Answer :
Answer:
The range of values containing 90% of the population of x is from 175.55 units to 208.45 units.
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]\u = 192, \sigma = 10[/tex]
Determine the range of values containing 90% of the population of x.
The lower end of this interval is the value of X when Z has a pvalue of 0.5 - 0.9/2 = 0.05
The upper end of this interval is the value of X when Z has a pvalue of 0.5 + 0.7/2 = 0.95
Lower end
X when Z has a pvalue of 0.05, so X when [tex]Z = -1.645[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.645 = \frac{X - 192}{10}[/tex]
[tex]X - 192 = -1.645*10[/tex]
[tex]X = 175.55[/tex]
Upper end
X when Z has a pvalue of 0.95, so X when [tex]Z = 1.645[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.645 = \frac{X - 192}{10}[/tex]
[tex]X - 192 = 1.645*10[/tex]
[tex]X = 208.45[/tex]
The range of values containing 90% of the population of x is from 175.55 units to 208.45 units.