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Recent problems solved by 'edjones'
edjones answered: 7568 problems
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1 solutions
Answer 312335 by edjones(7569)   on 2011-05-28 01:35:39 (Show Source):
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Probability-and-statistics/454919: When playing bingo, 75 balls are placed in a bin and balls are selected at random. Each ball is marked with a letter and number as indicated by the following chart.
Playing Bingo When playing bingo, 75 balls are placed in a bin and balls are selected at random. Each ball is marked with a letter and number as indicated in the following chart.
B I N G O
1-15 16-30 31-45 46-60 61-75
For example, there are balls marked B1, B2, up to B15; I16, I17, up to I30; and so on. Now, assuming one bingo ball is selected at random, determine
42.
the probability that it does not contain the letter G.
1 solutions
Answer 312330 by edjones(7569) on 2011-05-28 00:45:42 (Show Source):
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Probability-and-statistics/454934: Suppose we want to determine the (binomial) probability (p) of getting 0 (zero) heads in 5 flips of a coin. Using Table 2 in Appendix B of the text, what values of n, x, and p would we use to look up this probability, and what would be the probability?
Table 2 states
n=5 x=0
0.01 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70
.951 .774 .590 .328 .168 .078 .031 .010 .002
1 solutions
Answer 312324 by edjones(7569) on 2011-05-28 00:27:27 (Show Source):
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Permutations/454591: 10 people are going on a camping trip in 3 cars that hold 5,2, and 4 passengers, respectively. How many ways is it possible to transport the 10 people to their campsite?
1 solutions
Answer 312133 by edjones(7569) on 2011-05-26 21:11:18 (Show Source):
You can put this solution on YOUR website!I'll let you do the calculations
5 seater has only 4 people : 10C4 * 6C4 * 2C2
4 seater has only 3 people : 10C5 * 5C3 * 2C2
2 seater has only 1 person : 10C5 * 5C4 * 1C1
ADD the three results for the answer.
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Ed
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Probability-and-statistics/454594: Three different gaming machines are in a line along a wall. The machines guarantee the following outputs with listed probabilities:
Machine A: P(1)=1/7 ; P(3)=3/7 ; P(7)=3/7
Machine B: P(4)=1
Machine C: P(3)=5/9 ; P(6)=4/9
Suppose that you want to select the game giving the highest average output over a long period of time. Which machine would you select? Why?
1 solutions
Answer 312132 by edjones(7569) on 2011-05-26 20:57:16 (Show Source):
You can put this solution on YOUR website!Expectation=n*p
A:
1/7 * 1 = 1/7
3 * 3/7 = 9/7
7 * 3/7 = 21/7
.
31/7 expectation = 4 3/7
.
B:
4*1=4 expectation
.
C:
3 * 5/9 = 15/9
6 * 4/9 = 24/9
.
39/9 expectation = 4 3/9
.
Select A because it has the highest expectation.
.
Ed
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Probability-and-statistics/454580: Bob takes an online IQ test and finds that his IQ according to the test is 134. Assuming that the mean IQ is 100, the standard deviation is 15, and the distribution of IQ scores is normal, what proportion of the population would score higher than Bob? Lower than Bob?
1 solutions
Answer 312126 by edjones(7569) on 2011-05-26 20:30:55 (Show Source):
You can put this solution on YOUR website!z=(134-100)/15
=2.267
By finding the area under the curve for the z we find that .0117 score higher and .9883 score lower in the population.
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Ed
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test/454581: The circulation of a newsletter decreased from 3,200 to 2,464. What was the percent of decrease in circulation?
1 solutions
Answer 312124 by edjones(7569) on 2011-05-26 20:22:35 (Show Source):
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Probability-and-statistics/454512: A co-ed soccer team has 9 women and 7 men on its roster. 6 players play on the field at one time.
a) What is the probability the starting lineup will have 3 men and 3 women on it?
b) What is the probability the starting lineup will have at least 5 men on it
c) What is the expected number of women on the team?
1 solutions
Answer 312123 by edjones(7569) on 2011-05-26 20:18:57 (Show Source):
You can put this solution on YOUR website!a) (9C3 * 7C3)/16C6 = .3671
.
b) (7C6 + (7C5 * 9C1))/16C6 = .02448
.
c)7/16 * 6 = 2.625 expected number of women on the team.
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Ed
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Probability-and-statistics/454513: Speed trap on the highway set by the O.P.P. shows that the mean speed of cars is 103.4 km/h with a standard deviation of 9.8 km/h. The posted speed limit on the highway is 100 km/h.
What percentage of drivers are technically driving under the speed limit?
1 solutions
Answer 312117 by edjones(7569) on 2011-05-26 19:51:11 (Show Source):
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Probability-and-statistics/454515: Speed trap on the highway set by the O.P.P. shows that the mean speed of cars is 103.4 km/h with a standard deviation of 9.8 km/h. The posted speed limit on the highway is 100 km/h.
Speeders caught traveling 25 km/h or more over the speed limit are subject to a $5000 fine. What percentage of speeders will be fined $5000?
