Questions on Algebra: Probability and statistics answered by real tutors!

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At Ajax Spring Water, a half-liter bottle of soft drink is supposed to contain a mean of 520 ml. The filling process follows a normal distribution with a known process standard deviation of 4 ml.
(a) Which sampling distribution would you use if random samples of 10 bottles are to be weighed? Why?
Mean of the sample means: 520 ml
St. Dev. of the sample means: 4/sqrt(10)
Both are prescribed by the Central Limit Theorem
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(b) Set up hypotheses and a two-tailed decision rule for the correct mean using the 5 percent level of significance.
Ho: mean = 520
Ha: mean is not 520
Reject Ho if z(mean of the sample) is > 1.96 or <-1.96
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(c) If a sample of 16 bottles shows a mean fill of 515 ml, does this contradict the hypothesis that the true mean is 520 ml?
z(515) = (515-520)/[4/sqrt(10)]= -3.9528...
Yes, it is evidence the mean is not 520.
============================================
Cheers,
Stan H.

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P=2L+2W
A=L*W
46=2L+2W
126=L*W OR L=126/W
46=2(126/W)+2W
46=252/W+2W
46=(252+2W*W)/W
46=(252+2W^2)/W
46W=252+2W^2
2W^2-46W+252=0
2(W^2-23W+126)=0
2(W-14)(W-9)=0
W-14=0
W=14 & L=9
W-9=0
W=9 & L=14
PROOF:
46=2*14+2*9
46=28+18
46=46
14*9=126
126=126

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You randomly select one card from a deck of 52 cards. Find the probabilty of selecting an ace or a 8?
--------------------------
# of ways to succeed: 8
# of possible outcomes: 52
P(ace or 8) = 8/52 = 2/13 = 0.153846...
Cheers,
Stan H.

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there are 4 aces therefore the probability is 4/52=1/13
there are 4 8's therefore the probability would be the same 1/13
:
if aces and 8's together it would be 8/52=2/13

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there are 3 numbers less than 4------->1,2,3 so 3/6 =1/2 so 1 out of 2 chance that you would get a number less than 4 on one roll of the die

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seperately there are 4 aces and 4 6's therefore the probability is the same 4/52=1/13 of choosing an ace or a 6
:
probability of choosing an ace or 6 on the same draw would be 8/52=2/13

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There are 26 black cards (13 spade and 13 club cards) to choose from. So this means that n=26. Since the amount per hand is 5 cards, this means that r=5




Since order does not matter, we must use the combination formula:


Start with the given formula



Plug in and



Subtract to get 21


Expand 26!



Expand 21!




Cancel



Simplify


Expand 5!




Multiply 26*25*24*23*22 to get 7,893,600



Multiply 5*4*3*2*1 to get 120



Now divide



So 26 choose 5 (where order doesn't matter) yields 65,780 unique combinations


So there are 65,780 possible ways to choose all black cards.

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Bill has been given a 20 question mulitple choice test. He hasnt attended class recently, and there are 5 answers to each question, but only one is correct...what is the probability that he will answer more than five questions on the test correctly?..explain your answer
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No, that is not the correct answer.
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It is a binomial problem with n= 20, p = 1/5; x>5
n is the number of trials
p is the probability of success on each trial
x is the number of successes
---------------------------------
P(x>5) = 1 - binomcdf(20,(1/5),4) = 0.37035...
This, from using a TI-83 calculator.
=======================================
Cheers,
Stan H.

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Well, I got a different number for the mean on the Goliath group:
Goliath mean = 1282.
On the other two groups, my answer was the same as yours.
But the question seems to be "Are the means the same for each group" and the answer is clearly, no they are not!

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Region 1995 1997 2000
Northe 51.4 51.6 53.6
Midwest 61.8 62.5 64.4
South 91.8 94.2 100.2
West 57.7 59.4 63.2
West in 1997:
59.4/267.7=.2219 or 22.19% probability they lived in the West.
Midwest in 1995:
61.8/262.7.2352 or 23.52%
North or Midwest in 1997:
114.1/267.7=.4262 or 42.62%.
Not from the South in 2000:
(281.4-100.2)/281.4=181.2/281.4=.6439 or 64.39%

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.3/1=.3 or 30% chance he'll get a hit.

