Resolviendo los planteamientos tenemos que : a) P(x = 7) = 8, 13%b) P(x = 10) = 12, 71%c) P(x≤12) = 78, 52%d) P(8≤x≤15) = 74, 52%◘Desarrollo : Para darle respuesta a cada uno de los planteamientos empleamos la aproximación de la distribución binomial a la normal : <img src="https://tex.z-dn.net/?f=P%28X%3Dx%29%3DP%28%5Cfrac%7Bx-0%2C5-%5Cmu%7D%7B%5Cdelta%7D%29%5Cleq%20Z%5Cleq%5Cfrac%7Bx%2B0%2C5-%5Cmu%7D%7B%5Cdelta%7D%29" />a) P(x = 7) : Datos : n = 1000p = 0, 01x = 7μ = n * p → 10σ = √np(1 - p) → 3, 15Sustituimos los valores en la ecuación : <img src="https://tex.z-dn.net/?f=P%28X%3D7%29%3DP%28%5Cfrac%7B7-0%2C5-10%7D%7B3%2C15%7D%29%5Cleq%20Z%5Cleq%5Cfrac%7B7%2B0%2C5-10%7D%7B3%2C15%7D%29" /><img src="https://tex.z-dn.net/?f=P%28X%3D7%29%3DP%28-1%2C11%7D%29%5Cleq%20Z%5Cleq%28-0%2C79%29%7D" /><img src="https://tex.z-dn.net/?f=P%28X%3D7%29%3DP%28Z%5Cleq-0%2C79%29-P%28Z%5Cleq%20-1%2C11%29" /><img src="https://tex.z-dn.net/?f=P%28X%3D7%29%3D%200%2C2147-0%2C1334" />P(X = 7) = 0, 0813 ≈ 8, 13%b) P(x = 10)<img src="https://tex.z-dn.net/?f=P%28X%3D10%29%3DP%28%5Cfrac%7B10-0%2C5-10%7D%7B3%2C15%7D%29%5Cleq%20Z%5Cleq%5Cfrac%7B10%2B0%2C5-10%7D%7B3%2C15%7D%29" /><img src="https://tex.z-dn.net/?f=P%28X%3D7%29%3DP%28-0%2C16%7D%29%5Cleq%20Z%5Cleq%280%2C16%29%7D" /><img src="https://tex.z-dn.net/?f=P%28X%3D7%29%3DP%28Z%5Cleq0%2C16%29-P%28Z%5Cleq%20-0%2C16%29" /><img src="https://tex.z-dn.net/?f=P%28X%3D7%29%3D%200%2C5635-0%2C4364" />P(X = 7) = 0, 1271 ≈ 12, 71%c) P(x≤12)<img src="https://tex.z-dn.net/?f=P%28X%20%5Cleq%20x%29%3DP%28Z%3C%5Cfrac%7Bx%2B0%2C5-%5Cmu%7D%7B%5Csigma%7D%29" />Sustituyendo : [img = 10][img = 11]P(x≤12) = 0, 7852 ≈ 78, 52d) P(8≤x≤15)[img = 12]Sustituyendo : [img = 13][img = 14][img = 15][img = 16]P(8≤x≤15) = 0, 7452 ≈ 74, 52%.