When performing a hypothesis test for the ratio of two population va.pdf
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When performing a hypothesis test for the ratio of two population variances, the upper critical F value is denoted FR. The lower critical F value, FL , can be found as follows: interchange the degrees of freedom, and then take the reciprocal of the resulting F value found in table A-5. FR can be denoted F?/2 and FL can be denoted F1-?/2. Find the critical values FL and FR for a two-tailed hypothesis test based on the following values: n1 = 9, n2 = 7, ? = 0.05 a. 0.2411, 4.1468 b. 0.3931, 4.1468 c. 0.2150, 4.8232 d. 0.2150, 5.5996 Solution c. 0.2150, 4.8232.
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When should we use partial derivatives and when should we use ordinary derivatives?
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Thank you.
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Three office workers take a coffee break. They chose one of their nu.pdf
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\"When I was young, people called me a gambler. As the scale of my operations
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tasty, you want IT so much. From time to time you can see a senior monkey has a bite at it, he
seems so happy. So you work harder and harder to get to the Coconut. As you climb on the
hierarchy tree, pushing other monkeys off on the way, you get closer and closer. Sometimes they
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Solution
We will need to multiply the probability of the event occurring times the probability
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Solution
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Solution
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Solution
APPLY MESH ANALYSIS IN LOOPS ,
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\"When I was young, people called me a gambler. As the scale of my operations
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playing hard- and a total burn out have contributed to turn me into a wanker. Literally. Well...
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[1/(12*12*12)] *12 =1/144.
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Average value of each throw = 3.5 Average value of 3 throws = 10.5 The nearer the
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Three coins are tossed, using ordered triples, do the following
NOTE:eg- if an outcome was 1st coin-heads, 2nd coin-tails and 3rd coin= heads then the ordered
triple would be (h, t, h)
a) write the set S for the sample space:
b)write the set E which is the event that exactly 1 head comes up
c) find p(E)
d) find O(E)
e) write the set F which is the event that at least 2 tails come up
f) find p(F)
g) find O(F)
Solution
a)
sample space S={(h,h,h),(t,t,t),(h,t,h),(t,h,h),(h,h,t),(t,t,h),(t,h,t),(h,t,t)}
b)
the set E which is the event that exactly 1 head comes up={(t,t,h),(t,h,t),(h,t,t)}
c)
p(E)=n(E)/n(S)=3/8
d)
O(E)=1-P(E)=1-3/8=5/8
e)
the set F which is the event that at least 2 tails come up={(t,t,h),(t,h,t),(h,t,t),(t,t,t)}
f)
p(F)=n(F)/n(S)=4/8=1/2
g)
O(F)=1-p(F)=1/2.
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Solution
The probability of drawing one spade is 13/52 or 1/4. The odds of drawing another
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11/50), you get 132/10200. This simplifies to 11/850..
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Tommy\'s age, Roland was one year older than the triple of Tommy\'s age (at that time).
When Tommy will be of Susie\'s age, then Roland will be 10 years younger than the double of
Susie\'s age(at that time). How old will Roland be, when Tommy\'s age will be the 3/4 of Susie\'s
age (of that time)?
Solution
let age of susie be s, tommy be t and roland be r. s+t+r = 42...eqn1 susie would be
tommy\'s age s-t yrs back age of tommy then would be t - (s-t) = 2t-s yrs rolands age s-t yrs back
= r-(s-t) = r+t-s now r+t-s = 1+3(2t-s) =>5t - 2s -r = -1 ....eqn2 tommy will be susie\'s age (s-t)
yrs later age of susie would be s+(s-t) = 2s-t age of roland would be r+(s-t) = r+s-t r+s-t = 2(2s-t)
-10 =>3s-r -t = 10...eqn3 solving eqn1, 2, 3 we get r = 20 s = 13 t = 9. 3 yrs later tommy\'s age
would be 3/4 times that of susie. roland\'s age = 20+3 = 23yrs.
Three office workers take a coffee break. They chose one of their nu.pdf
Three office workers take a coffee break. They chose one of their numbers at random to pay for
the coffee as follows. All three flip a coin simultaneously and the one having a different outcome
pays for the coffee. If all coins come up the same, they flip the coins again. How long does it
take to determine who pays for the coffee?
