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Recent problems solved by 'stanbon'
stanbon answered: 57894 problems
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2430..2459 , 2460..2489 , 2490..2519 , 2520..2549 , 2550..2579 , 2580..2609 , 2610..2639 , 2640..2669 , 2670..2699 , 2700..2729 , 2730..2759 , 2760..2789 , 2790..2819 , 2820..2849 , 2850..2879 , 2880..2909 , 2910..2939 , 2940..2969 , 2970..2999 , 3000..3029 , 3030..3059 , 3060..3089 , 3090..3119 , 3120..3149 , 3150..3179 , 3180..3209 , 3210..3239 , 3240..3269 , 3270..3299 , 3300..3329 , 3330..3359 , 3360..3389 , 3390..3419 , 3420..3449 , 3450..3479 , 3480..3509 , 3510..3539 , 3540..3569 , 3570..3599 , 3600..3629 , 3630..3659 , 3660..3689 , 3690..3719 , 3720..3749 , 3750..3779 , 3780..3809 , 3810..3839 , 3840..3869 , 3870..3899 , 3900..3929 , 3930..3959 , 3960..3989 , 3990..4019 , 4020..4049 , 4050..4079 , 4080..4109 , 4110..4139 , 4140..4169 , 4170..4199 , 4200..4229 , 4230..4259 , 4260..4289 , 4290..4319 , 4320..4349 , 4350..4379 , 4380..4409 , 4410..4439 , 4440..4469 , 4470..4499 , 4500..4529 , 4530..4559 , 4560..4589 , 4590..4619 , 4620..4649 , 4650..4679 , 4680..4709 , 4710..4739 , 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56970..56999 , 57000..57029 , 57030..57059 , 57060..57089 , 57090..57119 , 57120..57149 , 57150..57179 , 57180..57209 , 57210..57239 , 57240..57269 , 57270..57299 , 57300..57329 , 57330..57359 , 57360..57389 , 57390..57419 , 57420..57449 , 57450..57479 , 57480..57509 , 57510..57539 , 57540..57569 , 57570..57599 , 57600..57629 , 57630..57659 , 57660..57689 , 57690..57719 , 57720..57749 , 57750..57779 , 57780..57809 , 57810..57839 , 57840..57869 , 57870..57899, >>Next
Miscellaneous_Word_Problems/664586: Roberto bought a tennis racquet. He gave the salesman a fifty-dollar bill. Roberto's change was $6.00 less than he paid for the racquet. How much did he pay for the racquet?
1 solutions
Answer 413368 by stanbon(57967)   on 2012-10-10 18:10:21 (Show Source):
You can put this solution on YOUR website!Roberto bought a tennis racquet. He gave the salesman a fifty-dollar bill. Roberto's change was $6.00 less than he paid for the racquet. How much did he pay for the racquet?
------------------------------
change = cost - 6
cost = 50 - change
------------------------
Rearrange:
change - cost = -6
change + cost = 50
-----
Add and solve for "change":
2*change = 44
change = 22
---
solve for "cost":
22 - cost = -6
cost = $28.00
==================
Cheers,
Stan H.
==================
|
Probability-and-statistics/664349: In your own words, describe the relationship between/among the following: Standard error and maximum error
I have answered this wrong more than once and I think what my instructor is wanting is an example involving equal signs with the standard error on the left and maximum error on the right. I am confused as to what he wants, if some one can help.
1 solutions
Answer 413327 by stanbon(57967) on 2012-10-10 14:10:41 (Show Source):
You can put this solution on YOUR website!In your own words, describe the relationship between/among the following: Standard error and maximum error
-----
Standard error is the standard deviation.
---
Max error is the number of standard deviations associated
with the significance level for the problem.
Cheers,
Stan H.
|
Probability-and-statistics/664342: For a particular sample of 63 scores on a psychology exam, the following results were obtained.
