Answer:
68.26% of the data would be between 16.4 and 21.6.
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 problem, we have that:
[tex]\mu = 19, \sigma = 2.6[/tex]
What percentage of the data would be between 16.4 and 21.6?
This is the pvalue of Z when X = 21.6 subtracted by the pvalue of Z when X = 16.4. So
X = 21.6
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{21.6 - 19}{2.6}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413.
X = 16.4
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{16.4 - 19}{2.6}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.1587
So 0.8413 - 0.1587 = 0.6826 = 68.26% of the data would be between 16.4 and 21.6.
