Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. 27% of the possible Z values are greater than .

27% of the possible Z values are greater than 0.613

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

27% of the possible Z values are greater than

The 100 - 27 = 73rd percentile, which is X when Z has a pvalue of 0.73. So X when the z-score is 0.613.

27% of the possible Z values are greater than 0.613

x^6

step-by-step explanation:

this is correct because:

x-6=0

0*x=0

so

x^6