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Tutors Answer Your Questions about Probability-and-statistics (FREE)
Question 241450: Is an ace considered a picture card in probability and statistics?
Answer by Alan3354(6097)   (Show Source):
Question 241395: I rolled a pair of number cubes 30 times. Heres a google spreadsheet webpage that shows what numbers I got:
http://spreadsheets.google.com/pub?key=tTcZez2zPqpi6D1a7RHTOcQ&output=html
There were many questions, but the ones I don't get are: Use your data to find the following experimental probabilities P(5) P(6) P(12) and P(36)
Make a chart to find the "SAMPLE SPACE" which I don't know how to do...
The other 5 questions I could understand.
Found 2 solutions by jsmallt9, stanbon:
Answer by jsmallt9(594)  (Show Source):
You can put this solution on YOUR website!(Note: I am going to explain this without having looked at your data.)
To find the probabilities using the experimental data just make the following fractions and reduce:
P(5) = (number of times you spun a 5)/30
P(6) = (number of times you spun a 6)/30
P(12) = (number of times you spun a 12)/30
P(36) = (number of times you spun a 36)/30
(If you are using standard number cubes (aka dice), then P(36) should be 0 since a spin of two dice cannot possibly add up to 36.)
As far as the chart goes, I am not sure if there is a specific kind of chart that should be used. You could use a bar chart where one axis is the number of spins and the other axis is the sum for that spin. I'd suggest that the sums be on the horizontal axis (aka x-axis) and the number of spins would be on the vertical axis (aka y=axis). Then for each sum you would draw a bar from the x-axis up to the hieight that matches the number of times you spun that number. For example, if you spun an 8 ten times, then you would draw a bar at 8 on the x-axis up to where 10 is on the y-axis. Do this for each number you spun. (There will be no bars for numbers you never spun.)
I hope this helps.
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!I rolled a pair of number cubes 30 times. Heres a google spreadsheet webpage that shows what numbers I got:
http://spreadsheets.google.com/pub?key=tTcZez2zPqpi6D1a7RHTOcQ&output=html
----------------
What numbers are on your cubes?
What "result" are you looking at from each roll?
Are you looking for the sum of the upward faces?, the product of the
up ward faces? or what?
----
I'm asking this so I can understand how you would get P(36)?
Cheers,
Stan H.
=========================
There were many questions, but the ones I don't get are: Use your data to find the following experimental probabilities P(5)
P(6) P(12) and P(36)
Make a chart to find the "SAMPLE SPACE" which I don't know how to do...
The other 5 questions I could understand.
Question 241409: Can anybody help me with this question? The question is:
Develop a 90 percent confidence interval for the population mean. Interpret the results.
The Greater Pittsburgh Area Chamber of Commerce wants to estimate the mean time workers who are employed in the downtown area spend getting to work. A sample of the 15 workers reveals the following number of minutes spent traveling.
29 38 38 33 38 21 45 34
40 37 37 42 30 29 35
Develop a 98 percent confidence interval for the population mean. Interpret the results.
Thank you.
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!Develop a 90 percent confidence interval for the population mean. Interpret the results.
The Greater Pittsburgh Area Chamber of Commerce wants to estimate the mean time workers who are employed in the downtown area spend getting to work. A sample of the 15 workers reveals the following number of minutes spent traveling.
29 38 38 33 38 21 45 34
40 37 37 42 30 29 35
------------
sample mean = 35.067
sample std = 6.017
-------
standard error = z*s/sqrt(n)
standard error = 1.645*6.017/sqrt(15) = 2.556
-------------------------------------------------------
90% CI: 35.067-2.556 < u < 35.067+2.556
90% CI: 32.511 u < 37.62
======================================================
Develop a 98 percent confidence interval for the population mean. Interpret the results.
Use the z-value for 98%: z = 2.3263
sample mean is the same as above
standard error = 2.3263*6.017/sqrt(15)=3.614
---
98% CI: 35.067 - 3.614 < u < 35.067 + 3.614
98% CI: 31.45 < u < 38.68
---------------------------------
Meaning: We have 98% confidence the population mean is between
31.45 and 38.68
==============================================
Cheers,
Stan H.
Question 241382: Suppose that the probability that a new medication will cause a bad side effect is 0.03. If this medication is given to 150 people, what is the probability that more than three of them will experience a bad side effect? Round answer to 4 decimal places.
