Answer:
a) 94.06% probability that a water specimen collected from the river has an alkalinity level exceeding 45 milligrams per liter.
b) 94.06% probability that a water specimen collected from the river has an alkalinity level below 55 milligrams per liter.
c) 50.98% probability that a water specimen collected from the river has an alkalinity level between 48 and 52 milligrams per liter.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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:
![\mu = 50, \sigma = 3.2](https://tex.z-dn.net/?f=%5Cmu%20%3D%2050%2C%20%5Csigma%20%3D%203.2)
a. exceeding 45 milligrams per liter.
This probability is 1 subtracted by the pvalue of Z when X = 45. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{45 - 50}{3.2}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B45%20-%2050%7D%7B3.2%7D)
![Z = -1.56](https://tex.z-dn.net/?f=Z%20%3D%20-1.56)
has a pvalue of 0.0594.
1 - 0.0594 = 0.9406
94.06% probability that a water specimen collected from the river has an alkalinity level exceeding 45 milligrams per liter.
b. below 55 milligrams per liter.
This probability is the pvalue of Z when X = 55.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{55 - 50}{3.2}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B55%20-%2050%7D%7B3.2%7D)
![Z = 1.56](https://tex.z-dn.net/?f=Z%20%3D%201.56)
has a pvalue of 0.9604.
94.06% probability that a water specimen collected from the river has an alkalinity level below 55 milligrams per liter.
c. between 48 and 52 milligrams per liter.
This is the pvalue of Z when X = 52 subtracted by the pvalue of Z when X = 48. So
X = 52
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{52 - 50}{3.2}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B52%20-%2050%7D%7B3.2%7D)
![Z = 0.69](https://tex.z-dn.net/?f=Z%20%3D%200.69)
has a pvalue of 0.7549
X = 48
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{48 - 50}{3.2}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B48%20-%2050%7D%7B3.2%7D)
has a pvalue of 0.2451
0.7549 - 0.2451 = 0.5098
50.98% probability that a water specimen collected from the river has an alkalinity level between 48 and 52 milligrams per liter.