like you did in Problem 16. This problem also comes from Section. You will need to use a table of z scores (supplied on the Final Exam) to do part c.

**Problem 17** Suppose the mean score on a test is 76 with a standard deviation of 8.

a. What is the z-score for a student who scored a 90 on the test?

**Solution**

b. What test score corresponds to a z-score of –0.35?

**Solution** In part a, you were given a data value and asked to find the z score. This problem does the opposite. Put the values into the z score formula:

c. If the test scores are normally distributed, then what percent of the students would score between 80 and 89 on the test?

**Solution** Graphically, we are trying to determine the portion of area under a bell curve from 80 to 89.

To get areas, we need to calculate the z values and read the area from the z table formula.

For a score of *x* = 80,

And for *x* = 89,

A z score of 0.5 corresponds to an area of 0.192 from the mean of 76 to 80. A z value of 1.63 corresponds to an area of .449 from the mean of 76 to 89. The area between 80 and 89 is the difference between these areas or .449 – .192 or .257.

]]>Problem 16 from Section 14.4 uses z score to decide what is more “impressive”.

**Problem 16** In 1949, Jackie Robinson hit .342 for the Brooklyn Dodgers. In 1973, Rod Carew hit .350 for the Minnesota Twins. In the 1940s, the mean batting average was .267 and the standard deviation was .0326. In the 1970s, the mean batting average was .261 and the standard deviation was .0317. Determine which batting average was more impressive. Explain/show how you determined your solution.

**Solution** To determine which average is more impressive, we need to discover which batting average is higher than the mean. To do this we need to consider the standard deviation which indicates how spread out the distribution of scores is. A z score helps us do this.

For Jackie Robinson, we use the mean and standard deviation for his decade to get

This means Jackie Robinson’s batting average is about 2.3 standard deviations above the mean.

For Rod Carew, the z score is

Rod Carew’s batting average is 2.81 standard deviations above average. Since he is farther above average than Jackie Robinson’s batting average, Rod Carew’s average is more impressive.

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**Problem 15** The following data are the number of reports of mishandled baggage per 1000 passengers for 10 US airlines during three months of 2010.

a. Construct a frequency distribution table. Use classes of width 1.0 (starting at 1.5). If a number is on a class boundary, count it in the higher class.

**Solution** Form the intervals in a table and then count the mishandled claims per 1000. Make sure that numbers on the boundary of an interval, like 3.5, go in the higher the interval. This is the reason why we write [3.5, 4.5).

b. Construct a histogram of the data. Include a title and label axes and units.

**Solution**

c. According to the data, what percentage of the time was the number of baggage complaints per 1000 less than 3.5?

**Solution** The total number of data points is 30, Of those data values, 18 are less than 3.5 so the percentage below 3.5 is ^{18}/_{30} or 60%.

*E* = *p*_{1} *v*_{1} + *p*_{2} *v*_{2} + … + *p _{n} v_{n}*

**Problem 14** If you pay $5 to play a game in which a pair of dice is rolled, calculate the expected value of the game given the following outcomes: if a total less than five comes up then you win $20, if a total of 7 or 11 comes up then you win $10 and for all other totals you win nothing.

**Solution** To find the expected value of the game, we need to know all of the possible outcomes and corresponding probabilities.

The expected value is the sum of each outcome times the corresponding probability,

E = 15( ^{6}/_{36 }) + 5( ^{8}/_{36 }) – 5( ^{22}/_{36 }) = ^{20}/_{36} ≈ 0.56

This means that each time you play the game, you can expect to win $0.56.

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**Problem 13** Assume that two cards are drawn without replacement from a standard 52-card deck, find the following:

a. What is the probability that two aces are drawn?

**Solution** The probability that the first card is an ace and the second card is an ace is the product of the probabilities along the branch containing the aces in a tree diagram

So

P(1^{st} card is ace and 2^{nd} card is ace) = ^{4}/_{52} ∙ ^{3}/_{51} = ^{12}/_{2652} = ^{1}/_{221}

b. What is the probability of drawing a second heart given that you have already drawn a heart?

**Solution** To find the probability that the second card is a heart, given that the first card is a heart, we examine a tree diagram.

Alternately, we can realize that if one heart has been drawn, there are twelve hearts left in the remaining 51 cards so

P(2^{nd} card is heart | 1^{st} card is heart) = ^{12}/_{51}

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For combinations where selections may not be repeated, we use

Many of you may choose to use technology to compute these values. But this won’t get you very far if you cannot distinguish between when you use permutations and when you use combinations.These types of problems are covered in Section 12.3 of your textbook.

**Problem 11** Graduating seniors must present four projects as part of a major requirement, and must select them from a list of 12 possible projects. In how many ways can this be done?

**Solution **The order in which the four projects are given do not make a difference and you can’t repeat the same project so this problem needs to be done with combinations. Since you are selecting 4 projects from a total of 12, the total number of ways is

**Problem 12** A computer password consists of two different letters of the alphabet followed by three different digits from 0 to 9. How many different passwords are possible?

**Solution **In a computer password, rearranging the letters or numbers results in a different password. Because of this we are counting permutations. The number of ways to rearrange the letters is P(26,2) and the number of ways to rearrange the digits is P(10,3). So the total number of passwords is

Notice that you could also do this problem with a slot diagram resulting in

]]>This content is covered in Section 9.5.

**Problem 10** Please make the conversions listed below (round off parts c and d to 3 decimal places):

a. 3.45 centimeters = __________ meters

**Solution**

b. 2.59 liters = __________ milliliters

**Solution**

c. 15 quarts = __________ liters

**Solution**

d. 150 kilograms = __________ pounds

**Solution**

**Problem 9** Calculate the finance charges for the following credit card account for April (which has 30 days) using the average daily balance method. March balance was $750 and the annual interest rate is 15.9%.

Date Transaction

April 5 Payment of $150

April 11 Charged $40 for a tank of gasoline

April 15 Charged $25 for lunch

April 24 Charged $120 for concert tickets

**Solution **To calculate the finance charge, we need to calculate the average daily balance. This requires us to keep track of the running balance.

Using the balances above and the number of days for each balance, we can calculate the average daily balance as

Using the monthly interest rate, we calculate the finance charge as

where the charge has is rounded to the nearest penny.

]]>This content is covered in Section 8.4 of your textbook.

**Problem 8** Gina is saving for her son’s college education by putting $350 each month into an ordinary annuity. If the annuity pays an annual interest rate of 3.25%, how much will she have saved in eight years?

**Solution **To solve this problem, we need to use the annuity formula

with *R* = 350, *i* = 0.0325, *m* = 12 and *n* = 8 ∙ 12:

where the amount has been rounded to the nearest penny.

]]>Don’t simply go to the annuity formula when you see payments are being made since the annuity formula and the amortization formula both have payments in them.

This content is covered in Section 8.5 of your textbook.

**Problem 7** Assume that you have taken out a 20-year amortized loan of $140,000 at an annual interest rate of 4.75%. What is the monthly payment on your loan?

**Solution** Start from the payment formula

with *P* = 140000, *r* = .0475, *m* = 12, *n* = 20∙12. Put in the numbers and solve for *R*:

where the payment is rounded to the nearest penny.

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