Question

The alkalinity level of water specimens collected from the Han River in Seoul, Korea, has a mean of 50 milligrams per liter and a standard deviation of 3.2 milligrams per liter. (Environmental Science & Engineering, Sept. 1, 2000.) Assume the distribution of alkalinity levels is approximately normal and find the probability that a water specimen collected from the river has an alkalinity level a. exceeding 45 milligrams per liter. b. below 55 milligrams per liter. c. between 48 and 52 milligrams per liter.

Answer 1

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 50, \sigma = 3.2$

a. exceeding 45 milligrams per liter.

This probability is 1 subtracted by the pvalue of Z when X = 45. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{45 - 50}{3.2}$

$Z = -1.56$

$Z = -1.56$ 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.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{55 - 50}{3.2}$

$Z = 1.56$

$Z = 1.56$ 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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{52 - 50}{3.2}$

$Z = 0.69$

$Z = 0.69$ has a pvalue of 0.7549

X = 48

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{48 - 50}{3.2}$

$Z = -0.69$

$Z = -0.69$ 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.