The probability of no shows on the ABC airlines is given in the table below. What is the expected number of no shows and the variance? No Shows 0 1 2 3 4 Select one: O a. 0.20,3.61 O b. 2.00,5.12 O c. 1.90,1.51 O d. 0.40,8.73 Probability 0.15 0.25 0.26 0.23 0.11
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| x | p(x) | x*p(x) | |
| 0 | 0.15 | 0 | 0 |
| 1 | 0.25 | 0.25 | 0.25 |
| 2 | 0.26 | 0.52 | 1.04 |
| 3 | 0.23 | 0.69 | 2.07 |
| 4 | 0.11 | 0.44 | 1.76 |
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- Refer to the sample data for polygraph tests shown below. If one of the test subjects is randomly selected, what is the probability that the subject is not lying? Is the result close to the probability of 0.4720.472 for a negative test result? Did the Subject Actually Lie? No (Did Not Lie) Yes (Lied) Positive test results 1414 4343 Negative test results 3737 1414 The probability that a randomly selected polygraph test subject was not lying is _____.Refer to the sample data for pre-employment drug screening shown below. If one of the subjects is randomly selected, what is the probability that the test result is a false positive? Who would suffer from a false positive result? Why? Pre-Employment Drug Screening Results Positive test result Negative test result Drug Use Is Indicated Drug Use Is Not Indicated Subject Uses Drugs 37 12 Subject Is Not a Drug User 19 32 The probability of a false positive test result is nothing. (Round to three decimal places as needed.)use a ti 84 device
- Talent Corp is a HR-consulting firm. The number of new clients that they have obtained each month has ranged from 0 to 6. The number of new clients has the probability distribution that is shown below. Number of New Clients Probability 0 0.05 1 0.10 2 0.15 3 0.30 4 0.25 5 0.10 6 0.05 Compute the expected number of new clients each month. Compute the variance and standard deviation for the expected number of new clients. What is the probability that there will be 2 or fewer clients next month?The following table represents a Discrete Probability Distribution for the number of days it takes John’s Garage to repair vehicles for its customers. Number of days to repair a vehicle 1 2 3 4 5 6 7 Probability 26.7% 33.6% 15.8% 13.7% 6.3% 2.4% 1.5% With the aid of a table, calculate the following mean variance, and standard deviationRefer to the sample data for pre-employment drug screening shown below. If one of the subjects is randomly selected, what is the probability that the test result is a false positive? Who would suffer from a false positive result? Why? Pre-Employment Drug Screening Results Positive test result Negative test result Drug Use Is Indicated Drug Use Is Not Indicated Subject Uses Drugs 42 11 Subject Is Not a Drug User 2 29 The probability of a false positive test result is (Round to three decimal places as needed.) Who would suffer from a false positive result? Why? O A. The person tested would suffer because he or she would be suspected of using drugs when in reality he or she does not use drugs. B. The employer would suffer because the person tested would not be suspected of using drugs when in reality he or she does use drugs. C. The person tested would suffer because he or she would not be suspected of using drugs when in reality he or she does use drugs. D. The employer would suffer…
- Refer to the sample data for polygraph tests shown below. If one of the test subjects is randomly selected, what is the probability that the subject is not lying? Is the result close to the probability of 0.443 for a negative test result? Did the Subject Actually Lie? No (Did Not Lie) Yes (Lied) O Positive test results 10 44 Negative test results 36 The probability that a randomly selected polygraph test subject was not lying is (Type an integer or decimal rounded to three decimal places as needed.) Is the result close to the probability, rounded to three decimal places, of 0.443 for a negative test result? because there is V a 0.050 absolute difference between the probability of a true response and the probability of a negative test result. Enter your answer in the answer box. Prise 6. LEGOHelp with B & C.We toss a symmetric coin five times. Let X denote the number of eagles tossed. Determine the distribution of the random variable X, calculate its expected value, variance, and standard deviation.
- A variable y of a finite population has the frequency distribution shown in the table. Suppose a member is selected at random from the population and let Y denote the value of the variable y for the member obtained. Complete parts (a) through (d) below. y 3 4 6. f 1 1 11 7 a. Determine the probability distribution of the random variable Y. 3 y P(Y = y) (Type integers or decimals. Do not round.) b. Use random-variable notation to describe the events that Y takes on the value 5, a value less than 5, and a value of at least 5. The event that Y takes on the value 5 can be represented as The event that Y takes on a value less than 5 can be represented as The event that Y takes on a value of at least 5 can be represented as { }. c. Find P(Y = 5), P(Y 5). %3D P(Y = 5) = (Type an integer or a decimal. Do not round.) P(Y<5) = (Type an integer or a decimal. Do not round.) P(Y 2 5) = (Type an integer or a decimal. Do not round.) LOBlood Types and Rh Factors In addition to being grouped into four types, human blood is grouped by its Rhesus (Rh) factor. Consider the figures below that show the distributions of these groups for Americans. A B АВ Rh + 37% 34% 10% 4% Rh - 6% 6% 2% 1% Source: www.infoplease.com Send data to Excel Choose 1 American at random. Find the following probabilities. Enter your answers as decimals rounded to 3 decimal places.What are the differences between the two types of random variables in statistics? B i U Font Family AAA = in O 0