First quartile = 47 Third quartile = 81 Standard deviation = 11 Range = 68
Mean = 70 Median = 71 Mode = 77 Midrange = 55
Answer each of the following:
I. What score was earned by more students than any other score? Why?
II. What was the highest score earned on the exam?
III. What was the lowest score earned on the exam?
IV. According to Chebyshev's Theorem, how many students scored between 48 and 92?
V. Assume that the distribution is normal. Based on the Empirical Rule, how many students scored between 59 and 81?
1 solutions
Answer 413255 by stanbon(57967) on 2012-10-10 09:52:50 (Show Source):
You can put this solution on YOUR website!For a particular sample of 63 scores on a psychology exam, the following results were obtained.
First quartile = 47 Third quartile = 81 Standard deviation = 11 Range = 68
Mean = 70 Median = 71 Mode = 77 Midrange = 55
Answer each of the following:
I. What score was earned by more students than any other score?:::77
Why? 77 is the Mode
---------------------
II. What was the highest score earned on the exam?:::
Middle score is 71 and (1/2)Range = 34
Ans: 71+34 = 105
----------------------
III. What was the lowest score earned on the exam?
Ans: 71-34 = 37
--------------------------------
IV. According to Chebyshev's Theorem, how many students scored between 48 and 92?
(48-70)/11 = -2
(92-70)/11 = +2
Cheb say "at least 1-(1/2)^2 = 75% of the data lies in that interval"
----------------------------
V. Assume that the distribution is normal. Based on the Empirical Rule, how many students scored between 59 and 81?
(59-70)/11 = -1
(81-70)/11 = 1
P(-1<= z <=1) = 0.6827 is approximately 68%
======================
Cheers,
Stan H.
======================
|
Probability-and-statistics/664270: on my debit card,,,,,the last 5 numbers if read upside down spell my girlfriends last name,,,,what are the odds of that?
1 solutions
Answer 413254 by stanbon(57967) on 2012-10-10 09:41:34 (Show Source):
You can put this solution on YOUR website!on my debit card,,,,,the last 5 numbers if read upside down spell my girlfriends last name,,,,what are the odds of that?
--------
There are 9*10^4 = 90000 5-digit numbers,
---
The odds of your having the number you have are 1:89999
===================
Cheers,
Stan H.
|
Probability-and-statistics/664335: Suppose we have a set of blood pressures with a mean of 70 Diastolic, and a sample standard deviation of 10 points. Assuming a normal distribution of Diastolic blood pressures, between what two values should 95% of all Diastolic blood pressures lie?
1 solutions
Answer 413249 by stanbon(57967) on 2012-10-10 09:31:51 (Show Source):
You can put this solution on YOUR website!Suppose we have a set of blood pressures with a mean of 70 Diastolic, and a sample standard deviation of 10 points. Assuming a normal distribution of Diastolic blood pressures, between what two values should 95% of all Diastolic blood pressures lie?
-----
Find the z-value with a left tail of 0.025
invNorm(0.025) = -1.96
Note: Correspondingly z = 1.96 has a right tail of 0.025
---
Find the corresponding blood prssure values using x = z*s + u
----
lower value: x = -1.96*10 + 70 = 50.4
upper value: x = +1.96*10 + 70 = 89.6
=========================================
Cheers,
Stan H.
|
Probability-and-statistics/664242: Can you please help me solve this? Assume that your alarm clock has a .87 probability of working. What is the probability that the alarm clock will fail on the morning of a final exam?
1 solutions
Answer 413247 by stanbon(57967) on 2012-10-10 09:25:03 (Show Source):
You can put this solution on YOUR website!Assume that your alarm clock has a .87 probability of working. What is the probability that the alarm clock will fail on the morning of a final exam?
-----
P(works)+P(doesn't work) = 1
---
P(fail) = 1 - P(works)
---
P(fail) = 1 - 0.87
---
P(fail) = 0.13
=====================
Cheers,
Stan H.