1 solutions
Answer 312113 by edjones(7569) on 2011-05-26 19:42:48 (Show Source):
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Probability-and-statistics/454517: Speed trap on the highway set by the O.P.P. shows that the mean speed of cars is 103.4 km/h with a standard deviation of 9.8 km/h. The posted speed limit on the highway is 100 km/h.
Police officers tend not to pull over drivers between 100 km/h and 110 km/h. What percentage of drivers is this?
1 solutions
Answer 312108 by edjones(7569) on 2011-05-26 19:25:49 (Show Source):
You can put this solution on YOUR website!z(below M)=(100-103.4)/9.8= -0.3469
z(above M)=(110-103.4)/9.8= 0.6735
.
Area under curve -0.3469 to 0 = 0.1357
Area under curve 0 to 0.6735 = 0.2497
0.1357+0.2497=.3854=38.54%
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Ed
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Probability-and-statistics/454518: Speed trap on the highway set by the O.P.P. shows that the mean speed of cars is 103.4 km/h with a standard deviation of 9.8 km/h. The posted speed limit on the highway is 100 km/h.
The top 2% of all drivers speeding are subject to losing their license. According to the data, what speed must a driver be traveling to lose his or her license?
1 solutions
Answer 312101 by edjones(7569) on 2011-05-26 19:08:28 (Show Source):
You can put this solution on YOUR website!A one tailed distribution with the area under alpha of .02 has a z of 2.05
9.8*2.05=20.1 (rounded)
103.4+20.1=123.5 km/h Speed the driver must be traveling to lose his license.
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Ed
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Probability-and-statistics/454537: On any given day the probability that a certain gas station will run out of gas is 1/10.
What is the probability the gas station will run out of gas at least two days in the next two weeks?
What is the expected number of times the gas station will run out of gas in the next 2 weeks?
1 solutions
Answer 312094 by edjones(7569) on 2011-05-26 18:38:37 (Show Source):
You can put this solution on YOUR website!p=.1, q=.9
q^14+.14q^13p=.5846 probability it wont run out more than once
1-.5846=.4154 the probability the gas station will run out of gas at least two days in the next two weeks.
.
14*.1=1.4 the expected number of times the gas station will run out of gas in the next 2 weeks.
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Ed
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Permutations/453776: at the conclusion of a soccer game whose two teams each include 11 players, each player on the winning team "gave five" to (slapped hands) each player on the losing team. each player on the winning team also gave five to each other player on the winning team. how many fives were given?
1 solutions
Answer 311723 by edjones(7569) on 2011-05-24 22:25:29 (Show Source):
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Permutations/453783: Four cards are drawn from a standard deck of 52 cards without replacement. Find the probability that the first card is a heart, the second is a club, and the third and fourth are diamonds.
1 solutions
Answer 311718 by edjones(7569) on 2011-05-24 22:07:35 (Show Source):
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Linear_Equations_And_Systems_Word_Problems/453755: Hello, I am a completely stumped as to how to solve this problem! I've been working on it for two days and can't seem to figure out the correct set-up. Any help would be greatly appreciated. Thank you.
The Fancy Fashions Store had $8000 available each month for advertising. Newspaper ads cost $400 each and no more than 20 can be run per month. Radio ads cost $200 each and no more than 30 can run per month. TV ads cost $1200 each, with a maximum of 6 available each month. Approximately 2000 women will see each newspaper ad, 1200 will hear each radio commercial, and 10,000 will see each TV ad.
a. How much of each type of advertising should be used if the store wants to maximize its ad exposure?
b. A marketing analyst is puzzled by the results of part a. More women see each newspaper ad than hear each radio commercial, he reasons, so it makes no sense to use radio commercials and no newspaper ads. How would you respond?
1 solutions
Answer 311714 by edjones(7569) on 2011-05-24 21:58:17 (Show Source):
You can put this solution on YOUR website!The cost of tv/viewer is 1200/10000=$0.12
For radio it is 200/1200=$0.17
For newspaper it is 400/2000=$0.20
.
Do 6 TV ads $7200 and 4 radio ads $800.
6*10000+4*1200=64,800 exposures.
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Ed
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Probability-and-statistics/453763: "Suppose a system can continue to operate successfully if one or both of its components do not fail. Suppose the probability of component #1 failing is .065 and the probability of component #2 failing is .045. Assuming the component failure is independent of any other component failing, determine the probability the system operates successfully."
I didn't get very far before getting stuck, but it looks like the system can operate with one or both components operating successfully. There are 4 possible outcomes of component operation:
1: Component #1 and Component #2 both work
2: Component #1 works, Component #2 does not work
3: Component #1 does not work, Component #2 works
4: Neither Component #1 or Component #2 work.
The first three circumstances are the only possible situations where the system will operate successfully. But I don't know what to do next...
1 solutions
Answer 311711 by edjones(7569) on 2011-05-24 21:38:06 (Show Source):
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Probability-and-statistics/453762: In the NASCAR racing industry ball bearings are automatically produced on a Kronar BBX machine and are known to be normally distributed. For one of the production lines, the mean diameter is set at 20.00 mm (millimeters). The standard deviation, over a long period of time, was computed to be 0.150 mm. What percent of the ball bearings on this line will have diameters of 20.27 mm or more?
1 solutions
Answer 311710 by edjones(7569) on 2011-05-24 21:33:13 (Show Source):
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