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?"

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.

Step-by-step 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!!!

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step-by-step explanation:

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.