=========================
|
Probability-and-statistics/664282: There are 9 letters; 5 consonants and 4 vowels. Three letters are chosen at random. What is the probability of choosing more than one vowel?
1 solutions
Answer 413246 by stanbon(57967) on 2012-10-10 09:19:45 (Show Source):
You can put this solution on YOUR website!There are 9 letters; 5 consonants and 4 vowels. Three letters are chosen at random. What is the probability of choosing more than one vowel?
----
P(choose more than one vowel) = 1 - P(choose no vowels)
----
= 1 - 5C3/9C3
----
= 1 - 0.1190
-----
= 0.8810
=================
Cheers,
Stan H.
|
Probability-and-statistics/664323: A study reported that in a random sampling of 100 women over the age of 35 showed that 8 of the women were married 2 or more times. Based on the study results, how many women in a group of 5,000 women over the age of 35 would likely be married 2 or more times?
1 solutions
Answer 413245 by stanbon(57967) on 2012-10-10 09:15:10 (Show Source):
You can put this solution on YOUR website!A study reported that in a random sampling of 100 women over the age of 35 showed that 8 of the women were married 2 or more times. Based on the study results, how many women in a group of 5,000 women over the age of 35 would likely be married 2 or more times?
-----
Solve as a proportion:
x/5000 = 8/100
---
x = (8/100)5000
x = 400
=================
Cheers,
Stan H.
|
Expressions-with-variables/664329: Use the emlimination method to solve the following system of equations.
2x-y+z=-3
2x+2y+3z=2
3x-3y-z=-4
I have gotten many answers for this question and somehow none of them are right. Thank you for the help.
1 solutions
Answer 413244 by stanbon(57967) on 2012-10-10 09:12:19 (Show Source):
You can put this solution on YOUR website!2x-y+z=-3
2x+2y+3z=2
3x-3y-z=-4
-----------------
I used a matrix function on my TI-84 to get:
x = 1
y = 3
z = -2
===========
Cheers,
Stan H.
================
|
Mixture_Word_Problems/664325: .How many pounds of dried apples worth $5 per pound should me mixed with how many pounds of dried apricots worth $9 per pound to makel0 pounds of mixture worth $77?
1 solutions
Answer 413241 by stanbon(57967) on 2012-10-10 09:09:02 (Show Source):
You can put this solution on YOUR website!How many pounds of dried apples worth $5 per pound should me mixed with how many pounds of dried apricots worth $9 per pound to makel0 pounds of mixture worth $77?
--------------
Equation:
x + y = 10 lbs
5x + 9y = 77
----------------------
5x + 5y = 50
5x + 9y = 77
-----
Subtract and solve for "y":
4y = 27
y = 27/4 = 6 3/4 lbs (amt. of apricots needed)
----
Solve for "x":
x + (27/4) = 40/4
x = 13/4 = 3 1/4 lbs (amt. of apples)
---------------------
Cheers,
Stan H.
|
Probability-and-statistics/664027: I. Use Chebyshev’s theorem to find what percent of the values will fall between 249 and 333 for a data set with a mean of 291 and standard deviation of 14.
II. Use the Empirical Rule to find what two values 99.7% of the data will fall between for a data set with a mean of 219 and standard deviation of 20
1 solutions
Answer 413188 by stanbon(57967) on 2012-10-09 22:56:51 (Show Source):
You can put this solution on YOUR website!I. Use Chebyshev’s theorem to find what percent of the values will fall between 249 and 333 for a data set with a mean of 291 and standard deviation of 14.
--------
(249-291)/14 = -3
(333-291)/14 = +3
----------------------------
The percent is at least 1-(1/3)^2 = 8/9 = 0.89 = 89% of the data.
