Home/ English/Mathematics/Given that z is a standard normal random variable, discover z for each instance (to 2 decimals). A. The area to the best of z is 001.

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## Given that z is a traditional normal random variable, find z for each situation (to 2 decimals). A. The location to the best of z is 001.

Given that z is a standard normal random variable, uncover z for each case (to 2 decimals). A. The location to the ideal of z is 001. B. The area to the appropriate of z is 0.045. C. The location to the ideal of z is 0.05. D. The location to the ideal of z is 0.2. a) Z = 2.33

b) Z = 1.7

c) Z = 1.65.

d) Z = 0.84.

Step-by-step explanation:

Z-score:

In a set with mean  , the zscore of a measure X is offered by: The Z-score steps exactly how many traditional deviations the meacertain is from the intend. After finding the Z-score, we look at the z-score table and uncover the p-value associated via this z-score. This p-worth is the probcapability that the value of the meacertain is smaller than X, that is, the percentile of X, which is likewise the area to the left of z. Subtracting 1 by the pvalue, we get the probcapacity that the value of the measure is greater than X, which is likewise the area to the best of z.

A. The location to the best of z is 0.01.

Z has a pworth of 1 – 0.01 = 0.99. So Z = 2.33.

B. The area to the ideal of z is 0.045.

Z has actually a pworth of 1 – 0.045 = 0.955. So Z = 1.7

C. The location to the right of z is 0.05.

Z has a pworth of 1 – 0.05 = 0.95. So Z = 1.65.

D. The location to the right of z is 0.2.

Z has actually a pworth of 1 – 0.2 = 0.8. So Z = 0.84.

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