Mathematics, 05.03.2020 17:43 deepspy599otchpd

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.

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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.

Answer from: Quest

A= r2 × 3.14 11×11 ↓ ↓ 121 × 3.14 = 379.94 → 379.9

Answer from: Quest

62926281916 inches of rain: )

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