=======================================================================
II. Use the Empirical Rule to find what two values 99.7% of the data will fall between for a data set with a mean of 219 and standard deviation of 20
--------
Solve 1 - (1/k)^2 = 0.997
----
= (1/k)^2 = 0.003
----
k^2 = 1/0.003
---
k^2 = 333.33
-----------------------
k = 18 +
---
lower limit: 219-18*20 = -141
upper limit: 219+18*20 = 579
=================================
Cheers,
Stan H.
======================
|
Probability-and-statistics/664062: A committee is formed by selecting 4 members from a group of 10 Republicans and 8 Democrats find the probability there is at least one member from each party on the committee
1 solutions
Answer 413184 by stanbon(57967) on 2012-10-09 22:30:14 (Show Source):
You can put this solution on YOUR website!A committee is formed by selecting 4 members from a group of 10 Republicans and 8 Democrats find the probability there is at least one member from each party on the committee
--------
P(1 <= x <= 4) = 1 - P(4 Dem or 4 Rep are chosen)
-----
= 1 - [(10C4 + 8C4)/18C4]
-----
= 1 - 0.0915
------------
= 0.9085
==============
Cheers,
Stan H.
|
Probability-and-statistics/664110: Say that a randomly chosen individual can identify the professor with probability 0.47. What is the probability that, of 9 chosen people, exactly 7 people recognize the professor?
1 solutions
Answer 413181 by stanbon(57967) on 2012-10-09 22:21:33 (Show Source):
You can put this solution on YOUR website!Say that a randomly chosen individual can identify the professor with probability 0.47. What is the probability that, of 9 chosen people, exactly 7 people recognize the professor?
--------
Binomial Problem with n = 9 ; p(can ID)=0.47
----------------
P(x = 7) = 9C7(0.47)^7*(0.53)^2 = binompdf(9,0.47,7) = 0.0512
====================
Cheers,
Stan H.
==================
|
test/664189: Ok. If hoshi walked 10 meters in 3 seconds...how many meters did she walk per second? Please help!
1 solutions
Answer 413175 by stanbon(57967) on 2012-10-09 22:10:53 (Show Source):
You can put this solution on YOUR website!If hoshi walked 10 meters in 3 seconds...how many meters did she walk per second?
-------
Solve the proportion:
x meters/1 second = 10 meters/3 seconds
--------------------------------------------
x = 10/3 = 3 1/3 meters
==================================
Cheers,
Stan H.
|
Geometry_proofs/664181: Please help me solve this problem!
Given: line k is parallel to line p
prove:m<1 is congruent to m<3
i need the statements and reasons! please help me! thank you
1 solutions
Answer 413170 by stanbon(57967) on 2012-10-09 22:04:44 (Show Source):
You can put this solution on YOUR website!I think you will have to better describe the problem.
What do you mean by m<1 and m<3 and what do the
congruence refer to?
Cheers,
Stan H.
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Travel_Word_Problems/664178: a car moved for half the distance at a speed of 55 mph and the other half the distance 75 mph. what is the average speed?
Show ALL work
1 solutions
Answer 413163 by stanbon(57967) on 2012-10-09 22:00:42 (Show Source):
You can put this solution on YOUR website!a car moved for half the distance at a speed of 55 mph and the other half the distance 75 mph. what is the average speed?
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1st half DATA:
rate = 55 mph ; distance = x miles ; time = x/55 hrs
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2nd half DATA:
rate = 75 mph ; distance = x miles ; time = x/75 hrs
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Average speed = (total distance)/(total time) = (2x)/(x/55+x/75)
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= (2x)/[(75x+55x)/55*75)] = (2*55*75x)/(120x) = 68.75 mph
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Cheers,
Stan H.
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Travel_Word_Problems/664158: Matt travels at a speed of 64km/hr, if Matt travelled at 78 km/hr he would arrive at his destination 10 Min earlier. What distance did Matt travel?
1 solutions
Answer 413156 by stanbon(57967) on 2012-10-09 21:48:40 (Show Source):
You can put this solution on YOUR website!Matt travels at a speed of 64km/hr, if Matt travelled at 78 km/hr he would arrive at his destination 10 Min earlier. What distance did Matt travel?
