vicente is 84.5 minutes with a standard deviation of 4.2 minutes. Find the probability that a vehicle from San Marino will arrive in San Vicente with a travel time between 58.6 minutes to 72.3 minutes. (Draw a box around your final answer.)
Test the hypothesis that the average content of containers of a particular lubricant is 10 liters if the contents of a random sample of 10 containers are 9.7,9.7,10.3,10.3,9.8,9.8,9.7,10.5,10.4,and10.4 liters.Use a 0.10 level of significance and assume that the distribution of contents is normal.
Fit a binomial distribution for which the Mean is 6, and the Variance is 4.
The formula for finding a z score for an individual value when there is a normal distribution is z = mean – value / Standard deviation True False
Sample data were collected in a study of calcium supplements and their effects on blood pressure. A placebo group and a calcium group began the study with measures of blood pressures for each group. What test would be most appropriate, if you have non-normal data? Name of Test _____________________________________________
Rent for two-bedroom apartments in Davis, CA, follow a normal distribution with mean of $2,200 and standard deviation of $300. A random sample of 36 two-bedroom homes results in a mean $2,075. What is the probability that the sample mean is greater than or equal to $2,075?
The full height of a certain species of tree has a normal probability distribution with a mean of 195.6 ft and a standard deviation of 7.5 ft. State the random variable. ¯x = the height of a certain species of tree ¯x = the mean height of a certain species of tree x = the height of a certain species of tree x = the diameter of a certain species of tree A tree of this type grows in my backyard, and it stands 187.4 feet tall. Find the probability that the height of a randomly selected tree is shorter than mine. Round to three decimal places.Which is the notation for the probability that the height is shorter than mine? P(x < 187.4) P(¯x < 195.6) P(¯x > 195.6) P(x > 195.6) P(¯x< 187.4) P(x < 195.6)
The distribution of average hours worked per week for random samples of 1000 workers in the U.S. has a mean of 40 and a standard error of 0.04 The distribution in this scenario is a: sampling distribution sample distribution population distribution
Because variance divided by the sample size gives you the estimated standard error, as the sample size ______, the estimated standard error will ______. a. increases ; decrease b. decreases ; decrease c. increases ; increase d. decreases ; increase
accidents in two weeks? Q6- Assume the probability that a maple tree at age 5 grows less than 150 cm is equal to 0.3. If the height of maple trees at age 5 are estimated to be normally distributed with mean u cm and variance 64 cm. find u.
A sample of 100 clients of an exercise facility was selected. Let X = the number of days per week that a randomly selected client uses the exercise facility. Frequency 2 1 15 30 3 27 4 10 8 6. 8 Find the 80th percentile.
(a) Find the standard error of the mean for each sampling situation (assuming a normal population). (Round your answers to 2 decimal places.) Standard Error a. o = 30, n = 4 b. o = 30, n = 16 c. o = 30, n = 64

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