5. Suppose that the age of a randomly selected business major can be considered tobe a random variable with mean 20 years and standard deviation 2 years. A group of64 business majors is selected at random and their average age is calculated.a) What is the approximate sampling distribution of and why?T is approximately normal with mean 20 and s.d. 2 by central limit theoremT is approximately normal with mean 20 and s.d.2√64T is approximately binomial with mean 20 and s.d.theoremby central limit theorem2√64by central limitT is approximately binomial with mean 20 and s.d. 2 by central limit theorem

We have given the following information Mean = 20 Standard deviation = 2 And n = 64

Answered: 5. Suppose that the age of a randomly…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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2. Suppose it is claimed that amongst the population of STM1001 students, the number of items bought online in the past year is normally distributed with a mean of u 20 and variance of o2 = 25. Further suppose that we wish to study the mean of the population by taking a random sample of n = 61 STM1001 students. Using this information, write down the distribution of the sample mean, X. Note: In each box, enter a single number rounded to four decimal places where relevant. (2 marks) X~ N( 20 0.4098 2 4. Again following on from the previous question, what is the probability that when we take a random sample of n = 61 STM1001 students, the sample mean would be either greater than 20.3521 or less than 19.6479? [int. orty will b own that
2. A population has member numbers 2, 4, 6, 8, 10, 12. Then 2 samples are taken at random and without replacement. If it is known that u =7 Find: - Population variance (o?) - The mean standard deviation of the sample (ox?)
6) A normal distribution has mean 500 and variance 2500. Find the area under the normal curv the mean to 600.
4. Researchers working for a sports drink company would like to know if a new blend if its popular sports drink can help improve marathon runners' final running time. Marathon times, Y, are known to follow a Normal distribution. Suppose that runners using the old blend sports drink have a mean finishing time of 300 minutes (µ = 300 minutes). A random sample of 20 runners was selected to use the new blend of sports drink in a marathon, finishing with an average time of 270 minutes and a standard deviation of 20 minutes (y = 270 min, s = 20 min). Can it be claimed that the new blend lowers finishing times? COr is it possible that random chance alone can explain the discrepancy? Carry out a hypothesis test of the true mean u using the 4- step procedure. Set your alpha-level to 0.05. %3D c. Which test statistic and sampling distribution under the null should we use? O i. z-score; standard Normal O ii. t-statistic; Student'st
4. In a population of 10,000 businesses, we want to estimate the average sales Y For that, we sample n = = 100 businesses using simple random sampling. Furthermore, we have at our disposal the auxiliary information 'number of employees', denoted by x, for each business. The data coming from the sample are: - - X = 50 employees (true mean for x ), - Ŷ = 5.2 × 10 Euros (average sales in the sample), - X = 45 employees (sample mean), - s² = 25 x 1010 (sample variance of y₁), - 2 - s² = 15 (sample variance of x^), x p=0.80 (sample linear correlation coefficient between x and y ), a. Calculate the ratio estimate of Ỹ. Is this estimate unbiased or biased? Explain. b. What is the ‘true' variance formula for this estimate? c. Calculate an estimate of the true variance. Is the variance estimate unbiased or biased? Explain. d. Give a 95% confidence interval for Y.
Number 16
2. Doctors volunteer their time on a daily basis in a clinic. The total number of patients served in a day at the clinic has mean 90 and variance 1890. Determine the probability that 120 or more patients can be served in a day at the clinic. Use the Central Limit Theorem with continuity correction. Use linear interpolation with the normal distribution to find the probability.
6.3.10 Question Help ▼ Three randomly selected households are surveyed. The numbers of people in the households are 1, 4, and 10. Assume that samples of size n =2 are randomly selected with replacement from the population of 1, 4, and 10. Construct a probability distribution table that describes the sampling distribution of the proportion of odd numbers when samples of sizes n = 2 are randomly selected. Does the mean of the sample proportions equal the proportion of odd numbers in the population? Do the sample proportions targe the value of the population proportion? Does the sample proportion make a good estimator of the population proportion? Listed below are the nine possible samples. 1,1 1,4 1,10 4,1 4,4 4,10 10,1 10,4 10,10 2 Construct the probability distribution table. Sample Proportion Probability (Type an integer or fraction.)
5. Suppose X is the number of successes in a random sample of size n from a large population with a proportion of success p. What are the mean and the standard deviation of the sampling distribution of X in terms of n and p? The mean is p(l-pland the standard deviation is p(1-p) b. The mean is npand the standard deviation is ap(1-p). The mean is p(1-pland the standard deviation is pp. C. d. The mcan is 7pand the standard deviation is rp.
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. (a) In words, define the random variable X. Edit Insert - Formats - B A く) (b) Find the ZSCore for a trial lasting 20 days. (Round your answer to three decimal places.) (c) If one of the trials is randomly chosen, find the probability that it lasted at least 20 days. (d) Shade the area corresponding to this probability in the graph below. (Hint: The x-axis is the z-score. Use your z-score from part (b), rounded to one decimal place).
when a population is uniformly distributed (flat), is the sampling distribution of xbar for random samples of 2 observations roughly normal, strongly skewed, or triangular?
The weight of a male bald eagle has approximately a normal distribution with mean ?= 7.4 pounds and a standard deviation of ?= 1.6 pounds. Suppose we take a random sample of n=64 male bald eagles. Let X̄ be the random variable representing the mean weight in pounds of the 64 sampled male bald eagles. Let Xtot be the random variable representing sum of the weights of the 84 sampled bald eagles. a) About what proportion of male bald eagles have weights between 7.0 and 7.8 pounds? b) About what proportion of male bald eagles have weights greater than 7.8 pounds? c) About how many of the 64 sampled male bald eagles would you expect to have weight greater than 7.8 pounds? (nearest integer)? d) About how many of the 64 sampled male bald eagles would you expect to have weight between 7.0 and 7.8 pounds? (nearest integer)? e) What is the standard deviation of the distribution of X̄ (in pounds)? f) What is the standard deviation of the distribution of Xtot (in pounds)? g) What is the…

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