Question

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 _____________.

Answer 1

Answer:

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 0, \sigma = 1$

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.

$Z = \frac{X - \mu}{\sigma}$

$0.613 = \frac{X - 0}{1}$

$X = 0.613$

27% of the possible Z values are greater than 0.613