Answer:
Following are the solution to this question:
Explanation:
In point a:
Consider Binomial ecxperiment for performance probability, p = 0.70.
Therefore, the chance of fail q = 0.30 (1 - 0.70)
N = 9 Sample size
The Binomial likelihood for each success amount could be found using the Excel function BINOMDSIT (x,n,p,0):
[tex]\text{Number of Successes}, x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Probability, P(X=x) \\\\[/tex]
[tex]0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P(X=0)=BINOMDIST(0, 9, 0.70, 0) =0.000020\\\\1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P(X=1)=BINOMDIST(1, 9, 0.70, 0) =0.000413\\\\2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P(X=2)=BINOMDIST(2, 9, 0.70, 0) =0.003858\\\\3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P(X=3)=BINOMDIST(3, 9, 0.70, 0) =0.021004\\\\4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P(X=4)=BINOMDIST(4, 9, 0.70, 0) =0.073514\\\\[/tex]
[tex]5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P(X=5)=BINOMDIST(5, 9, 0.70, 0) =0.171532\\\\6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P(X=6)=BINOMDIST(6, 9, 0.70, 0) =0.266828\\\\7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P(X=7)=BINOMDIST(7, 9, 0.70, 0) =0.266828\\\\8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P(X=8)=BINOMDIST(8, 9, 0.70, 0) =0.155650\\\\9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P(X=9)=BINOMDIST(9, 9, 0.70, 0) =0.040354\\\\[/tex]
To obtain the necessary Histograms, use X values as just an input X range as well as the probability value values as an output Y range in Excel table in attachment please find it.
In point b:
They recognize that its mean and standard variance of its binomial experiment was,
[tex]Mean = n\times p[/tex]
[tex]= 9 \times 0.70\\\\=6.3\\\\[/tex]
[tex]\sigma =\sqrt{ npq} \\\\[/tex]
[tex]= \sqrt{9 \times 0.70 \times 0.30} \\\\ =\sqrt{1.89} \\\\ =1.375[/tex]
The average is 6.3, so the expected number of friends to whom addresses were found is 6.
In point c:
They must establish the value of n in addition to have [tex]P(X \geq 2)=0.97[/tex]
This assertion of probability is replaceable as:
[tex]\to P(X \geq 2) =1-P(X <2 ) =1-P(X \leq 1 )=0.97[/tex]
[tex]\to P(X \leq 1 )=(1-0.97)=0.03[/tex]
It can be found with the Excel Aim Search feature They must therefore apply approximately 5 identities to be 97% confident that two addresses are found at least.
