Answer:
Explanation:
Garrett achieves a field goal is independent of each kick. Hence X no of kicks is binomial with p = 0.83, and q =0.17
No of trials 3
Hence X can take values as 0 ,1,2,3 and
![P(X=r) = 4Cr(0.83)^r (0.17)^(4-r)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/e5cl5ezg0y4wpaig28bxsab0a27m10asd4.png)
, r=0,1,2,3
Using the above we obtain prob distribution as
X 0 1 2 3
P(X) 0.0049 0.0720 0.3513 0.5718
P(X>=2) =P(X=2,3,4) =0.9231
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2) Z value for company A =
![(260-272)/(5.4) =-2.22](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ha20p5atdu0bg8x7vwtt19kdx48ol7mmog.png)
Z value for company B =
![(260-249)/3.8 =2.89](https://img.qammunity.org/2020/formulas/mathematics/middle-school/oolhru0smpepdsf6kcibbs7c17wexv1kdh.png)
From z value probablility is more for company A as
P(Z>-2.22) >P(Z>2.89)