I think I know how to compute for "exactly three students," but I cannot figure out how to do "more than three students"
Here's how I approached "exactly three":
p = 0.03, q = 1 - p = 0.97, n = 150
P(r, n) = nCr (p)^r * (q)^(n - r)
P(3, 150) = 150C3 * 0.03^3 * 0.97^147 = 0.1691
Since I am trying to figure out how to do these types of problems on my own, please include the steps.
Thank you!
Found 2 solutions by stanbon, edjones:
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!Suppose that the probability that a new medication will cause a bad side effect is 0.03. If this medication is given to 150 people, what is the probability that more than three of them will experience a bad side effect? Round answer to 4 decimal places.
I think I know how to compute for "exactly three students," but I cannot figure out how to do "more than three students"
---------------------
You should have or you need a cumulative binomial chart or a calculators
with a Stat function.
------
Your problem:
Binomial with n = 150, p = 0.03, 4 <= x <= 150
------------------
Your problem:
P(4 <= x <= 150) = 1 - P(0 <= x <= 3)
= 1 - [P(x= 0) + P(x=1) + P(x=2) + P(x=3)]
= 1 - [0.01037 + 0.048106..+ 0.110843 + 0.16912...)
= 1 - [0.33844
= 0.66156...
========================
Stan H.
Answer by edjones(3298)  (Show Source):
You can put this solution on YOUR website!You are correct so far as you have gone.
Now do the same procedure for exactly 2, 1 and 0 students.
Find the sum of the 4 probabilities and subtract it from 1 and you have your answer.
.
Ed
Question 241296: 1)A sales promoter offers a special price for a scanner with a printer. The probabilities that each functions satisfactorily over three years are 0.68 and 0.72 respectively. What is the probability that at the end of the three years:
i) Both scanner and printer will function satisfactorily?
ii) Only the printer will be functioning?
iii) At least one of the two machines will be functioning? State any assumptions made.
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2)240 students were asked to taste three types of cola in a random order and were not told their identity. The contingency table below shows the counts of preferred colas classified by the subject specialism of students.
Cola A Cola B Cola C
Mathematics 13 23 44
Economics 36 24 10
Statistics 24 8 8
Management 11 28 11
Use These data to investigate the association, if any, between a student's subject specialism and preferred cola-type. Interpret any association which exists, testing at more than one significance level.
--------------------------------------------------------------------------------
3)The times required by three workers to perform an assembly-line task were recorded on five randomly selected occasions. Here are the times, to the nearest minute.
Hank Joseph Susan
8 8 10
10 9 9
9 9 10
11 8 11
10 10 9
Assuming the normality of populations under consideration, at the 5% significance level,
do the data provide sufficient evidence to conclude that a difference exists between the times required by three workers to perform an assembly-line task?
Your kind assistance is highly appreciated
Bahman Mennati
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!1)A sales promoter offers a special price for a scanner with a printer. The probabilities that each functions satisfactorily over three years are 0.68 and 0.72 respectively. What is the probability that at the end of the three years:
i) Both scanner and printer will function satisfactorily?
Ans: (0.68*0.72) = 0.4896..
ii) Only the printer will be functioning?
Ans: 0.72*(1-0.68) = 0.2304..
iii) At least one of the two machines will be functioning? State any assumptions made.
Ans: At least one = 1 - P(none)
At least one = 1 - (1-0.72)(1-0.68) = 0.9104
-------------------------------------------------------------------------
2)240 students were asked to taste three types of cola in a random order and were not told their identity. The contingency table below shows the counts of preferred colas classified by the subject specialism of students.
Cola A Cola B Cola C
Mathematics 13 23 44
Economics 36 24 10
Statistics 24 8 8
Management 11 28 11
Use These data to investigate the association, if any, between a student's subject specialism and preferred cola-type. Interpret any association which exists, testing at more than one significance level.
Ho: The row and column factors are independent.
H1: They are dependent
I ran an ANOVA-test on the data and got:
test statistic: F = 0.05486...
p-value: 0.9469
Since the p-value is greater even then 10%, do not reject
Ho at the 10% significance level.
The row and column factors are independent.
---------------------------------------------------------
--------------------------------------------------------------------------------
3)The times required by three workers to perform an assembly-line task were recorded on five randomly selected occasions. Here are the times, to the nearest minute.