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a) Find P( 8,5)

Start with 8 and multiply by 1 less each time until 
you have 5 factors,

like this:

b) Find C( 8,5)

Cancel the 's and the 's:



Cancel the  and the 





Edwin

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5/2=(y+8)/(y-4)
Cross-multiply to get:
5(y-4) = 2(y+8)
5y-20 = 2y+16
3y = 36
y = 12
==================
Cheers,
Stan H.

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Warning: Stanbon's
(b) and (c) are incorrect.
Edwin's solution:
A box contains 74 brass washer, 86 steel washers and 40 aluminium washers. Three washers are drawn at random from the box without replacement.
a) Determine the probability that all three are steel washers.

There are 86 steel washers.
There are  or  washers.  So the
desired probability is

 86 choose 3
------------
200 choose 3

b) Determine the probability that there are no aluminium washers drawn, when three washers are drawn at random from the box without replacement.

There are 74+86 or 160 non-aluminum washers.  (This can also be calculated
by subtracting the 40 aluminum washers from the total 200, i.e., 200-40=160.)

 160 choose 3
--------------  =
 200 choose 3

c) Find the probability that there are two brass washers and either a steel or
an aluminium washer when three are drawn at random, without replacement.

There are 74 brass washers.
There are 86+40 or 126 washers that are either steel or aluminum.
So the desired probability is

 (74 choose 2) + (126 choose 1)
--------------------------------  =
       200 choose 3



Edwin

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A box contains 74 brass washer, 86 steel washers and 40 aluminium washers. Three washers are drawn at random from the box without replacement.
a) Determine the probability that all three are steel washers.
# of ways to succeed = 86C3
# of possible outcomes = 200C3
P(success) = 86C3/200C3 = 0.0779...
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b) Determine the probability that there are no aluminium washers drawn, when three washers are drawn at random from the box without replacement.
# of ways to succed = 114C3/200C2
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c) Find the probability that there are two brass washers and either a steel or an aluminium washer when three are drawn at random, without replacement.
# of ways to succeed:
Draw to brass washers in 74C2 ways
Draw a steel or aluminum washer 126 ways
Total # of ways to succeed 74C2*126
Prob of success: [74C2*126]/[200C3] = 0.25912
=========================
Cheers,
Stan H.

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(a)A U C is
{2,4,5,8,9}ie its the set of all elements that are in A and B
(b)Aintersection B becomes
{8,9} ie its the set of all alike elements in Aand B
(c)your question seems to be incomplete since there should be a set of elements say A or B that you should complement with U.
Assuming the question was A'(A complement)then we would have;
{1,3,5,6,7,10,11,12,13,14,15,16,17}ie all elements in U that aren't in A
(d)B U A U C
first find B U A which is {2,4,5,8,9}
then(B U A) U C which becomes {2,4,5,8,9,11,12,17}and hence the answer.

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Let f(n) = the number of zeros at the end of n! when n! is multiplied out.
Find:
f(12) = 12! = 12*11*10*9*8*7*6*5*4*3*2*1
Find the number of times 10 is a factor.
--------------------------
Change factors to prime number form:
f(12) = 2^2*3 * 11 *2*5*3^2*2^3 *7*2*3 *5* 2^2 *3*2*1
f(12) = 2^10 * 3^4 * 5^2 * 7
The prime-factor form has two tens which is 2^2*5^2, so only two zeroes at the end
of 12!
---------------

f(34)= (2*17)*(2^4)*(3*5)*(2*7)*13*(12!)
12! contributes 2 zeroes
The other factors contribute 2*5 or one zero
So, f(34) has a total of three zeroes at the tail end.
=================================================
Cheers,
Stan H.

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Will someone please help me.
1. A red die and a blue die are tossed. What is the probability that the red die shows an odd number and the blue die shows a 1 or 2?
---
P(red odd) = 3/6 = 1/2
P(blue 1 or 2) = 2/6 = 1/3
--
P(red odd AND blue 1or 2) = (1/2)(1/3) = 1/6
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2. From a group of 6 men and 4 women, a committee of 3 is to be selected at random. Find P(at least 2 women).
P(2 or 3 women) = P(2 women/1 man) + P(3 women/0 men)
But P(2 women and 1 man) = [4C2*6C1/10C3] = [6*6/120] = 3/10
And P(3 women/0 men) = 4C3/10C3 = 4/120 = 1/30
Therefore, Answer = (3/10) + (1/30) = 1/3
=========================================================================
Cheers,
Stan H.

