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Recent problems solved by 'stanbon'
stanbon answered: 57135 problems
Jump to solutions: 0..29 , 30..59 , 60..89 , 90..119 , 120..149 , 150..179 , 180..209 , 210..239 , 240..269 , 270..299 , 300..329 , 330..359 , 360..389 , 390..419 , 420..449 , 450..479 , 480..509 , 510..539 , 540..569 , 570..599 , 600..629 , 630..659 , 660..689 , 690..719 , 720..749 , 750..779 , 780..809 , 810..839 , 840..869 , 870..899 , 900..929 , 930..959 , 960..989 , 990..1019 , 1020..1049 , 1050..1079 , 1080..1109 , 1110..1139 , 1140..1169 , 1170..1199 , 1200..1229 , 1230..1259 , 1260..1289 , 1290..1319 , 1320..1349 , 1350..1379 , 1380..1409 , 1410..1439 , 1440..1469 , 1470..1499 , 1500..1529 , 1530..1559 , 1560..1589 , 1590..1619 , 1620..1649 , 1650..1679 , 1680..1709 , 1710..1739 , 1740..1769 , 1770..1799 , 1800..1829 , 1830..1859 , 1860..1889 , 1890..1919 , 1920..1949 , 1950..1979 , 1980..2009 , 2010..2039 , 2040..2069 , 2070..2099 , 2100..2129 , 2130..2159 , 2160..2189 , 2190..2219 , 2220..2249 , 2250..2279 , 2280..2309 , 2310..2339 , 2340..2369 , 2370..2399 , 2400..2429 , 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56970..56999 , 57000..57029 , 57030..57059 , 57060..57089 , 57090..57119 , 57120..57149, >>Next
Travel_Word_Problems/581351: During the first part of a trip, a canoeist travels 44 miles at a certain speed. The canoeist travels 15 miles on the second part of the trip at a speed 5 mph slower. The total time for the trip is 2 hrs. What was the speed on each part of the trip? 1 solutions
Answer 371586 by stanbon(57203) on 2012-03-02 18:36:29 (Show Source):
You can put this solution on YOUR website!During the first part of a trip, a canoeist travels 44 miles at a certain speed. The canoeist travels 15 miles on the second part of the trip at a speed 5 mph slower. The total time for the trip is 2 hrs. What was the speed on each part of the trip?
----------
1st Part DATA:
distance = 44 miles ; rate = x mph ; time = d/r = 44/x hrs
-------------------
2nd part DATA:
distance = 15 miles ; rate = (x-5)mph ; time = d/r = 15/(x-5)
-------------------
Equation:
time + time = 5 hrs
44/x + 15/(x-5) = 5
------
Multiply thru by x(x-5) to get:
44(x-5) + 15x = 5x(x-5)
44x - 220 + 15x = 5x^2-25x
------
59x-220 = 5x^2-25x
5x^2 - 69x + 220 = 0
Usable solution:
x = 8.8 mph (1st part rate)
x-5 = 3.8 mph (2nd part rate)
===============================
Cheers,
Stan H.
|
Probability-and-statistics/581320: Using data from Boston, Massachusetts, a test of independence is run on the claim that ice cream sales per month and the number of car wrecks per month are independent. The claim is rejected.
Using number of car wrecks as the x variable and ice cream sales as the y variable, an r value of r=0.923 is then computed and shown to exceed the critical value for this data. The data is double checked and verified. This shows that car wrecks cause ice cream sales.
NOTE: While this specific r-value is made up, this general pattern has been shown in several real world data sets involving ice cream sales and number of car wrecks in major cities on the east coast of the United States.
***I saw that this question has been asked by another student on Algebra.com however I really did not understand the justification that was given. I am supposed to explain what is wrong and why it is wrong. 1 solutions
Answer 371574 by stanbon(57203) on 2012-03-02 17:14:19 (Show Source):
You can put this solution on YOUR website!Using data from Boston, Massachusetts, a test of independence is run on the claim that ice cream sales per month and the number of car wrecks per month are independent. The claim is rejected.
Using number of car wrecks as the x variable and ice cream sales as the y variable, an r value of r=0.923 is then computed and shown to exceed the critical value for this data. The data is double checked and verified. This shows that car wrecks cause ice cream sales.
NOTE: While this specific r-value is made up, this general pattern has been shown in several real world data sets involving ice cream sales and number of car wrecks in major cities on the east coast of the United States.
-----
It's another example of correlation. It is not an example of causation.
----
Linear correlation and the associated r-value are mathematical ideas that
have nothing to do with causation.
