A salesperson makes four calls per day. A sample of 100 days gives the following frequencies of sales volumes.Number of SalesX0120314234TotalObserved FrequencyRecords show sales are made to 30% of all sales calls. Assuming independent sales calls, the number of sales per day should follow a binomial probability distribution. The binomial probability function presented in Chapter 5 isf(x) =n!x!(n-x)!*p*(1 − p)^ - *.ExpectedFrequencies(days)For this exercise, assume that the population has a binomial probability distribution with n = 4, p = 0.30, and x = 0, 1, 2, 3, and 4.(a) Compute the expected frequencies for x = 0, 1, 2, 3, and 4 by using the binomial probability function.30x = 2x = 3O x = 4Onone of the above332836100Do any of the categories have an expected frequency that is less than five? (Select all that apply.)Ox=0Ux=1 Before conducting a goodness of fit test, we must combine categories if necessary to satisfy the requirement that the expected frequency is five or more for all categories. What categories should we combine in order to satisfy this requirement?O x = 3 and x = 4O x = 2 and x = 3O x = 0 and x = 1O x = 0, x = 1, and x = 3O We don't need to combine any categories.(b) Use the goodness of fit test to determine whether the assumption of a binomial probability distribution should be rejected. Use a = 0.05. Because no parameters of the binomial probability distribution were estimated from the sample data, the degrees of freedom are k-1 when k is the number of categories.State the null and alternative hypotheses.O Ho: The population does not have a binomial distribution.H₂: The population has a binomial distribution.O Ho: The sample does not have a binomial distribution.H₂: The sample has a binomial distribution.O Ho: The population has a binomial distribution.H₂: The population does not have a binomial distribution.OH: The sample has a binomial distribution.H₂: The sample does not have a binomial distribution.Find the value of the test statistic. (Round your answer to three decimal places.)Find the p-value. (Round your answer to four decimal places.)p-value =State your conclusion.O Reject Ho. We conclude that the assumption of a binomial distribution can be rejected.O Reject Ho. We conclude that the assumption of a binomial distribution cannot be rejected.O Do not reject Ho. We conclude that the assumption of a binomial distribution can be rejected.O Do not reject Ho. We conclude that the assumption of a binomial distribution cannot be rejected.

Let X be the random variable that no. of sales. The observed frequencies are given as, x O 0 30…

Answered: (b) Use the goodness of fit test to…

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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A tube of listerine tartar control toothpaste 4.2 ounces. As people use the toothpaste the amount remaining in any tube is random. Assume the amount of toothpaste remaining in the tube follows a uniform distribution. What is the probability that there is more than 1.5 ounces remaining in the tube?
Children being exposed to unsafe levels of lead exposure is thought to be to an environmental factor increasing the chance of a child being diagnosed with ADHD. During the 1990s about 0.3% of children were exposed to unsafe levels of lead (usually from lead pipes). 3. If we let X = the number of children in our study with exposure to unsafe levels of lead we can use the binomial distribution to model the probability distribution of X. Explain why using the definition of Binomial from your notes. (Recall we randomly selected the 425 children and the probability a child is exposed is 0.3%. 4. The researchers found that 5 children in the study had unsafe levels of lead in their blood, Find the probability that in a random sample of 425 children we find at least 5 that have exposure to unsafe levels of lead. 5. Is it unusual to find at least 5 that have exposure to unsafe levels of lead? Explain showing work.
Is the expected value (mean) of the probability distribution of a random variable always one of the possible values of x from the distribution? ExplainWhy not
About 69% of people who are murdered knew the person who committed the murder. Suppose that Detective Williams's case load currently has 49 unsolved murders. Use a nomal distribution approximation to the binomial distribution to find the following probabilities: a. What is the probability that at least 30 of the victims among Detective Williams's case files knew their murderer? b. What is the probability that between 30 and 39 of the victims knew their murderer?
About 40% of all US adults will try to inflate an insurance claim to cover the deductible. Suppose that you work at an insurance company and just received 128 insurance claims. You are interested in the number of inflated claims. Is it fair to say this represents a binomial setting? Explain. Would it be reasonable to use the normal approximation? Explain. What is the probability that more than 80 of the claims are inflated?
The number of automobiles sold by a dealership in a day is an example of The binomial random variable. b. A continuous random variable. c. The normal random variable. d. A discrete random variable.
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I. Boston Children’s Hospital estimates that 90% of Americans have had chickenpoxby the time they reach adulthood.32(a) Suppose we take a random sample of 100 American adults. Is the use of the binomial distributionappropriate for calculating the probability that exactly 97 out of 100 randomly sampled Americanadults had chickenpox during childhood? Explain.(b) Calculate the probability that exactly 97 out of 100 randomly sampled American adults had chickenpoxduring childhood.(c) What is the probability that exactly 3 out of a new sample of 100 American adults have not hadchickenpox in their childhood?(d) What is the probability that at least 1 out of 10 randomly sampled American adults have had chickenpox?(e) What is the probability that at most 3 out of 10 randomly sampled American adults have not hadchickenpox?
A recent study showed that 71% of adults need some type of corrective lens (glasses or contacts). A. If I grab a 6 random people what is the probability that fewer than 4 need corrective lenses? B. If I take a random sample of 2000 people what is the mean and standard deviation of the number of people who need corrective lens? C. Use the normal approximation to the binomial to find the probability that I get 1375 or fewer people who need corrective lenses out of a random sample of 2000. Illustrate what you are finding on a sketch of the normal curve and show your work.

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