Answer :
Answer:
The score that separates the lower 5% of the class from the rest of the class is 55.6.
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 question:
[tex]\mu = 73, \sigma = 10.6[/tex]
Find the score that separates the lower 5% of the class from the rest of the class.
This score is the 5th percentile, which is X when Z has a pvalue of 0.05. So it is X when Z = -1.645.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.645 = \frac{X - 73}{10.6}[/tex]
[tex]X - 73 = -1.645*10.6[/tex]
[tex]X = 55.6[/tex]
The score that separates the lower 5% of the class from the rest of the class is 55.6.