Question

You consider yourself a bit of an expert at playing rock-paper-scissors and estimate that the probability that you win any given game is 0.4. In a tournament that consists of playing 60 games of rock-paper-scissors let X be the random variable that is the of number games won. Assume that the probability of winning a game is independent of the results of previous games.

You should use the normal approximation to the binomial to calculate the following probabilities. Give your answers as decimals to 4 decimal places.

a)Find the probability that you win at least 26 of the games.

P(X ≥ 26) =

b)Find the probability that you win less than 22 games.

P(X < 22) =

c)Find the probability that you win between 20 and 30 games.

P(20 ≤ X ≤ 30) =

Answer #1

Solution :

Given that p = 0.4 , n = 60

=> q = 1 - p = 0.6

=> mean μ = n*p = 60*0.4 = 24

=> standard deviation σ = sqrt(npq)

= sqrt(60*0.4*0.6)

= 3.7947

a)

=> P(x >= 26) = P(x > 25.5)

= P((x - μ)/σ > (25.5 - 24)/3.7947)

= P(Z > 0.3953)

= 1 − P(Z < 0.3953)

= 1 − 0.6554

= 0.3446

b)

=> P(x < 22) = P((x - μ)/σ < (22 - 24)/3.7947)

= P(Z < -0.5271)

= 1 − P(Z < 0.5271)

= 1 − 0.7019

= 0.2981

c)

=> P(20 <= x <= 30) = P(19.5 < x < 30.5)

= P((19.5 - 24)/3.7947 < (x - μ)/σ < (30.5 - 24)/3.7947)

= P(-1.1859 < Z < 1.7129)

= 0.8394

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