On a normally distributed anxiety test with mean 48 and standard deviation 4, approximately what anxiety test score would put someone in the top 5 percent? Group of answer choices

Anxiety score close to 54.58.

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:

Approximately what anxiety test score would put someone in the top 5 percent?

We have to find the 100 - 5 = 95th percentile, which is X when Z has a pvalue of 0.95. So X when Z = 1.645.

Anxiety score close to 54.58.

what i would say the best answer is a. 40.5

question: use the following table of a school bus during morning pickups to calculate its average speed between 2.8 km and 4.2 km

answer: 42 km/h

explanation: 1.4/0.033 42.42424 = 42 km/h

question answered by

(jacemorris04)