Hank Joseph Susan
8 8 10
10 9 9
9 9 10
11 8 11
10 10 9
Assuming the normality of populations under consideration, at the 5% significance level,
do the data provide sufficient evidence to conclude that a difference exists between the times required by three workers to perform an assembly-line task?
Note: The data posted is too confusing to decipher.
Cheers,
Stan H.
Question 241364: Given a 4 sided cube, with the numbers 1. 2. 3. 4. on each side, what would be the probability of rolling 2, 4 sided cubes and getting a sum of 5? whats the sample space and favorable outcome?
Answer by Edwin McCravy(2922) (Show Source):
You can put this solution on YOUR website! Given a 4 sided cube, with the numbers 1. 2. 3. 4. on each side, what would be the probability of rolling 2, 4 sided cubes and getting a sum of 5? whats the sample space and favorable outcome
Here is the sample space, where the left number within each
pair of parentheses represents the number rolled on the first
die and the second represents that on the second die.
(1,1) (1,2) (1,3) (1,4)
(2,1) (2,2) (2,3) (2,4)
(3,1) (3,2) (3,3) (3,4)
(4,1) (4,2) (4,3) (4,4)
Next I will color the ones red which have a sum of 5:
(1,1) (1,2) (1,3) (1,4)
(2,1) (2,2) (2,3) (2,4)
(3,1) (3,2) (3,3) (3,4)
(4,1) (4,2) (4,3) (4,4)
That's 4 out of 16, so the probability is  which reduces to 
Edwin
Question 241336: From 4 cards bearing the numbers -2, -1, 0, 1, one card will be drawn at random. Let X be the square of the number on the card. Find the expectaion of X. I have worked on this for hours but never seem to come up with a an answer that matches any of these: a 1, c 2 c 1.75 or d 1.5.
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!From 4 cards bearing the numbers -2, -1, 0, 1, one card will be drawn at random. Let X be the square of the number on the card. Find the expectaion of X. I have worked on this for hours but never seem to come up with a an answer that matches any of these: a 1, c 2 c 1.75 or d 1.5.
-----------
Values of "X": 4, 1, 0, 1
Probability for each is (1/4)
----------------
Expected value of X: (1/4)(4+1+0+1) = 6/4 = 3/2 = 1.5
===========================================================
Cheers,
Stan H.
Question 241340: John tosses a coin three times in succession. He wins $1 for head on the first toss, $2 for head on the second toss and $3 for head on the third toss. For each tail he losses $1.5. Find John's expected winnings. I saw the problem where the expected value was 81.25 cents but that is not one of my available answers which are 0.75, 1, 0.68 and -0.75.
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!John tosses a coin three times in succession. He wins $1 for head on the first toss, $2 for head on the second toss and $3 for head on the third toss. For each tail he losses $1.5. Find John's expected winnings.
Head on 1st toss: (1/2)*1 = 0.5
Head on 2nd toss: (1/2)(-1.5)+(1/2)2 = -0.75+1 = 0.25
Head on 3rd toss: (1/2)(-1.5)+(1/2)(-1.5)+(1/2)3 = -0.75-0.75+1.5 = 0
-------------------------
Expected value = 0.75
---------------------------
Cheers,
Stan H.
===========================================
I saw the problem where the expected value was 81.25 cents but that is not one of my available answers which are 0.75, 1, 0.68 and -0.75.
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Question 241210: the probability that 2 men in a group of 5 will still be working in 30 years is 3.5
What is the probability that 3 men in a group of 5 will still be working in 30 years?
can you also provide the formula with the answer?
thanks
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!the probability that 2 men in a group of 5 will still be working in 30 years is 0.35
------
I find the underlying p-value for this result is p = 0.4
You should be able to find that with your binomial table.
-----------
What is the probability that 3 men in a group of 5 will still be working in 30 years?
---
So P(x=3) = 5C3(0.4)^3(0.6)^2 = 0.2304
==============================================
Cheers,
Stan H.
Question 241019: A box contains 3 red, 3 blue and 4 white socks. In how many ways can 8 socks be pulled out of the box, one at a time, if order is important?
Answer by nyc_function(260)  (Show Source):
You can put this solution on YOUR website!Permutation: A set of objects in which position (or order) is important.
So, we are talking about permutation.
total socks (P) chosen socks can be written as 10P8.
We now use the formula nPr = n!/(n - r)!, where n is the total number of objects and r is the number of objects chosen (what we want).
