Question

At a particular chess club, it is quite common for a chess game to take over two hours to complete. Suppose that the lengths of these games are normally distributed with a mean of 153 minutes and a standard deviation of 45 minutes. Games which last in the longest 1% of all contests are given special recognition on an "Endurance Board" for all club members to see. How long would a game need to last to qualify for the "Endurance Board?"

Answer 1

A game would need to be at least 257.67 minutes long to qualify for the Endurance Board.

257.67 minutes = 257 minutes, 40 seconds = 4 hours, 17 minutes, 40 seconds.

Explanation:

Games that are given special recognition on the Endurance Board are the games that last in the longest 1% of all games.

If X is the random variable that represents the time a chess game takes before it is completed.

X is said to be normally distributed with

Mean = μ = 153 minutes

Standard deviation = σ = 45 minutes

Let games that last the longest 1% of the time last for a minimum of x' minutes.

P(X > x') = 1% = 0.01

P(X ≤ x') = 1 - P(X > x') = 1 - 0.01

P(X ≤ x') = 0.99

Indicating that such games are longer than 99% of all chess games.

This is a normal distribution problem

Let the z-score for these type of longest games with a minimum duration of x' minutes be z'.

P(X ≤ x') = P(z ≤ z') = 0.99

From the normal distribution table, z' = 2.326

z-score of any value is given as the value minus the mean divided by the standard deviation.

z = (x - μ)/σ

So,

z' = (x' - μ)/σ

2.326 = (x' - 153)/45

x' = (2.326×45) + 153

x' = 104.67 + 153 = 257.67 minutes = 257 minutes, 40 seconds.

Hope this Helps!!!