Respuesta :
Answer:
77.4% probability that a data value is between 36 and 41
Step-by-step explanation:
Problems of normally distributed samples are 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, we have that:
[tex]\mu = 38, \sigma = 2[/tex]
What is the probability that a data value is between 36 and 41?
This is the pvalue of Z when X = 41 subtracted by the pvalue of Z when X = 36.
X = 41
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{41 - 38}{2}[/tex]
[tex]Z = 1.5[/tex]
[tex]Z = 1.5[/tex] has a pvalue of 0.933
X = 36
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{36 - 38}{2}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.159
0.933 - 0.159 = 0.774
77.4% probability that a data value is between 36 and 41
