The sum could be greater than 1 but not equal to 1.
Hmm. I think true in this case
Given a table <span>representing
the probability distribution of the number of times the John Jay wifi
network is slow during a week. We call the random variable x.
Part A:
The total value of p(x) = 1.
Thus, </span><span>
.08 + .17 + .21 + k + .21 + k + .13 = 1
0.8 + 2k = 1
2k = 1 - 0.8 = 0.2
k = 0.2 / 2 = 0.1
Therefore,
the value of k is 0.1Part B:
The expected value of x is given by
Therefore,
the expected value of x is 3.01Part C:
</span><span>The expected value of
is given by
Therefore,
the expected value of is 12.45</span>
Part D:
The variance of x is given by
Therefore,
the variance of x is 3.39.
Part E
<span>The standard deviation of x is given by
Therefore,
the standard deviation of x is 1.84.
Part F:
The variance of ax, where a is a constant is given by
Thus, the variance of 3x is given by
Therefore,
the variance of 3x is 30.51.
Part G:
The probability that the network has no more that 4 slow times in one week is given by
Since, the </span>network slowness is independent from week to week, the <span>probability that if we look at 5 separate weeks, the network has no more than 4 slow times in any of those weeks is given by
Therefore, </span>
the probability that if we look at 5 separate weeks, the network has no more than 4 slow times in any of those weeks is 0.27Part H:
The variance of x^2 is given by
Thus,
Therefore,
the <span>
variance of the random variable is 141.37</span>
Answer:
The answer is C.
Step-by-step explanation:
You first have to take 50 and divide it by 3 to get 16.6 repeating. From there, you just have to turn it into a fraction which would be between 1 half and 3 fourths.
MK is the perpendicular bisector ò JL so MJL will be isosceles triangle, then JM = ML = 15, i guess