======================================
Cheers,
Stan H.
|
Probability-and-statistics/581282: Construct the five number summary and a box plot for the following gas mileasges per gallon of various company cars: 22 18 13 28 30 12 38 22 30 39 20 18 14 16 10.
When I did the math I got my 5 numbers to be 10 14 19 30 39... Is this correct?! And also could I get help with drawing the box?! Thank you :) 1 solutions
Answer 371573 by stanbon(57203) on 2012-03-02 17:09:19 (Show Source):
|
Probability-and-statistics/581338: What are the odds that two people would pick the same seven
books from a list of 150 to read? 1 solutions
Answer 371571 by stanbon(57203) on 2012-03-02 16:58:40 (Show Source):
You can put this solution on YOUR website!What are the odds that two people would pick the same seven
books from a list of 150 to read?
----
P(they will) = 1/150C7
----
P(they won't) = (150C7 - 1)/150C7
-----
odds they do = P(they will)/P(they don't) = 1:(150C7 -1)
========================
Cheers,
Stan H.
|
Probability-and-statistics/581303: how many ways can two republicans, one democrat, and one independent be chosen from nine republicans, five democrats, and two independents to fill 4 positions? 1 solutions
Answer 371570 by stanbon(57203) on 2012-03-02 16:54:12 (Show Source):
You can put this solution on YOUR website!how many ways can two republicans, one democrat, and one independent be chosen from nine republicans, five democrats, and two independents to fill 4 positions?
--------
Ans: 9C2*5C1*4C1 = 36*5*4 = 720
=============
Cheers,
Stan H.
==============
|
Probability-and-statistics/581324: If the claim says that the population mean is greater than 200 and the sample mean is 215, we can say that the claim is true even without a formal test
what is wrong with this statement
1 solutions
Answer 371564 by stanbon(57203) on 2012-03-02 15:58:04 (Show Source):
You can put this solution on YOUR website!If the claim says that the population mean is greater than 200 and the sample mean is 215, we can say that the claim is true even without a formal test
what is wrong with this statement
-----
Ho: u <= 200
Ha: u > 200 (claim)
----
The test results support the claim only if we reject Ho.
---
We reject Ho at the 95% significance level only if 215 is more than 1.96 standard deviations above the mean.
===========
Cheers,
Stan H.
|
Angles/581325: If P is the measure of an angle, 180-P is the measure of the supplement of the angle. If the measure of an angle equals three times the measure of its supplement, what is the measure of the angle? 1 solutions
Answer 371562 by stanbon(57203) on 2012-03-02 15:47:06 (Show Source):
You can put this solution on YOUR website!If the measure of an angle equals three times the measure of its supplement, what is the measure of the angle?
----
Let the angle be "x":
Equation:
x = 3(180-x)
x = 540-3x
4x = 540
x = 135 degrees
========================
Cheers,
Stan H.
|
Travel_Word_Problems/581323: A cyclist rode the first 16 mile-portion of his workout at a constant speed. For the 12 mile cool down portion of his workout, he reduced his speed by 2 miles per hour. Each portion took the same time. Find his speed during the first portion and the second cool down portion. 1 solutions
Answer 371560 by stanbon(57203) on 2012-03-02 15:42:50 (Show Source):
You can put this solution on YOUR website!A cyclist rode the first 16 mile-portion of his workout at a constant speed. For the 12 mile cool down portion of his workout, he reduced his speed by 2 miles per hour. Each portion took the same time. Find his speed during the first portion and the second cool down portion.
----
1st portion DATA:
distance = 16 miles ; rate = x mph ; time = d/r = 16/x hrs
----
2nd portion DATA:
distance = 12 miles ; rate = (x-2)mph ; time = 12/(x-2) hrs
-------
Equation:
time = time
16/x = 12/(x-2)
16x-32 = 12x
4x = 32
x = 8 mph (rate of 1st portion)
---
x-2 = 6 mph (rate on 2nd portion)
===============
Cheers,
Stan H.
|
Probability-and-statistics/581258: Let the random variable X follow a normal probability distribution with a mean 10 and a standard deviation of 2.
Using the above information, find x such that P(X>x)=0.05; find x such that P(X
What is the breakdown of this problem? Help!
Thank You! 1 solutions
Answer 371538 by stanbon(57203) on 2012-03-02 12:03:01 (Show Source):
You can put this solution on YOUR website!Let the random variable X follow a normal probability distribution with a mean 10 and a standard deviation of 2.
Using the above information, find x such that P(X>x)=0.05; find x such that P(X What is the breakdown of this problem? Help!
------
Draw a normal curve with horizontal axis "x".
Shade a right tail area that has 5% of the area under the curve
and to the right.
