Question

The time it takes for a certain athlete to complete 500 jumps on the rope follows a normal distribution with an expected value of 2.5 minutes and a standard deviation of 0.16 minutes.

(a) Calculate the probability that in the next attempt the athlete takes less than 2.5 minutes to complete the 500 jumps 
(b) Assuming that the athlete can perform at least 10 repetitions with the same stamina, what is the probability that he can complete 2000 jumps in less than 9.5 minutes? (5 min)
(c) If 10 athletes with the same characteristics perform the exercise simultaneously, what is the probability that less than 4 complete the 500 jumps before 2.25 minutes? 
(d) If the athlete wants to have at least a 10% probability of completing the 500 jumps in less than 2.2 minutes, how much should he lower his average?

Answer 1

We are given the distribution here as:

a) The probability that in the next attempt the athlete takes less than 2.5 minutes to complete the 500 jumps is computed here as:
P(X < 2.5)

As 2.5 minutes is the mean here and the normal distribution is symmetric about its mean, therefore the probability P(X > 2.5) = P(X < 2.5) = 0.5
Therefore 0.5 is the required probability here.

b) As we are given the time taken for 500 jumps as X, therefore Y = 4X would be the time taken for 2000 jumps. Therefore the distribution for Y here would be given as:

The probability now is computed here as:
P(Y < 9.5)

Converting it to a standard normal variable, we have here:

Getting it from the standard normal tables, we have here:

Therefore 0.2173 is the required probability here.

c) The probability that one of them complete before 2.25 minutes is computed here as:
P(X < 2.25)

Converting it to a standard normal variable, we have here:

Getting it from the standard normal tables, we have here:

The probability that less than 4 out of the 10 athletes complete the jumps before 2.25 minutes is computed here as:

P(Y < 4) = P(Y = 0) + P(Y = 1) + P(Y = 2) + P(Y = 3)

Therefore 0.9981 is the required probability here.

d) From standard normal tables, we have here:
P(Z < -1.282) = 0.1

Therefore, the mean here is computed as:

Therefore 1.99488 minutes is the required mean here.