10P8 = 10!/(10 - 8)!
Can you finish now?
Question 240984: Compute the test statistic used to test the null hypothesis that p1 = p2.
In a vote on the Clean Water bill, 53% of the 205 Democrats voted for the bill while 51% of the 230 Republicans voted for it.
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!Compute the test statistic used to test the null hypothesis that p1 = p2.
In a vote on the Clean Water bill, 53% of the 205 Democrats voted for the bill while 51% of the 230 Republicans voted for it.
---------------------------------------------------------
Ho: p1-p2 = 0
----
ts = (0.53-0.51)/sqrt[(0.53*0.47)/205 + (0.51*0.49/230)]
= 0.02/0.0480
= 0.4169
==============
Cheers,
Stan H.
Question 240856: Find Z when:
The claim is that the proportion of drowning deaths of children attributable to beaches is more than 0.25, and the sample statistics include n = 657 drowning deaths of children with 30% of them attributable to beaches.
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!Find Z when:
The claim is that the proportion of drowning deaths of children attributable to beaches is more than 0.25, and the sample statistics include n = 657 drowning deaths of children with 30% of them attributable to beaches.
----------------
Ho: p = 0.25
H1: p > 0.25
------------------
z(0.30) = (0.30-0.25)/sqrt[0.25*0.75/657] = 2.9597
======================================================
Cheers,
Stan H.
Question 240949: In statistics
Using the t value table
How do u find the lower tail area of .05 with 50 degrees of freedom?
Answer by Edwin McCravy(2922) (Show Source):
You can put this solution on YOUR website!
The area IS .05!!! I think you must mean how do you find the
t-score which has the area of .05 to the left of it.
If that's what you want, you find the value 50 under the df-column
then find the column that is headed 0.05 and the value in line with
50 in the df-column is t=1.676. That's the value of t that has
area .05 to the RIGHT of it. But since you want the t-score that
has area .05 to the LEFT of it, you simply make that value negative
because the t-curve is symmetrical.
Answer: t = -1.676
Edwin
Question 240887: what is the probability of having exactly 3 girls out of five children?
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!what is the probability of having exactly 3 girls out of five children?
---------------------------
P(x = 3) = 5C3*(1/2)^3*(1/2)^2 = 10(1/32) = 10/32 = 5/16
=========================================================
Cheers,
Stan H.
Question 240859: Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $110,000. This distribution follows the normal distribution with a standard deviation of $40,000.
a. If we select a random sample of 50 households, what is the standard error of the mean?
b. What is the expected shape of the distribution of the sample mean?
c. What is the likelihood of selecting a sample with a mean of at least $112,000?
d. What is the likelihood of selecting a sample with a mean of more than $100,000
e. What is the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000?
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $110,000. This distribution follows the normal distribution with a standard deviation of $40,000.
a. If we select a random sample of 50 households, what is the standard error of the mean?
The standard error depends on the level of significance.
If alpha is 5%, SE = 2.0096*40,000/sqrt(50) = 11369.0143
-------------------------------------------------------------
b. What is the expected shape of the distribution of the sample mean?
Normal with mean = 110,000 and std = 40,000/sqrt(25) = 5656.85
--------------------------------------------------------------
c. What is the likelihood of selecting a sample with a mean of at least $112,000?
t(112,000)= (112,000-110,000)/[40,000/sqrt(50)] = 0.3536..
-----------------
P(x >= 112000) = P(t >= 0.3536 with df = 49) = 0.3626
================================================================
d. What is the likelihood of selecting a sample with a mean of more than $100,000
Same process as "c"
----------------------------
e. What is the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000?
Combine the t-values from "c" and "d".
===========================================
Cheers,
Stan H.
Question 240765: accordin to goverment data, the probability that an adult was never married is 15%, in a random survey of 10 adults, what is the probability that at least 8 were married?
Found 3 solutions by solver91311, edjones, stanbon:
Answer by solver91311(5072)  (Show Source):
Answer by edjones(3298) (Show Source):
You can put this solution on YOUR website!Let m=married and n=never married
.85+.15=1
m^10+10m^9n+45m^8n^2 (derived from (m+n)^10)
.85^10 + 10*.85^9*.15 + 45*.85^8*.15^2
=.82
.
Ed
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!): according to goverment data, the probability that an adult was never married is 15%, in a random survey of 10 adults, what is the probability that at least
8 were married?