------
Draw a normal curve with horizontal axis "z".
Find the z-value that has a right tail of 5%
z = invNorm(0.95) = 1.6449
------------
Find the correspondina "x" value:
x = zs + u
x = 1.6449*2+10 = 13.2898
==============================
Cheers,
Stan H.
|
Probability-and-statistics/581206: For a standard normal distribution, find the percentage of data that are more than 2 standard deviations below the mean or more than 3 standard deviations above the mean. 1 solutions
Answer 371537 by stanbon(57203) on 2012-03-02 11:54:16 (Show Source):
You can put this solution on YOUR website!For a standard normal distribution, find the percentage of data that are more than 2 standard deviations below the mean or more than 3 standard deviations above the mean.
----
P(z<2 OR z>3) = normalcdf(-100,2)+normalcdf(3,100) = 0.9786
===============
Cheers,
Stan H.
==================
|
Probability-and-statistics/581254: Let the random variable X follow a normal probability distribution with a mean 10 and a standard deviation of 2.
Show your manual work of finding P(X>15).
What is known is it'll be 1 - # because the direction the sign goes, but other than that, I am not sure. Please help!
Thank You! 1 solutions
Answer 371536 by stanbon(57203) on 2012-03-02 11:50:00 (Show Source):
You can put this solution on YOUR website!Let the random variable X follow a normal probability distribution with a mean 10 and a standard deviation of 2.
Show your manual work of finding P(X>15).
----
z(15) = 5/2
P(x > 15) = P(z > 5/2) = normalcdf(5/2,100) = 0.0062
----------------------
Cheers,
Stan H.
=================
|
Probability-and-statistics/581251: Let the random variable X follow a normal probability distribution with a mean 10 and a standard deviation of 2. Find P(X<15).
What I do know is it will not be 1 - # because of the sign, but I am not exactly sure where to start. Help!
Thank You 1 solutions
Answer 371535 by stanbon(57203) on 2012-03-02 11:47:39 (Show Source):
You can put this solution on YOUR website! Let the random variable X follow a normal probability distribution with a mean 10 and a standard deviation of 2. Find P(X<15).
-----
Find the z-value of x = 15.
z(15)= (15-10)/2 = 5/2
----
P(x < 15) = P(z < 5/2) = normalcdf(-100,5/2) = 0.9938
=====================
Cheers,
Stan H.
|
Probability-and-statistics/581238: Fill in the P(X=x) values in the table below to give a legitimate probability distribution for the discrete random variable X, whose possible values are 0, 2, 4, 5, and 6. The table states that 0=0.29, 2=0.16, and 5=0.18. What are the values for 4 and 6? 1 solutions
Answer 371534 by stanbon(57203) on 2012-03-02 11:44:37 (Show Source):
You can put this solution on YOUR website!If the letters A, B, C, D, and E are to be used in a five-letter parking tag, how many different tags are possible if repetitions are not permitted?
---
Your table was not posted.
Whatever the missiing values are, all the probabilities must add up to 1.
Cheers,
Stan H.
|
Probability-and-statistics/580816: If the letters A, B, C, D, and E are to be used in a five-letter parking tag, how many different tags are possible if repetitions are not permitted? 1 solutions
Answer 371530 by stanbon(57203) on 2012-03-02 10:29:15 (Show Source):
You can put this solution on YOUR website!If the letters A, B, C, D, and E are to be used in a five-letter parking tag, how many different tags are possible if repetitions are not permitted?
---
choose 1st letter: 5 ways
choose 2nd letter: 4 ways
...
choose 5th letter: 1 way
-----------------------------
# of tags: 5! = 120
Cheers,
Stan H.
|
Probability-and-statistics/581209: Scores on a test have a mean of 63 and Q3 is 82. The scores have a distribution that is approximately normal. Find the standard deviation. Round your answer to the nearest tenth. 1 solutions
Answer 371526 by stanbon(57203) on 2012-03-02 10:06:11 (Show Source):
You can put this solution on YOUR website!Scores on a test have a mean of 63 and Q3 is 82. The scores have a distribution that is approximately normal. Find the standard deviation. Round your answer to the nearest tenth.