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1. The time a group of high school students arrive home from school each day was found to be normally distributed. The mean time was 3:15 PM and the times had a standard deviation of 15 minutes. What is the probability that a student chosen at random arrives home from school before 2:30 PM?
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Find the z-score of 2:30 which is 45 minutes before 3:15.
z(2:30) = (2:30 - 3:15)/15 = -45/15 = -3
P(x < 2:30) = P(z<-3) = 0.00135
==============================================
2. During a clothing sale, 1/4 of the store merchandise is reduced in price. Find the probability that 3 of 5 randomly-selected shirts have reduced prices.
I have to show all my work. Will you please help me.
---
This is a binomial problem.
n = 5, p = 1/4, x = 3
P(x=3) = 5C3(1/4)^3*(3/4)^2 = 10*(9/4^5) = 0.08789..
=================================
Cheers,
Stan H.

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Probability=success/total outcomes.
P=(hearts+diamonds+spade king+club king)/52
P=(13+13+1+1)/52
P=28/52
P=7/13 or .538 or 53.8% success in drawing a red card or a king.

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All possible numbers : 1,2,3,4,5,6
Even numbers: 2,4,6
.
.
.
Number of all possible numbers = 6
Number of even numbers = 3
.
.
.
Probability of even number = (Number of even numbers)/(Number of all possible numbers)
Probability of even numbers =3/6=1/2

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X = no. of aces drawn which can either be 0, 1, 2, or 3.

So make this chart:



Now we must calculate the probabilities to go in
the bottom row of the chart.

We first calculate the probability of getting 0 aces.
This means we draw 3 non-aces.  There are 48 non-aces.
So the number of ways to choose three non-aces is


The denominator is 

So the probability of getting 0 aces is 

Next we calculate the probability of getting 1 ace.
This means we draw 2 non-aces and 1 ace. There are 
48 non-aces. So the number of ways to choose two 
non-aces is  and the number of ways to
choose the ace is  ways.  That's 
 ways.

The denominator again is 

So the probability of getting exactly 1 ace is 


Next we calculate the probability of getting 
exactly 2 aces. This means we draw 1 non-ace and 
2 ace3. There are 48 non-aces. So the number of 
ways to choose one non-aces is  and 
the number of ways to choose 2 aces is  
ways.  That's 
 ways.

The denominator is as before 

So the probability of getting exactly 2 aces is 


Finally we calculate the probability of getting 3 aces.
The number of ways to choose 3 aces is  ways.  

The denominator is 

So the probability of getting exactly 3 aces is 


So we fill in the chart:

 

Now to find the expectation, we add the products of each
value of x times its probability:



That rounds to choice C, 

Edwin

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sam can do 1/4 job per day
lisa can do 1/6 job per day
tom can do 1/2 job per day
so together
:t/4+t/6+t/2=1(job)
multiply terms by 12
:
3t+2t+6t=12
:
11t=12
:
days to do the job together

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It's 8*7*6*5*4, = 6720 possibilities.
But since the order in which they're chosen doesn't matter, you have to divide by 5*4*3*2*1, which is 120. See that? It doesn't matter is person A is chosen 1st, 2nd or 5th. In math terms, that is "without order."
The result is 6720/120 = 560.
There are 560 possibilities.

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1/2(1/2)=1/4 chance of getting at least one head

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1. 8/16(8/16)=1/4 probability
2. 8/16(7/15)=7/30 probability

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In a single die, the chances on any number 1 - 6 are equally probable.
Each face has a 1/6 chance of being rolled.
The probability of rolling a 1,2,3 or 4 is the sum of the individual odds.
1/6 + 1/6 + 1/6 + 1/6 = 4/6 = 2/3

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{2},{5},{8}
{2,5},{2,8},{5,8}
{2,5,8}

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Think of it this way.
How many choices do you have for the first seat? 8 people, so 8 possibilities.
How many are left after the first person is in seat 1? 7 people
How many choices do you have for the second seat? 7 people, so 7 possibilities.
Keep doing this for each seat.
End result - 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 8!
have a look at this video for more examples of similar stuff
http://collabvsl.wetpaint.com/page/Basic+Probability

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1/(50*49*48*47*46*45*44*43)
1/2.16469*10^13
4.6196*10^-14 ans.