------------------
This is a binomial problem with n=10, p = 0.15, 8<=x<=10
-------------------------------
Ans: P(8<= x <=10) = 1 - binomcdf(10,0.15,7)
= 0.0000086651...
Cheers,
Stan H.
Question 240758: how many different arrangements can be made using two of theletters of the word texas if no letter is to be used more than once?
Answer by edjones(3298) (Show Source):
Question 240767: six cards are to be drawn or delt, without replacement, from a deck of 52 cards. In how many ways can this be done if
A. all of the cards to be drawn are clubs?
B. All are to be jacks or queens?
C. None are to be aces?
D. All are to be spades or all not spades?
Answer by edjones(3298) (Show Source):
You can put this solution on YOUR website!nCr=n!/((n-r)!r!)
.
A) 13C6/52C6=1716 / 20358520=.00008429 (13 clubs)
.
B) 8C6/52C6=28 / 20358520=.000001375 (8 Jacks or Queens)
.
C) 48C6/52C6=12271512 / 20358520=.6028
.
D) I changed this question to what I thought you meant.
(13C6 + 39C6) / 52C6 = (1716 + 3262623)/20358520 =.1603
.
Ed
Question 240763: Powerball jackpot can be won by picking 5 numbers out of 59 numbers then also picking 1 powerball number from a seperate 39 numbers. The order of the first 5 numbers doesn't matter.
A. how many ways are there to pick the first five numbers?
B. How many ways are there to pick the powerball number?
C. How many ways are ther to pick all six numbers? ( the powerball number comes from a different set of numbers than the first five)
Answer by solver91311(5072) (Show Source):
Question 240685: If you roll two six-sided fair dice, what is the probability the you will roll a sum less than 9?
a. 1 : 8 b. 8 : 36 c. 13 : 18 d. 9 : 12
Answer by nyc_function(260) (Show Source):
Question 240666: Last q before my exam: Football fans claim that it takes 35 minutes to get out of the stadium after a game when over 30000 fans attend the game.
To determine if the complains are justified , a random sample of 100 fans is taken. The average time it takes these 100 fans to exit the parking is 27 minutes and sample standard deviatian is 7 minutes.
a) Estimate the mean departe time= 27 minutes?
b) Construct a 95% CI for the mean departure time and interpret the result?
Thx soo much for all your help
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!Football fans claim that it takes 35 minutes to get out of the stadium after a game when over 30000 fans attend the game.
To determine if the complains are justified , a random sample of 100 fans is taken. The average time it takes these 100 fans to exit the parking is 27 minutes and sample standard deviatian is 7 minutes.
a) Estimate the mean departe time = 27 minutes?
b) Construct a 95% CI for the mean departure time and interpret the result?
----------
sample mean: 27 min
------
standard error: SE = t*s/sqrt(n)
t = invT(0.025,99) = 1.9842
SE = 1.9842*7/sqrt(100) = 1.389
---
95% CI: 27-1.389 < u < 27+1.389
95% CI: 25.61 < u < 28.389
---------------------------------
Meaning: We are 95% confident the mean departure time is between
25.61 min and 28.389 min
==================================
Cheers,
Stan H.
Question 240627: if you toss a coin 3 times, what is the probability that you will
get 3 heads?
Answer by rapaljer(3622) (Show Source):
You can put this solution on YOUR website!The probablility of getting a head on a single coin toss is 1 out of 2, or 1/2.
If you toss the coin three successive times, each toss is INDEPENDENT of the previous tosses, so the probability is the PRODUCT of the probabilities, which is
Dr. Robert J. Rapalje, Retired
Question 240563: 65 mi/h = ? ft/min
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!65 mi/h = ? ft/min
--------------------------
[65 miles/ hour]
---
= (65*5280]/[1*1*60] ft/min
---
= (343200/60] ft/min
----
= 5720 ft/min
====================
Cheers,
Stan H.
Question 240502: A party host gives a door prize to one guest chosen at random. There are 47 men and 40 women at the party. What is the probability that the prize goes to a woman?
1. glenbenn50@yahoo.com
Answer by jim_thompson5910(13794) (Show Source):
You can put this solution on YOUR website!Since there are 47 men and 40 women, this means that there are 47+40=87 people total.
