Answer:
84.1% of all values in a normal distribution have z ≤ 1.00.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
The percent of all values in a normal distribution for which z ≤ 1.00.
This is the pvalue of Z = 1.
Z = 1 has a pvalue of 0.8413.
Converting to percentage, to the nearest tenth.
84.1% of all values in a normal distribution have z ≤ 1.00.
