**QUESTION 1**

### The probability distribution of a random variable *X* is

*x*–2–1012*P ( X = x )*

Compute the mean, variance, and standard deviation of *X*.

### a.

### b.

### c.

### d.

**1 points **

**QUESTION 2**

### A 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.3

### b.9

### c.1

### d.7

### e.5

**1 points **

**QUESTION 3**

### Find the variance of the probability distribution for the histogram:

a.Var ( *X*) = 4.2625

### b.Var ( *X*) = 4.65

### c.Var ( *X*) = 4.28

### d.Var ( *X*) = 4.0125

**1 points **

**QUESTION 4**

### The 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 points **

**QUESTION 5**

### Rosa 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 A

ReturnsProbability-40.280.3100.5

Mutual Fund B

ReturnsProbability-20.260.680.2

Compute (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 points **

**QUESTION 6**

### The number of Americans without health insurance, in millions, from 1995 through 2002 is summarized in the following table.

Year19951996199719981999200020012002Americans40.541.443.644.740.239.241.143

What is the standard deviation of Americans without health insurance in the period from 1995 through 2002?

### a. million

### b. million

### c. million

### d. million

### e. million

**1 points **

**QUESTION 7**

### The 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 least

### b.At least

### c.At least

### d.At least

**1 points **

**QUESTION 8**

### A 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, 150160170180190220250270320

### a.

### b.

### c.

### d.

### e.

**1 points **

**QUESTION 9**

### The following table gives the scores of 30 students in a mathematics examination.

Scores90-9980-8970-7960-6950-59Students381351

Find 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 points **

**QUESTION 10**

### A 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 0.8

### b.At least 0.75

### c.At least 0.5

### d.At least 0.25

### e.At least 0.04

Two

**QUESTION 1**

### Find 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 ) = 0.0869

### b.*P* (- 1.36 < *Z* < 1.75 ) = 0.8730

### c.*P* (- 1.36 < *Z* < 1.75 ) = 0.9599

### d.*P* (- 1.36 < *Z* < 1.75 ) = 1.0468

**1 points **

**QUESTION 2**

### Suppose *X* is a normal random variable with and . Find the value of .

### a.0.9050

### b.0.8996

### c.0.8945

### d.0.8818

### e.0.9857

**1 points **

**QUESTION 3**

### Find the value of the probability of the standard normal variable *Z* corresponding to this area.

*P* ( *Z* > 2.31 )

a.*P* ( *Z* > 2.31 ) = 0.0084

### b.*P* ( *Z* > 2.31 ) = 0.0104

### c.*P* ( *Z* > 2.31 ) = 0.0136

### d.*P* ( *Z* > 2.31 ) = 0.9896

**1 points **

**QUESTION 4**

### Suppose *X* is a normal random variable with and . Find the value of *.*

### a.0.498

### b.0.726

### c.0.495

### d.0.4333e.0.72

**1 points **

**QUESTION 5**

### Find 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 ) = 0.3499

### b.*P* (0.3 < *Z* < 1.85 ) = 1.5857

### c.*P* (0.3 < *Z* < 1.85 ) = 0.9678

### d.*P* (0.3 < *Z* < 1.85 ) = 0.6179

Three

**QUESTION 1**

### According 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 points **

**QUESTION 2**

### To 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 points **

**QUESTION 3**

### Use 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 points **

**QUESTION 4**

### Use 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 points **