00:01
In question number 8, in point a, given that n equal to 6, and p equal to 0 .30, the expected value of binomial distribution is the product of the number of trails in under probability p, so it should be equal to n by p equal to 6 by 0 .30 equal to 1 .80.
00:25
The highest bars of the histogram should be about the expect value.
00:36
Thus, the second and the third bar should be the highest, which corresponds with the graph 2.
01:00
Note that the first bar of the histogram corresponds with r equal to 0, and the second corresponds with r equal to 1 and the third bar corresponds with r equal to 2 and at c in point b if it was given that in equal to 6 and p equal to 0 .50 so the expected value will be np equal to 6 by 0 .50 equal to 3.
01:58
So the highest bars of the histogram should be about the expect value.
02:04
Thus the fourth bar, which corresponds with r equal to 3, should be the highest, which corresponds with graph 1.
02:26
Note that the first bar of the histograms, the instagram corresponds with r equal to 0 and the second corresponds with r equal to 1 and the third corresponds with r equal to an atc.
02:40
In point c, if it's given that n equal to 6 and p equal to 0 .65, so the expected value of a binomial distribution is the product of the number of trails and under probability b.
03:01
So it will be equal to np equal to 6 by 0 .65 approximately equal to 4.
03:12
The highest bars of the histogram should be about the expected value, thus the fifth bar, which corresponds with r equal to z, equal to 4, should be the highest, which corresponds with graph triple i.
03:40
That the first bar of the histogram corresponds with r equal to 0, the second corresponds with r equal to 1 and the third with r equal to 2...