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To find the answer you need to add together the numerator(top number) and the denomenator (bottom number). So the answer is 9+8= 17, so the probablity is 9:17
Hope this helped!!:)

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White seeds:
7/17*6/16*5/15=210/4080=7/136 chances (or 5.147%) of the 3 being white seeds.
6/17*5/16*4/15=120/4080=4/136 chances (or 2.941%) of the 3 being yellow.
4/17*3/16*2/15=24/4080=8/1360 chances (or 0.588%) of the 3 being red.
7/136+4/136+8/1360
(70+40+8)/1360
118/1360=.086765 or 8.6765% success in all 3 being the same color.

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8*6*5=240 different mixes are possible.
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10*9*8*7=5,040 ways 4 words out of 10 can be listed on a 4 question test.

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You forgot the "r!" in the denominator.
C(n,r) = n!/(r!(n-r)!)
C(12,10) = 12!/(10!(12-10)!)
C(12,10) = 12!/(10!(2)!)
C(12,10) = (11*12)/(1*2)
C(12,10) = 132/2
C(12,10) = 66

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There are 8 possibilities for the eldest three children with them listed in order from oldest to youngest.
BBB
BBG
BGB
BGG
GBB
GBG
GGB
GGG
Only one of the eight possibilities has three boys so the probability of three boys is 1 / 8

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If r = 1.00, then this means that there is a perfect positive correlation. So as values in X increase, values in Y also increase (at the same rate as X)

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The range of values for a coefficient of correlation is -1 to 1

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We don't have your book. You need to post the problem.

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If the letters A, B, C, D, E, and F are used in a five-letter code, how many different codes are possible if repetitions are not permitted?
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For the 1st letter, you can choose one of 6. For the 2nd, one of the remaining 5, then 4, 3, 2. So the possibilities are 6*5*4*3*2*1, or 6! (6 factorial).
That's 720.

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An automobile dealer wants to construct a pie graph to represent types of cars that were sold in Jule. He sold 72 cars, 16 of which were convertibles. The convertibles will represent how many degrees in the circle?
.
The entire circle represents 72
and the "slice of pie" is represented by a portion:
16/72 * 100
= 0.22 * 100
= 22.222%
.
22.222% of 360 degrees
= .22222 * 360
= 80 degrees

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You didn't supply any numbers but here's, in general, how to find the probabilities.
(a) P(female is finance major)=number of female finance majors/total females.
(b) P(finance major is female)=number of female finance majors/total number of finance majors.

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You are talking about a "combination".
.
C(n,r) = n!/r!(n-r)!
.
C(18,1) = 18!/1!(18-1)!
C(18,1) = 18!/(17)!
C(18,1) = 18

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I believe you are referring to a "permutation".
.
P(n,r) = n!/(n-r)!
.
So, if you have
P(7,2) = 7!/(7-2)!
P(7,2) = 7!/5!
P(7,2) = (1*2*3*4*5*6*7)/(1*2*3*4*5)
P(7,2) = (6*7)
P(7,2) = 42

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How may FIVE card poker hands are there?
----------------------
The 1st card can be one of 52, the 2nd will be 1 out of 51, etc, so we get 52*51*50*49*48 = 311,875,200.
But, drawing 3 4 5 6 7 is the same hand as drawing 3 7 5 6 4 (the same 5 cards drawn in a different order), so we have to divide by 5*4*3*2*1 = 5! = 120.
That gives the number of different hands = 2,598,960.
-------------------------
This is based on the assumption that the same card in a different suit makes a different hand. For example, a full house with 3 Kings, the 10 of Hearts and the 10 of Spades, is viewed as different from a hand with any of the 5 cards exchanged for the same card of a different suit. It has no effect on where it's a winning or losing hand in poker.
Whether this assumption is valid is not specified.

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Half the cards in a 52 card deck are red (either hearts or diamonds)
There are 26 choices for the 1st card
There are 25 choices for the 2nd card
There are 24 choices for the 3rd card
There are 23 choices for the 4th card
There are 22 choices for the 5th card
Note that each time you draw a card, you reduce the
number of available red cards by 1
Now how many possible hands are there?
is the answer