Respuesta :
Answer:
a) 13.35% probability that the concentration exceeds 0.70
b) 1.32% probability that the concentration is at most 0.40
c) The largest 5% of all concentration values are above 0.748 mg/cm3.
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 have that:
[tex]\mu = 0.6, \sigma = 0.09[/tex]
(a) What is the probability that the concentration exceeds 0.70?
This is 1 subtracted by the pvalue of Z when X = 0.7. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.7 - 0.6}{0.09}[/tex]
[tex]Z = 1.11[/tex]
[tex]Z = 1.11[/tex] has a pvalue of 0.8665
1 - 0.8665 = 0.1335
13.35% probability that the concentration exceeds 0.70
(b) What is the probability that the concentration is at most 0.40?
pvalue of Z when X = 0.4. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.4 - 0.6}{0.09}[/tex]
[tex]Z = -2.22[/tex]
[tex]Z = -2.22[/tex] has a pvalue of 0.0132
1.32% probability that the concentration is at most 0.40
(c) How would you characterize the largest 5% of all concentration values?
They are above the 100-5 = 95th percentile, which is the value of X when Z has a pvalue of 0.95, so X when Z = 1.645.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.645 = \frac{X - 0.6}{0.09}[/tex]
[tex]X - 0.6 = 1.645*0.09[/tex]
[tex]X = 0.748[/tex]
The largest 5% of all concentration values are above 0.748 mg/cm3.
