Respuesta :
Answer:
Standard score of -0.25.
A data value of 89 is approximately in the 40th percentile.
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 = 90, \sigma = 4[/tex]
Give the standard score and approximate percentile for a data value of 89.
Z when X = 89
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{89 - 90}{4}[/tex]
[tex]Z = -0.25[/tex]
[tex]Z = -0.25[/tex] has a pvalue of 0.4013.
So a data value of 89 is approximately in the 40th percentile.
