1. A tax accountant has three choices for the method of treating one of a client’s deductions. After this choice has been made, there are only two choices for how a second deduction can be treated. What is the total number of possibilities for the treatment of the two deductions?

By the Fundamental Counting Principle, the number of possibilities = 3 * 2 = 6

2. A committee consists of eight members, each of whom may or may not show up for the next meeting. Assuming that the members will be making independent decisions on whether or not to attend, how many different possibilities exist for the composition of the meeting?

Since each member has two choices (attending or not attending) which he/she can make exercise independently of others, there are a total of 2^8 = 256 possible compositions.

3. The weights of cans of Monarch pears follow the normal distribution with a mean of 1,000 grams and a standard deviation of 50 grams. Calculate the percentage of the cans that weigh:
a. Less than 860 grams.
b. Between 1,055 and 1,100 grams.
c. Between 860 and 1,055 grams.

m = 1000, s = 50; z = (x - m)/s
(a) z = (860 - 1000)/50 = -2.8
P(x < 860) = P(z < -2.8) = 0.0026
About 0.26% of the cans weigh less than 860 g
(b) x = (1055 - 1000)/50 = 1.1 and (1100 - 1000)/50 = 2
P(1055 < x < 1100) = p(1.1 < z < 2) = 0.1129
About 11.29% of the cans weigh between 1055 g and 1100 g
(c)x = (860 - 1000)/50 =-2.8 and (1055 - 1000)/50 =1.1
P(860 < x < 1055) = p(-2.8 < z < 1.1) = 0.8618
About 86.18% of the cans weigh between 1055 g and 1100 g

4. The number of passengers on the Carnival Sensation during one-week cruises in the
Caribbean follows the normal distribution. The mean number of passengers per cruise is
1,820 and the standard deviation is 120.
a. What percent of the cruises will have between 1,820 and 1,970 passengers?
b. What percent of the cruises will have 1,970 passengers or more?
c. What percent of the cruises will have 1,600 or fewer passengers?
d. How many passengers are on the cruises with the fewest 25 percent of passengers?

 = 1820,  = 120; z = (x – )/
(a) z = (1970 – 1820)/120 = 1.25
Percentage of cruises that will have between 1820 and 1970 passengers =
100  Area under the Standard Normal Curve between z = 0 and z = 1.25, which is 100  0.3944 = 39.44%
(b) Percentage of cruises that will have 1970 passengers or more =
100  Area under the Standard Normal Curve to the right of z = 1.25, which is 100  0.1056 = 10.56%
(c) z = (1600 – 1820)/120 = -1.83
Percentage of cruises that will have 1600 or fewer passengers = 100  Area under the Standard Normal Curve to the left of z = -1.83, which is 100  0.0336 = 3.36%
(d) From the Area Tables, corresponding to 0.25, we get z = -0.6745
x =  +  z = 1820 + 120(-0.6745) = 1739 passengers.