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Slower speed DATA:
rate = 64 km/hr ; distance = x km ; time = x/64 hrs
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Faster speed DATA:
rate = 78 km/hr ; distance = x km ; time = x/78 hrs
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Equation:
time - time = 1/6 hr
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x/64 - x/78 = 1/6
--------
6*78x - 6*64x = 64*78
86x = 64*78
----
x = 59.43 km
=================
Cheers,
Stan H.
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Travel_Word_Problems/663803: If you are going 65 mph, how long would it take to go 1/10th of a mile?
What if you were going 55 mph?
1 solutions
Answer 412949 by stanbon(57967) on 2012-10-09 15:06:46 (Show Source):
You can put this solution on YOUR website!If you are going 65 mph, how long would it take to go 1/10th of a mile?
What if you were going 55 mph?
----
time = distance/rate
----
time = (1/10)/65 = 1/650 hr = 0.001538.. hr. = 5.54 seconds
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time = (1/10)/55 = 1/550 hr = 0.0.001818..hr. = 6.55 seconds
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Cheers,
Stan H.
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Money_Word_Problems/663800: Sydney earns $2300 a month. To qualify for a mortgage, her monthly payments must be 35% of he monthly income. Her monthly mortgage payments must be less than what amount in order to qualify for te mortgage?
Please show me how to organize and solve the problem.
1 solutions
Answer 412946 by stanbon(57967) on 2012-10-09 15:02:26 (Show Source):
You can put this solution on YOUR website! Sydney earns $2300 a month. To qualify for a mortgage, her monthly payments must be 35% of he monthly income. Her monthly mortgage payments must be less than what amount in order to qualify for te mortgage?
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monthly payment must be < 0.35*2300 = $805.00
==========================
Cheers,
Stan H.
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Graphs/663796: Find the value of y if the line through the points (2,y) and (4,6) has slope m=-2
1 solutions
Answer 412945 by stanbon(57967) on 2012-10-09 14:59:59 (Show Source):
You can put this solution on YOUR website!Find the value of y if the line through the points (2,y) and (4,6)
has slope m=-2
----
Solve:
(y-6)/(2-4) = -2
y-6 = 4
y = 10
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Cheers,
Stan H.
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Probability-and-statistics/663773: A confidence interval estimate for the population mean is given to be (35.85, 44.80). If the standard deviation is 12.298 and the sample size is 41, answer each of the following (show all work):
(A) Determine the maximum error of the estimate, E.
(B) Determine the confidence level used for the given confidence interval.
1 solutions
Answer 412944 by stanbon(57967) on 2012-10-09 14:57:56 (Show Source):
You can put this solution on YOUR website!A confidence interval estimate for the population mean is given to be (35.85, 44.80). If the standard deviation is 12.298 and the sample size is 41, answer each of the following (show all work):
(A) Determine the maximum error of the estimate, E.
Note: The width of the interval ALWAYS equals 2*E
---
2E = (44.8-35.85) = 8.95
E = 4.475
==============================
(B) Determine the confidence level used for the given confidence interval.
4.475 = z*12.298/sqrt(41)
z = 2.3299
Conf. Level = 1 - 2*normalcdf(2.3299,100) = 1-2% = 98%
=====================
Cheers,
Stan H.
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Probability-and-statistics/663785: I tried reading the text more than once and even tried the hints that were given but....thanks in advance
By measuring the amount of time it takes a component of a product to move from one workstation to the next, an engineer has estimated that the standard deviation is 3.15 seconds. Please answer the following and show all work: A) how many measurements should be made in order to be 99% certain that the maximum error of estimation will not exceed 1.0 seconds? B) what sample size is required for a maximum error of 2.0 seconds?
