Binomial Distribution and Z Score questions: "The fastest 8% of players..."

firelan

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Hi guys, im stuck on these 2 questions can you please help me solve them.

Suppose that there is a new video game that is quite challenging. Players will be graded according the times they took to finish the course. Assuming the times to finish the course follows a normal distribution of 20 minutes with a standard deviation of 5 minutes.


A) The fastest 8% of players will qualify to receive on Gold Medal by the program. What time qualify individuals for such award?


B) If the time taken to complete the task is less than 9 minutes, then the individual will be automatically given a code to download an advanced version for free. If there are 20 players, what are the chances that there is at least two people will earn the code?
 
Hi guys, im stuck on these 2 questions can you please help me solve them.

Suppose that there is a new video game that is quite challenging. Players will be graded according the times they took to finish the course. Assuming the times to finish the course follows a normal distribution of 20 minutes with a standard deviation of 5 minutes.
This corresponds to a standard normal distribution with z= (x- 20)/5.


A) The fastest 8% of players will qualify to receive on Gold Medal by the program. What time qualify individuals for such award?
What z corresponds to the top 8%? Equivalently, what z, in the standard normal distribution, has probability less than or equal to 0.92? Solve (x- 20)/5= z, with that value of z, for x.


B) If the time taken to complete the task is less than 9 minutes, then the individual will be automatically given a code to download an advanced version for free. If there are 20 players, what are the chances that there is at least two people will earn the code?
Find the probability, using the standard normal distribution, of (9- 20)/5= -11/5. Using that probability as "p", the probability of 'success" and, of course, q= 1- p as the probability of "failure" in a binomial distribution, with n= 20, find the probability of 0 and 1 "successes" and subtract the sum of those from 1.
 
This corresponds to a standard normal distribution with z= (x- 20)/5.



What z corresponds to the top 8%? Equivalently, what z, in the standard normal distribution, has probability less than or equal to 0.92? Solve (x- 20)/5= z, with that value of z, for x.



Find the probability, using the standard normal distribution, of (9- 20)/5= -11/5. Using that probability as "p", the probability of 'success" and, of course, q= 1- p as the probability of "failure" in a binomial distribution, with n= 20, find the probability of 0 and 1 "successes" and subtract the sum of those from 1.

why am I finding the probability of 0 and 1 successes, why am I not finding the failure in order to get the complement for the successes of "at least two".
 
why am I finding the probability of 0 and 1 successes, why am I not finding the failure in order to get the complement for the successes of "at least two".
You wrote If there are 20 players, what are the chances that there is at least two people will earn the code? So if we proceed with successes we need to compute p(2) + p(3) + ... + p(20). It is easier to compute the complement which is 1 - p(0) + p(1).

But you prefer to work with failures that is not wrong. If there are 2 or more successes then what would be the possible number of failures?

For 2 successes there would be 18 failures. For 3 successes there would be 17 failures. ... For 20 successes there would be 0 failures.

So we need to compute p(0 failures) + p(1 failures) + ... + p(18 failures) = 1 - p(19 failures) - p(20 failures).

So sure you can compute 1 - p(19 failures) - p(20 failures).

Is that what you meant? Or are you thinking that p( at least 2 success) = 1 - p(0 failures) - p(1 failures)?
 
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