Solution
\"When I was young, people called me a gambler. As the scale of my operations
increased I became known as a speculator. Now I am called a banker. But I have been doing the
same thing all the time\" -- Sir Ernest Cassel I\'m an Investment Wanker. I used to be a banker
but more than a decade in the Orc Race, tens of millions earned and lost for the bank, having to
survive several boom & bust cycles, thousands of days and nights spent working hard -and
playing hard- and a total burn out have contributed to turn me into a wanker. Literally. Well...
almost literally, let\'s say that today I could wank all day at the bank but I prefer to spend my
time on matters as futile as blogging, kung fu, History, finding out how I can become Superman,
learning new skills and Speculation. My transformation that began in early 2008 was at first kind
of forced. But since I\'ve understood the way the Orc Race really works and today for nothing on
Earth would I willingly run it back again. Running the Orc Race is like monkeys climbing on a
coconut tree. You shit on the monkeys below and when you look above all you can see is
assholes. At the very very top of the tree you can see IT, the Coconut. IT looks delicious, so
tasty, you want IT so much. From time to time you can see a senior monkey has a bite at it, he
seems so happy. So you work harder and harder to get to the Coconut. As you climb on the
hierarchy tree, pushing other monkeys off on the way, you get closer and closer. Sometimes they
even let you touch IT. In some very rare cases, when you really have outperformed and killed a
lot of other monkeys, you even can have a tiny tiny piece of IT. The only thing you want then is
the whole Coconut, so you work harder and harder, kill more and more monkeys until the day
you\'re the one who falls from the tree pushed off by a younger and stronger monkey. But here\'s
the sad truth: no monkey can have the whole Coconut, because it belongs to the owner of the
tree. The irony is, except for the case of a few billionaires who have been both ultra competent
and extremely lucky in their career, most of the time the Coconut tree belongs to millions of
shareholders, that\'s other monkeys climbing on a tree. In other words, nobody gets the Coconut..
Three independent events have probabilities 14,13 and 12. What is.pdf
Three independent events have probabilities 1/4,1/3 and 1/2. What is the probability that exactly
1 event will occur?
Solution
We will need to multiply the probability of the event occurring times the probability
of the events not occurring. For this we will need the complement of the probability of an event
occurring which is simply P(not occurring) = 1 - P(occurring) Probability of the only the first
event occurring: (1/4)(2/3)(1/2) = 1/12 Probability of the only the second event occurring:
(3/4)(1/3)(1/2) = 1/8 Probability of the only the third event occurring: (3/4)(2/3)(1/2) = 1/4
Probability of only one event occurring = 1/12 + 1/8 + 1/4 = 11/24.
Three points X,Y,Z are selected at random on a line segment L. What .pdf
Three points X,Y,Z are selected at random on a line segment L. What is the probability that Y
lies between X and Z?
Please show all steps of proof or work. THANKS!
Solution
sample space =xyz,xzy,yxz,yzx,zxy,zyx probability that Y lies between X and
Z=2/6=1/3.
Three points X1, X2, and X3 are selected at random from an interva.pdf
Three points X1, X2, and X3 are selected at
random from an interval of length L. Find the probability that none of the
three points are within a distance a of the left end where a < L.
Solution
P(a point lies within a distance \'a\' from the left end) = a/L
P(a point doesn\'t within a distance \'a\' from the left end) = (L-a)/L
P(none)= [(L-a)/L]^3.
Three Phase Circuits Why are we multiplying by 5 to the rms voltage.pdf
Three Phase Circuits: Why are we multiplying by 5 to the rms voltage source value?
Solution
APPLY MESH ANALYSIS IN LOOPS ,
THEN FIND CURRENTS
SO THE CURRENT IN THE CIRCUIT IS 5 A .
SO WE ARE MULTIPLYING WITH 5 A TO FIND POWER .
the magnitude of current in the circuit is 5 A , so to find.
Three office workers take a coffee break. They choose one of their n.pdf
Three office workers take a coffee break. They choose one of their numbers at random to pay for
the coffee as follows. All three flip a coin simultaneously and the one having the different
outcome pays for the coffee. If all coins come up the same, then they will flip the coins again.
How long does it take to determine who will pay for the coffee?
