Respuesta :
Answer:
0.0414
Step-by-step explanation:
Each error is uniform between -0.5 and 0.5, so the mean error is 0, and the variance is (b-a)²/12 = (0.5-(-0.5))²/12 = 1/12
If we sum 50 numbers, the errors will sum with each other, and the resultant mean and variance will be summed, because the errors are independent. The mean of the sum of 50 number is 0*50 = 0, and the variance in 50/12.
The central limit theorem states that the sum of identically distributed random variables has distribution approximately normal. In this case, if we call X the sum of the 50 random numbers, then X has distribution approximately N(μ = 0,σ = √(50/12)). If we divide X with its standard deviation √(50/12), we obtain (approximately) a standard normal random variable. Lets call Y = X/√(50/12). Y distribution is approximately N(0,1). Y is called the standarization of X.
The values of the cummulative distribution of the standard Normal random variable, denoted by Ф, are tabulated; you can find those values in the attached file. We want the error to be greater than 3. We will calculate the complementary event: the probability for the error to be between -3 and 3, and substract from 1 that result
P(-3 ≤ X ≤ 3) = P( -3/√(50/12) ≤ X/√(50/12) ≤ 3/√(50/12)) = P(-3/√(50/12) ≤ Y ≤ 3/√(50/12)) = Ф(3/√(50/12)) - Ф(-3/√(50/12))
Since the density function of a normal random variable centered at 0 is symmetric, then Ф(-3/√(50/12)) = 1- Ф(3/√(50/12)), as a result
P(-3 ≤ X ≤ 3) = Ф(3/√(50/12)) - Ф(-3/√(50/12)) = 2 Ф(3/√(50/12)) - 1 = 2 * Ф(2.04) - 1 = 2*0.9793 - 1 = 0.9586
hence, the probability for the error to be greater thar 3 is 1-0.9586 = 0.0414
