MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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- There are two types of cellular phones, handheld phones (H) that you carry and mobile phones ( M) that are mounted in vehicles. Phone calls can be classified by the traveling speed of the user as fast (F) or slow ( W). Monitor a cellular phone call and observe the type of telephone and the speed of the user. The probability model for this experiment has the following information: P[F] = 0.5, P[HF] = 0.2, P[MW] = 0.1. What is the sample space of the experiment? Find the following probabilities P[ W], P[MF], and P[ H].arrow_forwardGiven the following probabilities, which event is least likely to occur? P(A) = 1/5 P(B) = 0.23 P(C) = 25% P(D) = 1/3arrow_forwardWhen functioning correctly, a machine produces switches with a 0.8% defective rate. Find the probability that a batch of 1,000 switches contains 14 or more defective switches (assuming that the machine functions correctly, i.e., the population defective rate is 0.8%). Suppose a batch of 1,000 switches were inspected when completed, and 14 defective switches were found. Based on your answer to part a, does this sample imply that the machine needs adjustment? Explain your answer.arrow_forward
- Many fire stations handle emergency calls for medical assistance as well as calls requesting firefighting equipment. A particular station says that the probability that an incoming call is for medical assistance is 0.84. This can be expressed as P(call is for medical assistance) = 0.84. (a) Give a relative frequency interpretation of the given probability. O Eighty-four percent of all calls to the station are for medical assistance. O None of the calls to the station are for medical assistance. O Sixteen percent of all calls to the station are for medical assistance. O All calls to the station are for medical assistance. O Fifty percent of all calls to the station are for medical assistance. (b) What is the probability that a call is not for medical assistance?arrow_forwardBased on information from MRI Network, some job applicants are required to have several interviews before a decision is made. The number of required interviews and the corresponding probabilities are: 1 (0.09); 2 (0.31); 3 (0.37); 4 (0.12); 5 (0.05); and 6 (0.05). Questions: 1. Create a table showing the probability distribution shown in the problem. 2. Create a probability histogram from the given problem.arrow_forwardFor a certain antidepressant, about 80% of patients experience some sort of side effect. A researcher takes a random sample of 23 patients and surveys whether they have experienced any side effects. Use the binomial distribution to compute the probability that exactly 16 of the patients in the sample have experienced at least one side effect. Identify the following information required to find the probability of patients who have experienced side effects. Provide your answer below: n = X = p= trials successes probability of side effectsarrow_forward
- A concern for many adults is that computers and the Internet are reducing privacy. A survey conducted by Peter D. Hart Research Associates for the Shell Poll was reported in the USA Today. According to the survey, 37% of adults are concerned that employers are monitoring phone calls. Use the binomial distribution formula to calculate the probability that out of five adults, exactly three are concerned that employers are monitoring their phone calls. Select one: O a. 1.85 Ob. 0.799 Oc. 0.201 d. 3.15arrow_forwardA diagnostic test for disease X correctly identifies the disease 90% of the time. False positives occur 8%. It is estimated that 5.41% of the population suffers from disease X. Suppose the test is applied to a random individual from the population. Compute the following probabilities. (It may help to draw a probability tree.) The percentage chance that the test will be positive = The probability that, given a positive result, the person has disease X =arrow_forwardAssume that a sample is used to estimate a population proportion p. Find the 99.5% confidence interval for a sample of size 88 with 16 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.arrow_forward
- The natural remedy echinacea is reputed to boost the immune system, which will reduce flu and colds. A 6-month study was undertaken to determine whether the remedy works. From this study, the following probability distribution of the number of respiratory infections per year (X) for echinacea users was produced: 3 4 1 2 P(X) 0.342 0.323 0.203 0.074 0.058 Find the following probabilities: A. An echinacea user has more than one infection per year B. An echinacea user has no infection per year C. An echinacea user has between one and three (inclusive) infections per yeararrow_forwardThe natural remedy echinacea is reputed to boost the immune system, which will reduce flu and colds. A 6-month study was undertaken to determine whether the remedy works. From this study, the following probability distribution of the number of respiratory infections per year (X) for echinacea users was produced: 3 4 P(X) 0.339 0.327 0.203 0.072 0.059 Find the following probabilities: A. An echinacea user has more than one infection per year B. An echinacea user has no infections per year C. An echinacea user has between one and three (inclusive) infections per yeararrow_forwardUse the given probability value to determine whether the sample results could easily occur by chance, then form a conclusion. A study of the effect of seatbelt use in head-on passenger car collisions found that drivers using a seatbelt had a 64.1% survival rate, while drivers not using a seatbelt had a 41.5% survival rate. If seatbelts have no effect on survival rate, there is less than a 0.0001 chance of getting these results. What do you conclude? The probability shows that the sample results ▼ could could not have easily occurred by chance. It appears that there ▼ is is not sufficient evidence to conclude that seatbelts do have an effect on survival rate.arrow_forward
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