Attn Greg: (If possible)
If a die rolled one time, classical probability would indicate that the probability of a “two” should be 1/6th. If the die rolled 60 times and comes up “two” only 9 times, does this suggest that the die is “loaded”? Why or why not?
We would expect that it would come up “two” 10 times. So, getting one fewer “two’s” does not seem unlikely. In fact, we can use the binomial distribution to determine that the probability of getting 9 “two’s” is 0.134. That is, not unlikely at all.
It has been reported that about 35% of U.S. adults attended a sports event during the previous year. What are the odds that a randomly selected U.S. adult attended a sports event during the year?
The probability is 0.35.
The following contingency table of frequencies is based on a 5-year study of fire fatalities in Maryland. For Purpose of clarity, columns and rows are identified by letters A-C and D-G respectively.
Blood Alcohol Level of Victum
A B C
Age 0.00% 0.01-0.09% >0.10%
D 0-19 142 7 6 155
E 20-39 47 8 41 96
F 40-59 29 8 77 114
G 60 AND OVER 47 7 35 89
265 30 159
Can you please tell me the question? It seems to have been cut off.
A survey of employees at a large company found the following relative frequencies for one-way distances they had traveled to arrive at work.
Number of Miles (One-Way)
A B C D E F
=5 6-10 11-15 16-20 21-30 =31
Relative Frequency 0.38 0.25 0.16 0.09 0.07 0.05
a. What is the probability that a randomly selected individual will have to travel 11 or more miles to work?
=1-0.38-0.25 = 0.37
b. What is the probability that a randomly selected individual will have to travel between 6 and 15 miles to work?
=0.25+0.16 = 0.41
c. Draw a Venn diagram that includes the relative frequency probabilities in the table.
Since these probabilities are mutually exclusive, these probabilities do not overlap. Hence, we would draw 6 non overlapping circles with relative frequencies as the probabilities in each circle.
- Using the letter identifications provided, calculate the following probabilities P(A or B or C); P(A or F); P(A’ or B); P(A or B or C’).
P(A or B or C) = 0.38+0.25+0.16 = 0.79
P(A or F) = 0.38 + 0.05 = 0.43
P(A’ or B) = 1-0.38 = 0.62
P(A or B or C’) = 1-0.16 = 0.84.
Using the information presented in the table in exercise 5.12, calculate the following.
Which table is this? The questions aren’t numbered.
A fair coin is tossed three times. What probability that the sequence will be heads, tails, heads?
P(HTH) = ½*1/2*1/2 = 1/8 = 0.125
Using the table in exercise 5.12, calculate the conditional probability of C given each of the age groups, or P(C\D), P(C\E), etc. Compare these probabilities and speculate as to which age groups seem more likely than others to have been (according to legal definition at that time, 0.10% blood alcohol content) intoxicated at the time they were victims.
Again… which table is this?
An investment counselor would like to meet with 12 of his clients on Monday, but he has time for only 8 appointments. How many different combinations of the clients could be considered for inclusion into his limited schedule for that day?
12 choose 8 = 12!/(8!4!) = 12*11*10*9/24 = 495
A roadside museum has 25 exhibits but enough space to display only 10 at a time. If order or arrangement is considered, how many possibilities exist for the eventual display?
25!/15! = 1.18617E+13
Indicate whether each of the following random variables is discrete or continuous.
a. The diameter of aluminum rods coming off a production line -- continuous
b. The number of years of schooling employees have completed -- discrete
c. The Dow Jones Industrial Average. -- continuous
d. The volume of milk purchased during a supermarket visit. -- continuous
Determine the mean, variance, and standard deviation of the following discrete probability distribution:
X 0 1 2
P(x) 0.60 0.30 0.10
x / p(x) / x*p(x) / (x-E(x))^2 / (X-E(X))^2*p(x)0 / 0.6 / 0 / 0.25 / 0.15
1 / 0.3 / 0.3 / 0.25 / 0.075
2 / 0.1 / 0.2 / 2.25 / 0.225
mean---> / 0.5 / 0.45 / <--- variance
0.6708 / <--- standard devaition
A contractor must pay a $40,000 penalty if construction of an expensive home requires more than 16 weeks. He will receive a bonus of $10,000 if the home is completed within 8 weeks. Based on experience with this type of project, the contractor feels there is a 0.2 chance the home will require more than 16 weeks for completion, and there is a 0.3 chance it will be finished within 8 weeks. If the price of the home is $350,000 before any penalty or bonus adjustment, how much can the buyer expect to pay for her new home when it is completed?
x*p(x)310000 / 0.2 / 62000
350000 / 0.5 / 175000
360000 / 0.3 / 108000
expected value----> / 345000
Hence the expected value is $345,000
Laura McCarthy, the owner of Riverside Bakery, has been approached by insurance underwriters trying to convince her to purchase flood insurance. According to Local meteorologist, there is a 0.01 probability that the river will flood next year. Riverside’s profits for the coming year depend on whether Laura buys flood insurance and whether the river floods. The profits (which take into consideration the $10,000 premium for the flood insurance) for the four possible combinations of Laura’s choice and river conditions are:
The river
Does not flood Floods
Insurance No Flood
Decision Insurance 200,000 (-1,000,000)
Get flood
insurance $190,000 $200,000
a. If Laura decides not to purchase flood insurance, use the appropriate discrete probability distribution to determine Riverside’s expected profit next year.
200000*.99 + -1000000*.01 = 188000
b. If Laura purchases the flood insurance, what will be Riverside’s expected profit next year?
190000*.99 + 200000+0.01 = 190100
- Given the results in parts (a & b), provide Laura with recommendation.
If she buys flood insurance, she can expect more profits. So she should buy it.
A city law-enforcement official has stated that 20% of the items sold by pawn shops within the city have been stolen. Ralph has just purchased 4 items from one of the pawn shops. Assuming that the official is correct, and for x = the number of Ralph’s purchases that have been stolen, determine the following:
A. P(x = 0) = 1*.8^4 = 0.4096
B. P(2 = x) = 6*.2^2*.8^2 = 0.1536
C. P(1 = x = 3) = 0. (It can’t be both 3 and 1 at the same time.
D. P(x = 2) = 6*.2^2*.8^2 = 0.1536 (same as part b)
There may be some typos in the above. You should check and get back to me.
The U.S. Department of Labor has reported that 30% of the 2.1 million mathematical and computer scientist in the United States are women. If 3 individuals are randomly selected from the occupation group, and x = number of females, determine
P(x = 0) = 1*.7^3 = 0.343
P(x = 1) = 3*.3*.7^2 = 0.441
P(x = 2) = 3*.3^2*.7 = 0.189
P(x = 3) = 1*.3^3 = 0.027