"Let\\;p=\\frac{156}{240}=0.65\\\\\nnp=0.65\\times 240=156, \n\\\\n(1-p)=0.35\\times 240=84,\\\\\nnp\\geq 5, n(1-p)\\geq5, \\\\\n\\text{so we can use the Normal Distribution}\\\\\n\\text{ to Approximate Binomial Probabilities} \n\\\\p(x \\leq 156)=p(z \\leq \\frac {156.5-240\\times 0.65}{\\sqrt{240(0.65)(0.35)}})\\\\\np(z \\leq 0.07)=0.5+p(0\\leq z\\leq 0.07)\\\\\n=0.5+0.0279=0.5279"
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