Create a histogram for the data set. Click and drag on the horizontal axis to adjust the bar heights.
2 answers:
TO create a histogram of this data you will create a frequency table with intervals of numbers. Each interval of numbers must have the same amount of numbers represented.
Then use the intervals as your labels on the x-axis. Graph a bar as high as the frequency for each interval and make sure it touches the next bar.
Please see attached picture.
Then use the intervals as your labels on the x-axis. Graph a bar as high as the frequency for each interval and make sure it touches the next bar.
Please see attached picture.
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Answer:
= { (4,1) , (3,2) , (2,3) , (1,4) }
Step-by-step explanation:
Let's start writing the sample space for this experiment :
{ (1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6) , (2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6) , (3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6) , (4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6) , (5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) , (6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6) }
Let's also define the event ⇒
: '' The sum of the two dice is 5 ''
We can describe the event by listing all the favorables cases from ⇒
= { (4,1) , (3,2) , (2,3) , (1,4) }
In order to calculate we are going to divide all the cases favorables to
over the total cases from
. We can do this because all 36 of these possible outcomes from
are equally likely. ⇒
⇒
Finally we are going to define the event ⇒
: '' The number of the first die is exactly 1 more than the number on the second die ''
⇒
= { (2,1) , (3,2) , (4,3) , (5,4) , (6,5) }
Now given two events A and B ⇒
P ( A ∩ B ) =
We define the conditional probability as
with
We need to find therefore we can apply the conditional probability equation :
(I)
We calculate at the beginning of the question. We only need
.
Looking at the sets and
we find that (3,2) is the unique result which is in both sets. Therefore is 1 result over the 36 possible results. ⇒
Replacing both probabilities calculated in (I) :
We find out that