1 solutions
Answer 412942 by stanbon(57967) on 2012-10-09 14:50:51 (Show Source):
You can put this solution on YOUR website!By measuring the amount of time it takes a component of a product to move from one workstation to the next, an engineer has estimated that the standard deviation is 3.15 seconds. Please answer the following and show all work:
---------------------
A) how many measurements should be made in order to be 99% certain that the maximum error of estimation will not exceed 1.0 seconds? ------
n = [z*s/E]^2
---
n = [2.5758*3.15/1.0]^2 = 66 when rounded up
=====================================================
B) what sample size is required for a maximum error of 2.0 seconds?
n = [2/5758*3.15/2]^2 = 17 when rounded up
=============================
Cheers,
Stan H.
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Number-Line/663687: three fourths of the distance from 5 to 13
1 solutions
Answer 412879 by stanbon(57967) on 2012-10-09 09:42:06 (Show Source):
You can put this solution on YOUR website!three fourths of the distance from 5 to 13
----
(3/4)(13-5) = (3/4)(8) = 6
----------------------
If you are looking for the number that is
(3/4) the distance it is 5+6 = 11
=====================
Cheers,
Stan H.
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Probability-and-statistics/663680: The following table describes the asset allocation of your portfolio with the corresponding return in percentage, X, and the proportion of that specific asset in the portfolio, P(X).
Money Market: X=10% , P(X)=.10
Short Term Securities: X=15% , P(X)=.10
Long Term Debt: X=25% , P(X)=.20
Large Cap Equity: X=25% , P(X)=.35
Small Cap Equity: X=15% , P(X)=.15
International Equity: X=10% , P(X)=.10
Calculate the E(X) for this portfolio.
Calculate the E{ [X-E(X)] 2} for this portfolio.
1 solutions
Answer 412878 by stanbon(57967) on 2012-10-09 09:39:30 (Show Source):
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Probability-and-statistics/663663: These problems I have been working on all weekend, and I do not understand. My grade and passing the course depends on this. HOWEVER, I have not received a single reply. WHY? Very unusual, from the tutors here! I need your help, if I could pay I would! It is that important.
Problem 1: Assume that the mean score on a certain aptitude test across the nation is 100,and that the standard deviation is 20 points. Find the probability that the mean aptitude test score for a randomly selected group of 150 8th graders is 100. (Points:3)
Solution what I am doing right or wrong: z=(100)=(100-100)/20=-0/20=
Problem2:.Assume that a sample is drawn and z(a/2) =1.96 and σ=35. Answer the following questions:
If the Maximum Error of Estimate is 0.02 for this sample, what would be the sample size? ( I got this far: N=zs/E)^2N= (1.96 *35/)(.02^2=) don’t know if I am doing it right)
Given that the sample SIZE is 400 with this same z(a/2) and σ, what would be the Maximum Error of Estimate? (My workings are E= zs/sqrt(n) E= 1.96*35/sqrt400=.606)
What happens to the Maximum Error of Estimate as the sample size gets smaller? (I have no idea)
What effect does the answer to C above have to the size of the confidence interval? (My work; As sample gets larger the ME gets smaller so the CI gets smaller.) (Points:8)
Problem:3 By measuring the amount of time it takes a component of a product to move from one workstation to the next, an engineer has estimated that the SD is 4.34 seconds. Answer all of the question by showing your work:
How many measurements should be made in order to be 90% CERTAIN that the ME of estimation will not exceed 2.0 seconds?
What sample size is required for a ME of 0.5 seconds? (Points:6)
(My work: a) 1-a=0.90 =2.0 ) TOTALLY LOST
Problem 3: A 99% confidence interval estimate for a population mean was computed to be (47.1, 65.9). Determine the mean of the sample, which was used to determine the interval
Estimate. (Must show all work) Points: 4) ( z=0.4761 0.6591) from table 3- I really don’t know) just guessing …
Problem:4: A study was conducted to estimate the mean amount spent on birthday gifts for a typical family having 2 children. A sample of 185 was taken, and the mean amount spent was $244.16. Assuming a SD equal to $38.98, find the 98% confidence interval for μ, the mean for all such families. (Show ALL work) (Points 4). (Don’t know where to start as I am very confused)
Please answer me this time around....