Solution
\"When I was young, people called me a gambler. As the scale of my operations
increased I became known as a speculator. Now I am called a banker. But I have been doing the
same thing all the time\" -- Sir Ernest Cassel I\'m an Investment Wanker. I used to be a banker
but more than a decade in the Orc Race, tens of millions earned and lost for the bank, having to
survive several boom & bust cycles, thousands of days and nights spent working hard -and
playing hard- and a total burn out have contributed to turn me into a wanker. Literally. Well...
almost literally, let\'s say that today I could wank all day at the bank but I prefer to spend my
time on matters as futile as blogging, kung fu, History, finding out how I can become Superman,
learning new skills and Speculation. My transformation that began in early 2008 was at first kind
of forced. But since I\'ve understood the way the Orc Race really works and today for nothing on
Earth would I willingly run it back again. Running the Orc Race is like monkeys climbing on a
coconut tree. You shit on the monkeys below and when you look above all you can see is
assholes. At the very very top of the tree you can see IT, the Coconut. IT looks delicious, so
tasty, you want IT so much. From time to time you can see a senior monkey has a bite at it, he
seems so happy. So you work harder and harder to get to the Coconut. As you climb on the
hierarchy tree, pushing other monkeys off on the way, you get closer and closer. Sometimes they
even let you touch IT. In some very rare cases, when you really have outperformed and killed a
lot of other monkeys, you even can have a tiny tiny piece of IT. The only thing you want then is
the whole Coconut, so you work harder and harder, kill more and more monkeys until the day
you\'re the one who falls from the tree pushed off by a younger and stronger monkey. But here\'s
the sad truth: no monkey can have the whole Coconut, because it belongs to the owner of the
tree. The irony is, except for the case of a few billionaires who have been both ultra competent
and extremely lucky in their career, most of the time the Coconut tree belongs to millions of
shareholders, that\'s other monkeys climbing on a tree. In other words, nobody gets the Coconut..
Three people who do not know each other are standing in line at a mo.pdf
Three people who do not know each other are standing in line at a movie theatre waiting for the
box office to open. What is the probability that they were all born in the same month?
Solution
[1/(12*12*12)] *12 =1/144.
Three people are working independently at solving a statistical prob.pdf
Three people are working independently at solving a statistical problem. The respective
probabilities that they will solve the problem are 1/4, 1/2, and 1/3.
What is the probability that the problem will be solved?
Solution
P(solved)=(1/4)*(1/2)*(1/3)=1/24.
Three dice are tossed. Which is the more likely event That the su.pdf
Three dice are tossed. Which is the more likely event? That the sum of the faces is 9 or the sum
of the faces is 10? Justify your answer.
Solution
Average value of each throw = 3.5 Average value of 3 throws = 10.5 The nearer the
sum is to the average, the more is its probability. So, P(10)>P(9).
Three coins are tossed, using ordered triples, do the followingNOT.pdf
Three coins are tossed, using ordered triples, do the following
NOTE:eg- if an outcome was 1st coin-heads, 2nd coin-tails and 3rd coin= heads then the ordered
triple would be (h, t, h)
a) write the set S for the sample space:
b)write the set E which is the event that exactly 1 head comes up
c) find p(E)
d) find O(E)
e) write the set F which is the event that at least 2 tails come up
f) find p(F)
g) find O(F)
Solution
a)
sample space S={(h,h,h),(t,t,t),(h,t,h),(t,h,h),(h,h,t),(t,t,h),(t,h,t),(h,t,t)}
b)
the set E which is the event that exactly 1 head comes up={(t,t,h),(t,h,t),(h,t,t)}
c)
p(E)=n(E)/n(S)=3/8
d)
O(E)=1-P(E)=1-3/8=5/8
e)
the set F which is the event that at least 2 tails come up={(t,t,h),(t,h,t),(h,t,t),(t,t,t)}
f)
p(F)=n(F)/n(S)=4/8=1/2
g)
O(F)=1-p(F)=1/2.
Three cards are dealt from a standard 52-card deck. 1 A. What is t.pdf
Three cards are dealt from a standard 52-card deck.
1 A. What is the probability that all three cards are spades?
Solution
The probability of drawing one spade is 13/52 or 1/4. The odds of drawing another
spade is 12/51. The odds of drawing a third spade is 11/50. If you multiply (1/4 * 12/51 *
11/50), you get 132/10200. This simplifies to 11/850..
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