ized no mal distr bution (with a mean of 0 and a standard devat on of 1), comple

Question

ized no mal distr bution (with a mean of 0 and a standard devat on of 1), complete parts (a) through (d). Click here to view page 1 of the cumulative standardized normal distribution table. Click here to view page 2 of the cumulative standardized normal distribution table. The probability that Z is less than 1.01 is 0.8438 Round to four decimal places as needed.) b. What is the probability that Z is greater than 0.29? The probability that Z is greater than 0.29 is Round to four decimal places as needed.) c. What is the probability that Z is less than -0.29 or greater than the mean? The probability that Z is less than-0.29 or greater than the mean is Round to four decimal places as needed.) d. What is the probability that Z is less than - 0.29 or greater than 1.01? The probability that Z is less than - 0.29 or greater than 1.01 is Round to four decimal places as needed.)

Explanation / Answer

b) P( Z > -0.29) = 1 - P( Z < -0.29) .....(1)

P( Z < -0.29) = 0.3859 ( From z table)

plug this value in equation (1)

P( Z > -0.29) = 1 -  0.3859 = 0.6141

c) P( Z < -0.29) = 0.3859

and P( Z > 0) = 0.5

therefore probability that Z is less than -0.29 or greater than the mean (that is 0) is 0.3859 + 0.5 = 0.8859

d) P( Z < -0.29) = 0.3859

P( Z > 1.01) = 1 - P( Z < 1.01) = 1 - 0.8438 = 0.1562

therefore required probability =  P( Z < -0.29) + P( Z > 1.01) = 0.3859 + 0.1562 = 0.5421

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