assume normal distribution, calculate standard deviation, 170 is 2 sd's from the mean value of 150.
If X is the distribution of awarded cash of one roll, E[X] = 3, Var(X) = 2, Std(X) = sqrt(2). Then for Y = 50X, we have E[Y] = 150, Var(Y) = 5000, Std(Y) = 50*sqrt(2) Then we can calculate the Z score of 170, and use the normal distribution graph to read the one tail probability, and calculate accordingly.
The mean and standard deviation (population) of 1 roll is 3 and sqrt(2). Multiplying by 50 gives a mean of 150 and std dev of 50*sqrt(2). Now you need a truncated normal distribution. What you could do is enforce that the integral of the normal distribution from 50 to 250 (min and max winnings) = 1, and the integrated the a scaled normal distribution from 170 to 250 for the probability of winning $170 or more.
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You want to be careful to articulate the assumptions you are making. Since we are taking the sum of a large number of independent, identically distributed random variables (the 50 different games we will play), our total winnings will be approximately normally distributed, by the central limit theorem. The mean and variance for this will be the number of times we play, 50, times the mean and variance of a single game. The mean of a single game is 3. The variance of a single game is 2. So our total winnings will be approximately normally distributed with mean 150 (3 * 50) and variance 100 (2 * 50). We can use the 68-95-99.7 rule, along with the fact that 170 is 2 standard deviations from the mean of 150, to see that the probability that we win more than $170 is approximately (1 - 0.95) / 2 = 0.025. The answer is 2.5%.
If you were in a room with three light switches, and you could not see out of the room and you could only leave the room once. The three corresponding light bulbs are in another room. How do you tell which light switches belong to which bulbs?
Assumptions: You have three inputs: X, Y, Z with two possible states (0,1). You have three outputs: A,B,C with two possible states (0,1). The inputs and outputs are independent of each other, but one input should be matched to one output. For example, if X --> A, then X(1) --> A(1). Note: we don't actually know whether the normal state of the switch is open or closed. For our purposes 0 is normal state, 1 is abnormal state. You have one chance to observe the outputs. You could flip one of the switches and observe, but this will only tell you about one of the connections and leave two unknowns. You could flip two switches, but this still leaves two unknowns. You could flip all three, but this leaves three unknowns. You could flip none of them, but that leaves three unknowns. Another possibility is to McGyver the light switches by placing three resistors of different values in series with the switch, verifying the switch is closed, and then observe the relative brightnesses of the bulbs. More creatively, you could use your one trip to leave the room to find the documentation and/or the persons responsible for this cruel experiment. I am sure this is wrong, but I was never a math or CS major. =D
(I mean you would put one resistor in series with each of the three switches, obviously; the greatest value resistor would result in the most dimming).
My answer to this is: Assign a value to each light switch (A,B,C) Start a timer for 3 minutes Flip switch A and B on After the 3 minutes is up, flip B off again. Go into the other room. The lighting bulb belongs to switch A The the other two light bulbs (one will be hot and one will be cold) The hot bulb belongs to switch B The cold bulb belongs to switch C.
you need 3 candle sticks, measure 30 min out of first candle and this allows to measure 15 min from the second candle , when first and second candles are done you have 3rd candle left to measure 45min
Actually, 2 sticks is enough. Light both ends of Stick A and one end of Stick B at the same time, when A runs out, it would be 30 minutes after, and B is half done. Then light both ends of Stick B and after another 15 minutes, B will be done. We get a total of 30+15 = 45 minutes.
I agree only 2 sticks are needed as outlined above. Does anyone have information regarding the type of questions in the Logic and reasoning IQ tests? Is it recognizing patterns? Like triangle circle square triangle circle what comes next? (This being a very basic question of course)
First of all, Now a days there is no need to advertisement for facebook because Facebook is totally up in the market but for improve the facebook according to my view first step i want create some division for adult and child ..i have one idea for child to grow more in facebook..Child(age <15) have to just login with their id and use different application and play different mind games and participate in different activity..they have some criteria in their profile like they can not see adult photoes, Adult games or etc.. THank u
Break the problem down. Best to start by eliminating other options to leave you with the main possibilities.
There are 400 million people in the U.S Assuming there is 5 people to one house there is 80million homes in. 50 States, 150 major cities. Outo f the 80 million homes we know that 35% are registered democratics out of the35% democrats only 10% are extreme liberals so 10% of 30 million is 3 million homes in the U.S. will have blue doors.
I would call a coffee break and address each of the colleagues separately to try to determine the root of the problem. I would then set up a resolution discussion in a bid to resolve it.
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There are 4.83 million people in Ireland I would assume that 50% of them are males I would assume that an average male goes to the barber once every month I would assume that each barber can deliver services to 3 people per hour I would assume that each barber works an average of 30*4*11 hours I would assume that an average barber has an occupancy ratio of 70% 4830000*0.5*1 barber services demand per month in Ireland (males) 30*4*3*0.7 barber services per barber and month There are around 9.000 barbers in Ireland
This was fine, I nailed this one pretty much, apart from some minor slip ups. Although the interviewer went into a LOT of detail on some parts of this, which was annoying, because I had displayed I knew the answer already. Not sure what they were looking for there.
To find largest k values from array of size N, there are two approaches: Efficient Sort: O(NlogN + k) Max Heap: O(N + klogN) Quickselect: O(kN) For k=1, quickselect wins. Otherwise Max heap probability wins.