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1) In a normal distribution, what is the proportion of are between the z scores -2.09 and +.53?
2) In a normal distributio, find the probability expressed as a percentage that is greater than a z score of 2.33

Respuesta :

nobillionaireNobley
1.) P(-2.09 < z < 0.53) = P(0.53) - P(-2.09) = P(0.53) - [1 - P(2.09)] = P(0.53) + P(2.09) - 1 = 0.70194 + 0.98169 - 1 = 0.68363 = 68.36%

2.) P(z > 2.33) = 1 - P(z < 2.33) = 1 - 0.9901 = 0.0099 = 0.99%

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