Let the original distribution function be F, then we need to perform:
"S=\\sum\\limits_{i=1}^n F(x_i)\\geq1060"
From the Central limit theorem, the sum function will also be a normal distribution with a mathematical expectation of 1000 and a standard deviation of 200. Then apply the formula:
"P(S>10600)=\u0424(\\infty)-\u0424(\\frac{1060-1000}{200})=\\\\=0.5-0.1179=0,3821"
Where Ф is the Laplace function, the values are taken from the table
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