A game is played with a spinner on a circle, like the minute hand on a clock. The circle is marked evenly from 0 to 100, so, for example, the 3:00 position corresponds to 25, the 6:00 position to 50, and so on. The player spins the spinner, and the resulting number is the number of seconds he or she is given to solve a word puzzle. If 100 players are selected randomly, what is the probability that the avearge time these players will get to solve the word puzzle will exceed 50 seconds?

The time a person is given corresponds to a uniform distribution with values between 0 and 100. The mean of this distribution is 0+100/2 = 50 and the variance is (100-0)²/12 = 833.3.

When we take 100 players we are taking 100 independent samples from this same random variable. The mean sample, lets call it X, has equal mean but the variance is equal to the variance divided by the length of the sample, hence it is 833.3/100 = 8.333.

As a consecuence of the Central Limit Theorem, the mean sample (taken from independant identically distributed random variables) has distribution Normal with parameters μ = 50, σ= 8.333. We take the standarization of X, calling it W, whose distribution is Normal Standard, in other words

[tex]W = \frac{X - \mu}{\sigma} = \frac{X - 50}{8.333} \simeq N(0,1)[/tex]

The values of the cummulative distribution of the Standard Normal distribution, lets denote it [tex] \phi [/tex] , are tabulated and they can be found in the attached file, We want to know when X is above 50, we can solve that by using the standarization

[tex]P(X > 50) = P(\frac{X-50}{8.33} > \frac{50-50}{8.33}) = P(W > 0) = \phi(0) = 1/2[/tex]