Annie is a basketball player who makes, on average, 65% of her free throws. Assu

Question

Annie is a basketball player who makes, on average, 65% of her free throws. Assume each shot is independent and the probability of making any given shot is .65. What's the probability Annie will miss three straight free throws before she makes one? (If you use a calculator to get your answer, write your answer in standard notation, not calculator notation.) During a season, Annie takes 140 free throws. What's the exact binomial probability she'll make at least 100 out of 140 of these throws? For part B, are the conditions that permit you to use a normal approximation to the binomial satisfied? Explain. Redo part B using a normal approximation, without continuity correction, to the binomial. Draw a sketch of the situation: draw the distribution and shade the area representing the probability you're finding. Redo part B using a normal approximation, with continuity correction, to the binomial.

Explanation / Answer

Binomial Distribution
PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
A.
Annie will miss 3 sraight free throws before she makes 1 - 0.65)^3 * 0.65 = 0.027868
B.
P( X > = 100 ) = 1 - P( X < 100) = 0.0644
c.
n*p>5, 140*0.65> 5 => 91>5
n*(1-p)>5, 140*0.35> 5 => 91>5
Can Use Normal Approximation                  
d.
Normal Approximation to Binomial Distribution
Mean ( np ) =140 * 0.65 = 91
Standard Deviation ( npq )= 140*0.65*0.35 = 5.6436
Normal Distribution = Z= X- u / sd                   
P(X < 100) = (100-91)/5.6436
= 9/5.6436= 1.5947
= P ( Z <1.5947) From Standard NOrmal Table
= 0.9446                  
P( X > = 100 ) = 1 - P( X < 100) = 1 - .9446= 0.0554

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