1/4, since the radius is also halved if the diameter is halved, meaning (radius/2)^2 is radius^2 * 1/4.
Events A and B are independent if the following holds true:
P(A ∩ B) = P(A) * P(B)
where P(A ∩ B) is the probability of A and B, P(A) is the probability of A, and P(B) is the probability of B.
Setting A = "snow" and B = "cold weather", and plugging in the above numbers will give you the answer.
(a) Data with the eight day's measurement.
Raw data: [60,58,64,64,68,50,57,82],
Sorted data: [50,57,58,60,64,64,68,82]
Sample size = 8 (even)
mean = 62.875
median = (60+64)/2 = 62
1st quartile = (57+58)/2 = 57.5
3rd quartile = (64+68)/2 = 66
IQR = 66 - 57.5 = 8.5
(b) Data without the eight day's measurement.
Raw data: [60,58,64,64,68,50,57]
Sorted data: [50,57,58,60,64,64,68]
Sample size = 7 (odd)
mean = 60.143
median = 60
1st quartile = 57
3rd quartile = 64
IQR = 64 -57 = 7
Answers:
1. The average is the same with or without the 8th day's data. FALSE
2. The median is the same with or without the 8th day's data. FALSE
3. The IQR decreases when the 8th day is included. FALSE
4. The IQR increases when the 8th day is included. TRUE
5. The median is higher when the 8th day is included. TRUE
Answer:
Let's analyze the possible sums of both dices, i will use the notation:
Dice1 + Dice 2 = sum.
also remember that we have 2 dices, with 6 options each.
So the total number of combinations is 6*6 = 36
we have 36 possible outcomes.
I will start at the extremes, the minimum that we can sum is 2, and the maximum is 12, then:
We can have 2 if:
1 + 1 = 2
only one permutation.
and 12 if:
6 + 6 = 12
Again, only one permutation.
so 2 and 12 have the same chance (1 out of 36)
now, to have 3 we can have:
2 + 1 = 3
or
1 + 2 = 3.
and to have 11
5 + 6 = 11
6 + 5 = 11
Again, 3 and 11 have the same probability (2 out of 36 options)
And now we can see a pattern.
4 and 10 will have the same chance.
5 and 9 will have the same chance
6 and 8 will have the same chance
7 is the only number that has an unique chance (and it has the largest one)
