Answer:
mean = 1 power failure
variance = 1 (power failure)²
Step-by-step explanation:
Since the mean is computed as
mean = E(x) = ∑ x * p(x) for all x
then for the random variable x=power failures , we have
mean = ∑ x * p(x) = 0 * 0.4 + 1* 0.3 + 2*0.2 + 3* 0.1 = 1 power failure
since the variance can be calculated through
variance = ∑[x-E(x)]² * p(x) for all x
but easily in this way
variance = E(x²) - [E(x)]² , then
E(x²) = ∑ x² * p(x) = 0² * 0.4 + 1²* 0.3 + 2²*0.2 + 3²* 0.1 = 2 power failure²
then
variance = 2 power failure² - (1 power failure)² = 1 power failure²
therefore
mean = 1 power failure
variance = 1 power failure²
I dislike this kind of math, pointless in my eyes
What is the upper quartile, Q3, of the following data set? 54, 53, 46, 60, 62, 70, 43, 67, 48, 65, 55, 38, 52, 56, 41
scZoUnD [109]
The original data set is
{<span>54, 53, 46, 60, 62, 70, 43, 67, 48, 65, 55, 38, 52, 56, 41}
Sort the data values from smallest to largest to get
</span><span>{38, 41, 43, 46, 48, 52, 53, 54, 55, 56, 60, 62, 65, 67, 70}
</span>
Now find the middle most value. This is the value in the 8th slot. The first 7 values are below the median. The 8th value is the median itself. The next 7 values are above the median.
The value in the 8th slot is 54, so this is the median
Divide the sorted data set into two lists. I'll call them L and U
L = {<span>38, 41, 43, 46, 48, 52, 53}
U = {</span><span>55, 56, 60, 62, 65, 67, 70}
they each have 7 items. The list L is the lower half of the sorted data and U is the upper half. The split happens at the original median (54).
Q3 will be equal to the median of the list U
The median of U = </span>{<span>55, 56, 60, 62, 65, 67, 70} is 62 since it's the middle most value.
Therefore, Q3 = 62
Answer: 62</span>
<em>*Because of BODMAS, 1/2x=x/2.</em>
1/2x+3/4=1
1/2x+3/4=4/4
1/2x=4/4-3/4
1/2x=1/4
x/2=1/4
x=2/4
x=1/2