----
Q3 has a left-tail of 0.75
---
Find the corresponding z-score:
invNorm(0.75) = 0.6745
----
Equation:
Use x = zs+u to find "s":
82 = 0.6745*s+63
0.6745s = 19
s = 19/0.6745
s = 28.1694
==================
Cheers,
Stan H.
|
Probability-and-statistics/581222: Please help I will pay for tutoring if I can get this answer right
The number of times the SAT was taken and the number of students are as follow
Number of times Number of students
1 721,769
2 601,325
3 166,736
4 22,299
5 6,730
Let x be a random variable indicating the number of times a student takes the SAT. Show the probability for this random vaiable.(to 4 decimals)
x=1
x=2
x=3
x=4
x=5
My answer was
1-7217.69
2-3006.62 1 solutions
Answer 371524 by stanbon(57203) on 2012-03-02 10:01:01 (Show Source):
You can put this solution on YOUR website!The number of times the SAT was taken and the number of students are as follow
Number of times Number of students
1 721,769
2 601,325
3 166,736
4 22,299
5 6,730
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Total number os students = 1518859
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Let x be a random variable indicating the number of times a student takes the SAT. Show the probability for this random vaiable.(to 4 decimals)
P(x=1) = 721769/1518859 = 0.4752
P(x=2) = 601,325/1518859 = 0.3959
etc. (Use the same method to find the following probabilities)
x=3
x=4
x=5
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Cheers,
Stan H.
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Length-and-distance/581195: find the coordinates of point E if C is the midpoint of AE and A= (-3,1) and C=(-2,4) 1 solutions
Answer 371506 by stanbon(57203) on 2012-03-02 08:04:52 (Show Source):
You can put this solution on YOUR website!find the coordinates of point E if
C is the midpoint of AE and A= (-3,1) and C=(-2,4)
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Let the coordinates of point E be (x,y)
Equations:
(x+-3)/2 = -2
(y+1)/2 = 4
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x = 2*-2+3 = -1
y = 2*4-1 = 7
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The coordinates of E are (-1,7)
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Cheers,
Stan H.
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Triangles/581189: SOLUTION A(2,3) B(-3,0) and C(3,-3) how would i find the midpoint of AB AC and BC? 1 solutions
Answer 371505 by stanbon(57203) on 2012-03-02 08:02:01 (Show Source):
You can put this solution on YOUR website!A(2,3) B(-3,0) and C(3,-3) how would i find the midpoint of AB AC and BC?
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Just add the x-values of the end points and divide by 2
Do the same for the y-values.
Example:
Let the midpoint of AB be (x,y)
Then:
x = (2+-3)/2 = -1/2
y = (3+0)/2 = 3/2
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Midpoint of AB is (-1/2,3/2)
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Cheers,
Stan H.
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Probability-and-statistics/580970: Hi there, PLease help :)
What would have to be the test score (rounded to once place to the right of a decimal) to qualify for admittance into a institution with a mean of normally distributed test scores of 82.5 and a standard deviation was 6.9, assuming admittance depended on being in the top 17 percent.
Thanks,
Fiona 1 solutions
Answer 371503 by stanbon(57203) on 2012-03-02 07:55:45 (Show Source):
You can put this solution on YOUR website!What would have to be the test score (rounded to once place to the right of a decimal) to qualify for admittance into a institution with a mean of normally distributed test scores of 82.5 and a standard deviation was 6.9, assuming admittance depended on being in the top 17 percent.
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Find the z-score with a left tail of 0.83
invNorm(0.83) = 0.9542
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That tells you the score you are looking for is 0.9542
standard deviations above the mean.
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Find that score:
x = 0.9542*6.9+82.5 = 89.08
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Cheers,
Stan H.
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Probability-and-statistics/580987: A research team at a university conducted a binomial study showing that approximately 10% of all businessmen who wear ties wear them so tight that they actually reduce blood flow to the brain, diminishing cerebral functions. At the board meeting of the 15 businessmen, all of whom wear ties, what is the probability that
a. exactly 6 ties are too tight?
b. at least one tie is too tight?
hint: P(at least one)=1-p(none)
c. at least 13 are not too tight? 1 solutions
Answer 371501 by stanbon(57203) on 2012-03-02 07:51:25 (Show Source):
You can put this solution on YOUR website!A research team at a university conducted a binomial study showing that approximately 10% of all businessmen who wear ties wear them so tight that they actually reduce blood flow to the brain, diminishing cerebral functions. At the board meeting of the 15 businessmen, all of whom wear ties, what is the probability that
a. exactly 6 ties are too tight?
binompdf(15,0.10,6) = 0.0019
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b. at least one tie is too tight?
1 - P(none are too tight) = 1 - 0.9^15 = 0.7941
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c. at least 13 are not too tight?
1 - binomcdf(15,0.9,12) = 0.8159
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Cheers,
Stan H.
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Probability-and-statistics/581130: The diameter of an electric cable is normally distributed, with a mean of 0.8 inch and a standard deviation of 0.01 inch. What is the probability that the diameter will exceed 0.82 inch? 1 solutions
Answer 371500 by stanbon(57203) on 2012-03-02 07:45:43 (Show Source):
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Probability-and-statistics/581144: The probability of getting a red fruit candy from an assorted bag is 0.1.