Probability prize goes to woman = # of women/# total = 40/87 = 0.4597
which is about a 45.97% chance
Question 240241: It is asserted that 80 percent of the cars approaching an individual toll booth in New Jersey are equipped with an E-ZPass transponder. Find the probability that in a sample of six cars:
What is the probability that 10 of the selected policyholders submitted a claim last year?
What is the probability that 10 or more of the selected policyholders submitted a claim last year?
What is the probability that more than 10 of the selected policyholders submitted a claim last year?
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!It is asserted that 80 percent of the cars approaching an individual toll booth in New Jersey are equipped with an E-ZPass transponder.
------------
Binomial with n=6 ; p = 0.8
-------------------------------------
Find the probability that in a sample of six cars:
What is the probability that 10 of the selected policyholders submitted a claim last year?
---
You cannot have 10 when the sample size is 6.
==================================================
Cheers,
Stan H.
What is the probability that 10 or more of the selected policyholders submitted a claim last year?
What is the probability that more than 10 of the selected policyholders submitted a claim last year?
Question 240434: A bag contains 6 marbles Three red marbles two blue marbles and one white marble If two marbles are selected at random determine the probability that both are red?
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!A bag contains 6 marbles Three red marbles two blue marbles and one white marble If two marbles are selected at random determine the probability that both are red?
-----------------------
ways to get 2 red: 3C2 = 3
---------------
ways to get 2 of any color: 6C2 = 15
----------------------------
P(both red) = 3/15 = 1/5
===============================
Cheers,
Stan H.
Question 240348: A pizza pub is open 7 days a week. The manager allows each of the three employees to select one day off a week. That is the probability all 3 select the same day given that all 3 select a day beginning with the same letter.
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!A pizza pub is open 7 days a week. The manager allows each of the three employees to select one day off a week. What is the probability all 3 select the same day given that all 3 select a day beginning with the same letter.
----------------
The only days beginning with the same letter are saturdaya and sunday.
----------------
P(all select the same | all select saturday or sunday)
----
P(all select saturday or sunday) = 2(1/2)^2 = 1/2
P(all select the same AND all select saturday or sunday) = (2/7)^2
-------------------
P(all select same | all select sat or sun) = (2/7)^2/(1/2) = 8/49
==================================
Cheers,
Stan H.
Question 240385: Normal distribution
A mean amount of profit per car sold for a sample of 25 dealers was 290$, with a standard deviation of 100$ Develop a 90% Confidence interval for th epopulation mean?
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!Normal distribution
A mean amount of profit per car sold for a sample of 25 dealers was 290$, with a standard deviation of 100$ Develop a 90% Confidence interval for th epopulation mean?
-------------
sample mean: $290
standard error: 1.645*100/sqrt(25) = 32.9
---------------------------
90% CI: 290-32.9 < u < 290+32.9
====================================
Cheers,
Stan H.
Question 240388: English people get a mean of 7 hrs sleep per night. A random sample of 36 students revealed the number of 6.5 hrs sleep per night. The standard deviation of the sample was 1.2 hrs. Is it reasonable to conclude that the students sleep less than english people. Set up your hyptheses and test it with 5% level of significance
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!English people get a mean of 7 hrs sleep per night. A random sample of 36 students revealed the number of 6.5 hrs sleep per night. The standard deviation of the sample was 1.2 hrs. Is it reasonable to conclude that the students sleep less than english people. Set up your hyptheses and test it with 5% level of significance
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Ho: u = 7
H1: u < 7
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critical value for alpha = 5%, df = 35 left-tail tes: t = -1.6896
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test stat: t(6.5)= (6.5-7)/[1.2/sqrt(36)] =-2.5
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Since the ts is in the reject interval, reject Ho.
You have some evidence the students require less sleep.
======================================
Cheers,
Stan H.
Question 240261: I have an idea where to start but would appreciate any guidance you can provide.
Four hundred people apply for three jobs. 110 of the applicants are women.
a) If three persons are selected at random, what is the probability that all are women? (Round the answer to six decimal places.)
(b) If three persons are selected at random, what is the probability that two are women? (Round the answer to six decimal places.)
(c) If three persons are selected at random, what is the probability that one is a woman? (Round the answer to six decimal places.)
(d) If three persons are selected at random, what is the probability that none is a woman? (Round the answer to six decimal places.)
(e) If you were an applicant, and the three selected people were not of your gender, should the above probabilities have an impact on your situation? Why?
(Yes),the probabilities indicate the presence of gender discrimination.
