Respuesta :
Answer:
a) 26.76% probability that a student in the psychology department has a score less than 480.
b) 69.73% probability that a student in the psychology department has a score between 480 and 730.
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 problem, we haev that:
[tex]\mu = 544, \sigma = 103[/tex]
a. Find the probability that a student in the psychology department has a score less than 480.
This is the pvalue of Z when X = 480. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{480 - 544}{103}[/tex]
[tex]Z = -0.62[/tex]
[tex]Z = -0.62[/tex] has a pvalue of 0.2676.
26.76% probability that a student in the psychology department has a score less than 480.
b. Find the probability that a student in the psychology department has a score between 480 and 730.
This probability is the pvalue of Z when X = 730 subtracted by the pvalue of Z when X = 480.
X = 730
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{730 - 544}{103}[/tex]
[tex]Z = 1.81[/tex]
[tex]Z = 1.81[/tex] has a pvalue of 0.9649.
X = 480
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{480 - 544}{103}[/tex]
[tex]Z = -0.62[/tex]
[tex]Z = -0.62[/tex] has a pvalue of 0.2676.
0.9649 - 0.2676 = 0.6973
69.73% probability that a student in the psychology department has a score between 480 and 730.