1 solutions
Answer 412877 by stanbon(57967) on 2012-10-09 09:37:38 (Show Source):
You can put this solution on YOUR website!Problem 1: Assume that the mean score on a certain aptitude test across the nation is 100,and that the standard deviation is 20 points. Find the probability that the mean aptitude test score for a randomly selected group of 150 8th graders is 100. (Points:3)
Solution what I am doing right or wrong: z=(100)=(100-100)/20=-0/20=0
------
Note: In a continuous distribution the probability of gettin ANY exact
score is zero.
----------------------------------
Problem2:.Assume that a sample is drawn and z(a/2) =1.96 and σ=35. Answer the following questions:
If the Maximum Error of Estimate is 0.02 for this sample, what would be the sample size? ( I got this far:
N = (zs/E)^2
N = (1.96 *35/0.02)^2
= 11,764,900
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Given that the sample SIZE is 400 with this same z(a/2) and σ, what would be the Maximum Error of Estimate?
E= zs/sqrt(n)
E= 1.96*35/sqrt(400) = 3.43
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What happens to the Maximum Error of Estimate as the sample size gets smaller?
E = z*s/sqrt(n)
Notice that E and sqrt(n) are indirectly related.
So, as n gets smaller, E must get larger.
Try it with numbers and you will see what I mean.
---------------------------------------------
What effect does the answer to C above have to the size of the confidence interval? (My work; As sample gets larger the ME gets smaller so the CI gets smaller.) (Points:8)
Correct
-------------------------
Problem:3 By measuring the amount of time it takes a component of a product to move from one workstation to the next, an engineer has estimated that the SD is 4.34 seconds. Answer all of the question by showing your work:
How many measurements should be made in order to be 90% CERTAIN that the ME of estimation will not exceed 2.0 seconds?
----
n = [z*s/E]^2 = [1.645*4.34/2]^2 = 13 when rounded up
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What sample size is required for a ME of 0.5 seconds? (Points:6)
Same formula with 0.5 replacing 2.
Ans: 204 when rounded up
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Problem 3: A 99% confidence interval estimate for a population mean was computed to be (47.1, 65.9). Determine the mean of the sample, which was used to determine the interval Estimate.
Note: The width of the CI is ALWAYS 2*ME
2*ME = 65.9-47.1 = 18.8
ME = 9.4
---
Note The sum of the CI limits is ALWAYS 2*(sample mean)
2(sample mean) = 65.9+47.1 = 113
sample mean = 56.5
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Problem:4: A study was conducted to estimate the mean amount spent on birthday gifts for a typical family having 2 children. A sample of 185 was taken, and the mean amount spent was $244.16. Assuming a SD equal to $38.98, find the 98% confidence interval for μ, the mean for all such families. (Show ALL work) (Points 4). (Don’t know where to start as I am very confused)
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x-bar = 244.16
ME = 2.3263*38.98/sqrt(185) = 6.67
----
98% CI: 244.16-6.67 < u < 244.16+6.67
=============================================
Cheers,
Stan H.
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Probability-and-statistics/663557: Most four-year automobile leases allow up to 77,000 miles. If the lessee goes beyond this amount, a penalty of 20 cents per mile is added to the lease cost. Suppose the distribution of miles driven on four-year leases follow the normal distribution. The mean is 65,000 miles and the standard deviation is 5,000 miles.
What percent of the leases will yield a penalty because of excess mileage?
