cud ny1 help this out...400 draws are made at random with replacement from 5 tickets that are marked -2, -1, 0, 1, and 2 respectively. Find the expected value of a)the number of times positive numbers appear b)the sum of the positive numbers drawn
a) 3/5 * 400 = 240 times
b ) the sum of positives???? could be either, 0 , 1 , 2 i guess there is a 3/5 chance of a positive, and a 1/5 chance of each number.
still thinking...
is zero positive?
that is what i m also confused about..
wikipedia says no In mathematics 0 is the integer immediately preceding 1. Zero is an even number,[32] because it is divisible by 2. **0 is neither positive nor negative.** By most definitions[33] 0 is a natural number, and then the only natural number not to be positive. Zero is a number which quantifies a count or an amount of null size. In most cultures, 0 was identified before the idea of negative things (quantities) that go lower than zero was accepted. The value, or number, zero is not the same as the digit zero, used in numeral systems using positional notation. Successive positions of digits have higher weights, so inside a numeral the digit zero is used to skip a position and give appropriate weights to the preceding and following digits. A zero digit is not always necessary in a positional number system, for example, in the number 02. In some instances, a leading zero may be used to distinguish a http://en.wikipedia.org/wiki/0_(number)
so whats the answer
do you want just the answer or do you want me to explain how to get the answer?
i want u to explain....and answer also..:|
explanation: it is looking for how many times you would expect to draw a positive number. Two of the 5 options are positive numbers, so you can expect to draw a positive number 2 every 5 times.
did I explain that well?
i didnt gt tht...so wud it be 2/5*400
yes, it would
and for the second part, you assume you get exactly as many 1s as 2s because you are equally likely to draw them
who knew zero was not positive...
and it's not negative either.
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