(No),because in the hiring process all outcomes are not equally likely.
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!Four hundred people apply for three jobs. 110 of the applicants are women.
a) If three persons are selected at random, what is the probability that all are women? (Round the answer to six decimal places.)
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ways to pick 3 women: 110C3
ways to pick 3 persons: 400C3
Prob. = 110C3/400C3 = 215820/10586800 = 0.020386..
==========================================================
(b) If three persons are selected at random, what is the probability that two are women? (Round the answer to six decimal places.)
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Binomial with n = 3, p= 110/400, x=2
Ans: 3C2(110/400)^2(290/400) or binompdf(3,110/400,2) = 0.164484...
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(c) If three persons are selected at random, what is the probability that one is a woman? (Round the answer to six decimal places.)
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Comment: Same process as "b".
Ans binompdf(3,110/400,1)= 0.433641..
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(d) If three persons are selected at random, what is the probability that none is a woman? (Round the answer to six decimal places.)
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Same as "b" and "c" with x = 0
Ans: 0.381078...
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(e) If you were an applicant, and the three selected people were not of your gender, should the above probabilities have an impact on your situation? Why?
(Yes),the probabilities indicate the presence of gender discrimination.
(No),because in the hiring process all outcomes are not equally likely.
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I'll leave that for you to decide.
Cheers,
Stan H.
Question 240141: Below are weights (in lbs) of glass discarded in one week by randomly selected households.
3.52 8.87 3.99 3.61 2.33 3.21 0.25 4.94
Construct a 95% confidence interval estimate of the mean weight of glass discarded by all households.
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!Below are weights (in lbs) of glass discarded in one week by randomly selected households.
3.52 8.87 3.99 3.61 2.33 3.21 0.25 4.94
Construct a 95% confidence interval estimate of the mean weight of glass discarded by all households.
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sample mean: 3.84
Sample std: 2.457
------------------------
standard error = ts/sqrt(n)
t = invT(0.025,7) = 2.3646
---
standard error = 2.3646*2.457/sqrt(7) = 2.196
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95% CI: 3.84-2.196 < u < 3.84+2.196
95% CI:1.644 < u < 6.036
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Cheers,
Stan H.
Question 240150: In a survey of 745 randomly selected adults, 589 said that it's morally wrong to not report all income on tax returns. Assume that you must conduct a survey to determine the mean income reported on tax returns and you have access to actual tax returns. How many randomly selected tax returns must you survey if you want to be 99% confident that the mean of the sample is within $500 of the true population? Assume the reported incomes have a standard deviation of $28,785.
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!In a survey of 745 randomly selected adults, 589 said that it's morally wrong to not report all income on tax returns. Assume that you must conduct a survey to determine the mean income reported on tax returns and you have access to actual tax returns. How many randomly selected tax returns must you survey if you want to be 99% confident that the mean of the sample is within $500 of the true population? Assume the reported incomes have a standard deviation of $28,785.
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n = [z*s/E]^2
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n = [2.576*28785/500]^2 = 21990.07
Round up to 21991
Cheers,
Stan H.
Question 240143: Do the following. (Round the answers to six decimal places.)
(a)Find the probability of being dealt an "aces over kings" full house (three aces and two kings).
(b)Find the probability of being dealt a full house.
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!Do the following. (Round the answers to six decimal places.)
(a)Find the probability of being dealt an "aces over kings" full house (three aces and two kings).
# of ways to pick three aces: 4C3 = 4
# of ways to pick 2 kings: 4C2 = 6
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Prob = (4*6)/52C5 = 0.000009234
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Probability of being dealt a full house.
Pick a card: 13 ways
Pick 3 of the card: 4 ways
Pick another card: 12 ways
Pick 2 of that card: 6
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Prob = {13*4*12*6)/52C5 = 3744/2598960 = 0.00144....
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Cheers,
Stan H.
Question 240123: If you want to estimate the mean amount of money spent by car owners on their last car purchase, how many car owners must you survey if you want 95% confidence that your sample mean is in error by no more than $750?(Assume that the standard deviation of amounts spent on car purchases is 14, 227.)
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!If you want to estimate the mean amount of money spent by car owners on their last car purchase, how many car owners must you survey if you want 95% confidence that your sample mean is in error by no more than $750?(Assume that the standard deviation of amounts spent on car purchases is 14, 227.)