<< This is just one sample problem for my homework. I cannot seem to understand the process to answering these questions. Please show all formulas and calculations so that I can learn from this >>
1 solutions
Answer 412846 by stanbon(57967) on 2012-10-08 22:39:30 (Show Source):
You can put this solution on YOUR website!Most four-year automobile leases allow up to 77,000 miles. If the lessee goes beyond this amount, a penalty of 20 cents per mile is added to the lease cost. Suppose the distribution of miles driven on four-year leases follow the normal distribution. The mean is 65,000 miles and the standard deviation is 5,000 miles.
What percent of the leases will yield a penalty because of excess mileage?
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z(77000) = (77000-65000)/5000 = 12/5 = 2.4
----
P(milage > 77000) = P(z > 2.4) = normalcdf(2.4,100) = 0.0082
========================
Cheers,
Stan H.
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Probability-and-statistics/663575: A major cell phone service provider has determined that the number of minutes that its customers use their phone per month is normally distributed with a mean equal to 445.5 minutes with a standard deviation equal to 177.8 minutes. The company is thinking of charging a lower rate for customers who use the phone less than a specified amount. If it wishes to give the rate reduction to no more than 12 percent of its customers, what should the cut off be?
1 solutions
Answer 412845 by stanbon(57967) on 2012-10-08 22:34:09 (Show Source):
You can put this solution on YOUR website!A major cell phone service provider has determined that the number of minutes that its customers use their phone per month is normally distributed with a mean equal to 445.5 minutes with a standard deviation equal to 177.8 minutes. The company is thinking of charging a lower rate for customers who use the phone less than a specified amount. If it wishes to give the rate reduction to no more than 12 percent of its customers, what should the cut off be?
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Find the z-value with with a left-tail of 12%.
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invNorm(0.12) = -1.1750
---
Find the x-value that corresponds:
x = z*s + u
---
x = -1.1750*177.8+445.5 = 236.59
====================================
Cheers,
Stan H.
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Probability-and-statistics/663558: The following table describes the asset allocation of your portfolio with the corresponding return in percentage, X, and the proportion of that specific asset in the portfolio, P(X).
Money Market: X=10% , P(X)=.10
Short Term Securities: X=15% , P(X)=.10
Long Term Debt: X=25% , P(X)=.20
Large Cap Equity: X=25% , P(X)=.35
Small Cap Equity: X=15% , P(X)=.15
International Equity: X=10% , P(X)=.10
Calculate the E(X) for this portfolio.
Calculate the E{ [X-E(X)] 2} for this portfolio.
<< This is just one sample problem for my homework. I cannot seem to understand the process to answering these questions. Please show all formulas and calculations so that I can learn from this >>
1 solutions
Answer 412841 by stanbon(57967) on 2012-10-08 22:22:11 (Show Source):
You can put this solution on YOUR website!Money Market: X=10% , P(X)=.10
Short Term Securities: X=15% , P(X)=.10
Long Term Debt: X=25% , P(X)=.20
Large Cap Equity: X=25% , P(X)=.35
Small Cap Equity: X=15% , P(X)=.15
International Equity: X=10% , P(X)=.10
Calculate the E(X) for this portfolio.
Procedure: Multiply each x value by its P(x) value
then add those products to get E(x)
------------------------------------------
Calculate the E{ [X-E(X)] 2} for this portfolio.
Multiply each (x-E(x))^2 by its P(x)
then add those products to get E[x-E(x))^2]
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cheers,
Stan H.
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Probability-and-statistics/663335: there are 15 teachers; 11 are women and 4 are men. Two are selected at random to meet the governor. What is the probability that both are the same gender?
1 solutions
Answer 412824 by stanbon(57967) on 2012-10-08 21:07:27 (Show Source):
You can put this solution on YOUR website!there are 15 teachers; 11 are women and 4 are men. Two are selected at random to meet the governor. What is the probability that both are the same gender?
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# of ways to succeed: 11C2 + 4C2 = 55+6 = 61
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# of possible outcomes: 15C2 = 105
----
P(2 men or 2 women) = 61/105
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Cheers,
Stan H.
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