Respuesta :
Answer:
A score of 62 is not within one standard deviation of the mean.
Step-by-step explanation:
When the distribution is normal, we use 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.
Mean score of 54 and a standard deviation of 6.
This means that [tex]\mu = 54, \sigma = 6[/tex]
Calculate to see if a score of 62 is within one standard deviation of the mean.
Is Z between -1 and 1 when X = 62?
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{62 - 54}{6}[/tex]
[tex]Z = 1.33[/tex]
So a score of 62 is not within one standard deviation of the mean.