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n = [z*s/E]^2
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n = [1.96*14227/750]^2
n = 1383.34
Rounding up: n = 1384
=========================
Cheers,
Stan H.
Question 240002: My favorite CD has 11 songs on it, and I have memorized the songs and the order in which they come. I recently loaded it onto my IPOD to listen to it. I wanted to listen to it in the right order, but I had accidentally set the player to shuffle-play the songs from that album in a random order. Assuming the "next to play" song was always equally likely among the choices not yet played, what is the probability that it will play the correct 5 songs in the correct order so that I don't notice the mistake (until after the fifth song)?
Answer by Alan3354(6097) (Show Source):
You can put this solution on YOUR website!If shuffle allows repeats, then it's 1 chance out of 11^5, = 1 of 161051.
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If it doesn't allow repeats until all songs are played (which is stated), the possible sequences are 11*10*9*8*7 = 55440 --> 1 chance in 55440.
Question 239886: from a deck of 52 ordinary playing cards, 2 cards are drawn without replacement. find the probability that both are hearts
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website! from a deck of 52 ordinary playing cards, 2 cards are drawn without replacement. find the probability that both are hearts
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# of ways to draw 2 hearts: 13C2 = 13*6 = 78
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# of possible outcomes: 52C2 = 51*26 = 1326
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Probability of drawing two hearts: 78/1326 = 0.059
======================================================
Cheers,
Stan H.
Question 240040: Find the dample size required to estimate the maen age of required drivers in the United States. Assume that we want 95% confidence that the sample mean is within 1/2 yera of the true mean asge of the population. Also assume that the standard deviation of the population is known to be 12 years.
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!Find the sample size required to estimate the mean age of required drivers in the United States. Assume that we want 95% confidence that the sample mean is within 1/2 year of the true mean age of the population. Also assume that the standard deviation of the population is known to be 12 years.
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n = [z*s/E]^2
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n = [1.96*12/(1/2)]^2 = 2212.76
Rounding up n = 2213
=======================================
Cheers,
Stan H.
Question 239821: There were 2 colas, 1 ginger ale, 5 cherry sodas, and 4 root beers in the cooler. What is the probability of pulling out a cola?
Answer by checkley77(7072) (Show Source):
Question 239789: What is the probability that a randomly selected two-digit number is divisible by at least one of its digits? Express your solution as a simplified common fraction.
Answer by vleith(1977) (Show Source):
You can put this solution on YOUR website!strictly speaking, unless the number is 00, every number will be divisible by at least one digit. May not be divisible without a remainder, but that was not the question. I say 100%
Question 239740: An initial deposit of $15,000 is made into an account and monthly deposits of $5,000 are made thereafter. Interest is pays at 8% per year. Set up a table to show how much money is in the account at the end of each month for the first 5 months
Answer by Theo(675)  (Show Source):
You can put this solution on YOUR website!Interest rate per month is calculated as .08/12 + 1 = 1.0066666667
In the beginning you deposit 15,000
Interest is compounded monthly has been assumed though not stated.
Payments are made at the end of each month has been assumed though not stated.
The money invested is multiplied by (1 plus the interest rate) and then the payment is added.
Here's how it works.
x = (1 + i) = 1.0066666667
This is rounded here but not in the calculator.
All the numbers below are truncated to the nearest dollar but the complete number is stored in the calculator and used in the calculations.
you invest 15000.
at the end of the first month you have:
15000 * x = 15100 + 5000 = 20100
at the end of the second month you have:
20100 * x = 20234 + 5000 = 25234
at the end of the third month you have:
25234 * x = 25402 + 5000 = 30402
at the end of the fourth month you have:
30402 * x = 30604 + 5000 = 35604
at the end of the fifth month you have:
35604 * x = 35842 + 5000 = 40842
The complete number is:
40842.27423
That's $40,842.27
I confirmed using a financial calculator that this number is correct.
Question 239414: Assume the likelihood that any flight on Northwest Airlines arrives within 15 minutes of the scheduled time is .90. We select four flights from yesterday for study.
1. What is the likelihood all four of the selected flights arrived within 15 minutes of the scheduled time?
2. What is the likelihood that none of the selected flights arrived within 15 minutes of the scheduled time?
3. What is the likelihood at least one of the selected flights did not arrive within 15 minutes of the scheduled time?
Answer by solver91311(5072) (Show Source):
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