Respuesta :
Geometric Distributions:Deals
with the number of trials
required until a single success.
Geometric random variable: the number of trials (y) it takes to get a success in a geometric setting.
When do you use it? To find first success or failure
Calculator Commands
When it takes more than n trials...1-geometcdf(p,n)
When it take n trials... geometpdf(p,x)
When it it takes a max of n trials...geometcdf(p,x)
a. geometcdf(.16,5) = .0797
b. binomcdf(10,.84,5) = .013
c. From par b. if the soccer player's succes rate were still 84%, the probability that he would score 5 or few goals in 10 penalty kicks is 0.0130. This is so low thatwe should be suspicious about whether he can still hit 84% of his shots. We have convincing that his penalty kick success rate has fallen below 84%
Geometric random variable: the number of trials (y) it takes to get a success in a geometric setting.
When do you use it? To find first success or failure
Calculator Commands
When it takes more than n trials...1-geometcdf(p,n)
When it take n trials... geometpdf(p,x)
When it it takes a max of n trials...geometcdf(p,x)
a. geometcdf(.16,5) = .0797
b. binomcdf(10,.84,5) = .013
c. From par b. if the soccer player's succes rate were still 84%, the probability that he would score 5 or few goals in 10 penalty kicks is 0.0130. This is so low thatwe should be suspicious about whether he can still hit 84% of his shots. We have convincing that his penalty kick success rate has fallen below 84%

a. Using the geometric probability: .84^4 * (1-.84) = 0.07966; this means that the probability that he makes his first * the probability he makes his second * the probability he makes his third * the probability he makes his fourth * the probability he misses his fifth.
b. P (X<= 5) = binomcdf (10, 0.84, 5) = 0.0130
c. From the answer in B, if the success rate of the player still continues at 84%, the probability that he will score 5 or few goals in the 10 penalty kicks will still remain at 0.0130. This is quite low, we should examine him about whether he can still hit at 84% of his shots. So given from this data, we could say that his penalty kick success rate has fallen below the given rate of 84%.
b. P (X<= 5) = binomcdf (10, 0.84, 5) = 0.0130
c. From the answer in B, if the success rate of the player still continues at 84%, the probability that he will score 5 or few goals in the 10 penalty kicks will still remain at 0.0130. This is quite low, we should examine him about whether he can still hit at 84% of his shots. So given from this data, we could say that his penalty kick success rate has fallen below the given rate of 84%.