Finite Math (20 Question)

QUESTION 1The probability distribution of a random variable X isx–2–1012P ( X = x )Compute the mean, variance, and standard deviation of X.a.b.c.d.1 pointsQUESTION 2A probability distribution has a mean of 57 and a standard deviation of 1.4. Use Chebychev’s inequality to find the value of c that guarantees the probability is at least 96% that an outcome of the experiment lies between 57 – c and 57 – c. (Round the answer to nearest whole number.)a.3b.9c.1d.7e.51 pointsQUESTION 3Find the variance of the probability distribution for the histogram:a.Var ( X) = 4.2625b.Var ( X) = 4.65c.Var ( X) = 4.28d.Var ( X) = 4.01251 pointsQUESTION 4The birthrates in the country for the years 1981-1990 are:Year1981198219831984198519861987198819891990Birthrate15.915.515.515.715.715.615.715.916.216.7(The birthrate is the number of live births/1,000 population.)Compute the mean, variance, and standard deviation of the random variable X.a.b.c.d.1 pointsQUESTION 5Rosa Walters is considering investing $10,000 in two mutual funds. The anticipated returns from price appreciation and dividends (in hundreds of dollars) are described by the following probability distributions:Mutual Fund AReturnsProbability-40.280.3100.5Mutual Fund BReturnsProbability-20.260.680.2Compute (in dollars) the mean and variance for each mutual fund.a. Mutual Fund A:Mutual Fund B:b. Mutual Fund A:Mutual Fund B:c. Mutual Fund A:Mutual Fund B:d. Mutual Fund A: Mutual Fund B:1 pointsQUESTION 6The number of Americans without health insurance, in millions, from 1995 through 2002 is summarized in the following table.Year19951996199719981999200020012002Americans40.541.443.644.740.239.241.143What is the standard deviation of Americans without health insurance in the period from 1995 through 2002?a. millionb. millionc. milliond. millione. million1 pointsQUESTION 7The mean annual starting salary of a new graduate in a certain profession is $43,000 with a standard deviation of $500. What is the probability that the starting salary of a new graduate in this profession will be between $39,500 and $46,500?a.At leastb.At leastc.At leastd.At least1 pointsQUESTION 8A survey was conducted by the market research department of the National Real Estate Company among 500 prospective buyers in a large metropolitan area to determine the maximum price a prospective buyer would be willing to pay for a house. From the data collected, the distribution that follows was obtained.Compute the standard deviation of the maximum price (in thousands of dollars) that these buyers were willing to pay for a house. Round the answer to the nearest integer.Maximum Price Considered, 150160170180190220250270320a.b.c.d.e.1 pointsQUESTION 9The following table gives the scores of 30 students in a mathematics examination.Scores90-9980-8970-7960-6950-59Students381351Find the mean and the standard deviation of the distribution of the given data. Hint: Assume that all scores lying within a group interval take the midvalue of that group.a.b.c.d.e.1 pointsQUESTION 10A probability distribution has a mean of 45 and a standard deviation of 1. Use Chebychev’s inequality to estimate the probability that an outcome of the experiment lies between 43 and 47.a.At least 0.8b.At least 0.75c.At least 0.5d.At least 0.25e.At least 0.04TwoQUESTION 1Find the value of the probability of the standard normal variable Z corresponding to the shaded area under the standard normal curve.P ( – 1.36 < Z < 1.75 )a.P (- 1.36 < Z < 1.75 ) = 0.0869b.P (- 1.36 < Z < 1.75 ) = 0.8730c.P (- 1.36 < Z < 1.75 ) = 0.9599d.P (- 1.36 < Z < 1.75 ) = 1.04681 pointsQUESTION 2Suppose X is a normal random variable with and . Find the value of .a.0.9050b.0.8996c.0.8945d.0.8818e.0.98571 pointsQUESTION 3Find the value of the probability of the standard normal variable Z corresponding to this area.P ( Z > 2.31 )a.P ( Z > 2.31 ) = 0.0084b.P ( Z > 2.31 ) = 0.0104c.P ( Z > 2.31 ) = 0.0136d.P ( Z > 2.31 ) = 0.98961 pointsQUESTION 4Suppose X is a normal random variable with and . Find the value of .a.0.498b.0.726c.0.495d.0.4333e.0.721 pointsQUESTION 5Find the value of the probability of the standard normal variable Z corresponding to the shaded area under the standard normal curve.P ( 0.3 < Z < 1.85 )a.P (0.3 < Z < 1.85 ) = 0.3499b.P (0.3 < Z < 1.85 ) = 1.5857c.P (0.3 < Z < 1.85 ) = 0.9678d.P (0.3 < Z < 1.85 ) = 0.6179ThreeQUESTION 1According to data released by the Chamber of Commerce of a certain city, the weekly wages (in dollars) of female factory workers are normally distributed with a mean of 575 and a standard deviation of 50. Find the probability that a female factory worker selected at random from the city makes a weekly wage of $550 to $625.a.The probability is 0.5438.b.The probability is 0.5328.c.The probability is 0.4297.d.The probability is 0.5339.1 pointsQUESTION 2To be eligible for further consideration, applicants for certain Civil Service positions must first pass a written qualifying examination on which a score of 70 or more must be obtained. In a recent examination it was found that the scores were normally distributed with a mean of 67 points and a standard deviation of 6 points. Determine the percentage of applicants who passed the written qualifying examination.a.18.79% applicants passed the written qualifying examination.b.19.52% applicants passed the written qualifying examination.c.30.85% applicants passed the written qualifying examination.d.24.74% applicants passed the written qualifying examination.1 pointsQUESTION 3Use the appropriate normal distribution to approximate the resulting binomial distribution.The manager of Madison Finance Company has estimated that, because of a recession year, 5% of its 400 loan accounts will be delinquent. If the manager's estimate is correct, what is the probability that 11 or more of the accounts will be delinquent?a.The probability is 0.9994.b.The probability is 0.9891.c.The probability is 0.8137.d.The probability is 0.9854.1 pointsQUESTION 4Use the appropriate normal distribution to approximate the resulting binomial distribution.A basketball player has a 75% chance of making a free throw. What is the probability of her making 80 or more free throws in 120 trials?a.The probability is 0.9744.b.The probability is 0.9822.c.The probability is 0.9864.d.The probability is 0.9898.1 pointsQUESTION 5Use the appropriate normal distribution to approximate the resulting binomial distribution.Colorado Mining and Mineral has 800 employees engaged in its mining operations. It has been estimated that the probability of a worker meeting with an accident during a 1-year period is .1. What is the probability that more than 80 workers will meet with an accident during the 1-year period?a.The probability is 0.4684.b.The probability is 0.4669.c.The probability is 0.8491.d.The probability is 0.4761.

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