Answer:
P{W>0}=0.5
P{W=0}=0.25
E{W}=0
Step-by-step explanation:
<u>Given:</u>
Gambles are independent i.e. each player is equally likely to win or lose 1 unit. OR each player has equal probability to win or lose 1 unit.
Let W denote the net winnings of a gambler whose strategy is to stop gambling immediately after his first win.
Then
<u>A.P{W>0}=?</u>
P{W>0}=0.5, because each player is equally likely to win or lose on first gamble. i.e there is equal chances for winning or losing on the first gamble.
<u>B.P{W<0}=?</u>
for P{W<0} we need to find P{W=0} first as;
P{W=0}=0.25
As there is equal probability to win or lose, after first win, if you want to finish gamble with no profit (equal number of lose and win) then if you losing, you have equal probability to win or lose so to finish your game with P=0 your probability is 0.25 (half of 0.5)
P{W<0} means net lose which is equal to total probability minus probability of profit and probability of net profit equal to zero.
i.e. P{W<0}=1-P{W=0}-P{W>0}
P{W<0}=1-0.25-0.5=0.25
<u>C.E{W}=?</u>
E{W}=P{W>0}*{W>0}+P{W<0}*{W<0}+P{W=0}*{W=0}
E{W}=0.5*(1)+0.5(-1)+0.25*(0) (for any value of W,P{W>0}*{W>0}+P{W<0}*{W<0}=0, because sum of same positive and negative numbers is zero)
E{W}=0
Recall that
Plug in 
and you get your series, so the given series has a value of 
.
Answer:
The mean is more useful in this case because it would give an average value of the accidents for example 3 accidents per year but the median would give the middle value which may be 5 or greater or much lesser than the average. It would not give an approximate value of occurrences.
Step-by-step explanation:
Mean is the averaage of all the values.
Median is the value of the data which gives an estimate of the middle value. Middle values can be different than the average values.
The mean is
1) rigorously defined by a mathematical formula.
2) based on all the observations of the data
3) affected by extreme values
The meadian is
1) computed for open end classes like income etc.
2) not rigorously defined
3) is located when the values are not capable of quantitative measurment.
4) is not affected by extreme values.
The mean is more useful in this case because it would give an average value of the accidents for example 3 accidents per year but the median would give the middle value which may be 5 or greater or much lesser than the average. It would not give an approximate value of occurrences.
-5 + 2x² = -6x
rearrange the equation to the form ax² + bx + c = 0
=> 2x² + 6x - 5
use the quadratic formula to solve for the value(s) of x
=>
=>
=>
<span>
</span>∴ x =
OR x =
x = - 1.64 ; x = - 10.36