There are 50 pieces of candy in a bag.
What is the probability that there are at least 4 pieces of red candy in the bag? 1 solutions
Answer 371497 by stanbon(57203) on 2012-03-02 07:42:53 (Show Source):
You can put this solution on YOUR website!The probability of getting a red fruit candy from an assorted bag is 0.1.
There are 50 pieces of candy in a bag.
What is the probability that there are at least 4 pieces of red candy in the bag?
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P(4<= x <=50) = 1 - binomcdf(50,0.1,3) = 0.7497
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Cheers,
Stan H.
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Probability-and-statistics/581179: A machine produces bolts with a mean diameter of 0.30 inches and a standard deviation of 0.01 inches. The diameters are approximately normally distributed. What is the probability that a randomly selected bolt will have a diameter greater than 0.32 inches?
My answer = 0.023.
Thanks!! 1 solutions
Answer 371496 by stanbon(57203) on 2012-03-02 07:40:26 (Show Source):
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Probability-and-statistics/581185: Mary purchased 9 plants, but only needed 6 plants to be planted. How many ways can she select the 6 plants to be planted?
1 solutions
Answer 371493 by stanbon(57203) on 2012-03-02 07:38:10 (Show Source):
You can put this solution on YOUR website!Mary purchased 9 plants, but only needed 6 plants to be planted. How many ways can she select the 6 plants to be planted?
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Ans: 9C6 = 9C3 = (9*8*7)/(1*2*3) = 63 ways
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Cheers,
Stan H.
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Probability-and-statistics/580957: At one high school, the mean time for running the 100-yard dash is 15.2 seconds with a standard deviation of 0.9 seconds. The time are very closely approximated by a normal curve. Find the percent of times that are:
More than 15.2 seconds
Between 14.3 and 16.1 seconds
Thank you. 1 solutions
Answer 371456 by stanbon(57203) on 2012-03-01 22:12:39 (Show Source):
You can put this solution on YOUR website!At one high school, the mean time for running the 100-yard dash is 15.2 seconds with a standard deviation of 0.9 seconds. The time are very closely approximated by a normal curve. Find the percent of times that are:
More than 15.2 seconds
z(15.2) = (15.2-15.2)/0.9 = 0
P(z > 15.2) = 0.50 = 50%
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Between 14.3 and 16.1 seconds
z(14.3) = (14.3-15.2)/0.9 = -1
z(16.1) = (16.1-15.2)/0.9 = 1
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P(14.3 < x < 16.1) = P(-1< z < 1) = 0.6827
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cheers,
Stan H.
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Probability-and-statistics/581108: A coin is tossed three times. Find the probability that the coin lands on tails, then heads,then tails. 1 solutions
Answer 371454 by stanbon(57203) on 2012-03-01 22:06:28 (Show Source):
You can put this solution on YOUR website!A coin is tossed three times. Find the probability that the coin lands on tails, then heads,then tails.
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# of ways to succeed: 1
# of possible outcomes: 2*2*2 = 8
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P(tht) = 1/8
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cheers,
Stan H.
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Probability-and-statistics/580977: Hi there, please help :)
consider finding the probabilty of getting 6 hearts out of 10 draws from a deck of cards if after each draw the card is returned to the deck.
A) Does this meet the conditions of a binomial experiment?
b) if so, what is the probability?
1 solutions
Answer 371451 by stanbon(57203) on 2012-03-01 22:01:47 (Show Source):
You can put this solution on YOUR website!consider finding the probabilty of getting 6 hearts out of 10 draws from a deck of cards if after each draw the card is returned to the deck.
A) Does this meet the conditions of a binomial experiment?
Yes, because the probability on each trial is 1/4
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B) if so, what is the probability?
Ans: 10C6(1/4)^6*(3/4)^4 = binompdf(10,0.25,6) = 0.0162
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Cheers,
Stan H.
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Probability-and-statistics/580990: if X is a binomial random variable with n=37 and p=.39, find
a. What is the expected value?
b. what is the standard deviation? 1 solutions
Answer 371449 by stanbon(57203) on 2012-03-01 21:57:54 (Show Source):
You can put this solution on YOUR website!if X is a binomial random variable with n=37 and p=.39, find
a. What is the expected value?
E(x) = mean = np = 37*0.39 = 14.43
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b. what is the standard deviation?
std = sqrt(npq) = sqrt(37.0.39*0.61) = 2.9669
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Cheers,
Stan H.
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