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Lesechka [4]
3 years ago
6

Please someone help me

Mathematics
1 answer:
Free_Kalibri [48]3 years ago
7 0
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Read 2 more answers
The amount of pollutants that are found in waterways near large cities is normally distributed with mean 8.6 ppm and standard de
Setler79 [48]

We assume that question b is asking for the distribution of \\ \overline{x}, that is, the distribution for the average amount of pollutants.

Answer:

a. The distribution of X is a normal distribution \\ X \sim N(8.6, 1.3).

b. The distribution for the average amount of pollutants is \\ \overline{X} \sim N(8.6, \frac{1.3}{\sqrt{38}}).

c. \\ P(z>-0.08) = 0.5319.

d. \\ P(z>-0.47) = 0.6808.

e. We do not need to assume that the distribution from we take the sample is normal. We already know that the distribution for X is normally distributed. Moreover, the distribution for \\ \overline{X} is also normal because <em>the sample was taken from a normal distribution</em>.

f. \\ IQR = 0.2868 ppm. \\ Q1 = 8.4566 ppm and \\ Q3 = 8.7434 ppm.

Step-by-step explanation:

First, we have all this information from the question:

  • The random variable here, X, is the number of pollutants that are found in waterways near large cities.
  • This variable is <em>normally distributed</em>, with parameters:
  • \\ \mu = 8.6 ppm.
  • \\ \sigma = 1.3 ppm.
  • There is a sample of size, \\ n = 38 taken from this normal distribution.

a. What is the distribution of X?

The distribution of X is the normal (or Gaussian) distribution. X (uppercase) is the random variable, and follows a normal distribution with \\ \mu = 8.6 ppm and \\ \sigma =1.3 ppm or \\ X \sim N(8.6, 1.3).

b. What is the distribution of \\ \overline{x}?

The distribution for \\ \overline{x} is \\ N(\mu, \frac{\sigma}{\sqrt{n}}), i.e., the distribution for the sampling distribution of the means follows a normal distribution:

\\ \overline{X} \sim N(8.6, \frac{1.3}{\sqrt{38}}).

c. What is the probability that one randomly selected city's waterway will have more than 8.5 ppm pollutants?

Notice that the question is asking for the random variable X (and not \\ \overline{x}). Then, we can use a <em>standardized value</em> or <em>z-score</em> so that we can consult the <em>standard normal table</em>.

\\ z = \frac{x - \mu}{\sigma} [1]

x = 8.5 ppm and the question is about \\ P(x>8.5)=?  

Using [1]

\\ z = \frac{8.5 - 8.6}{1.3}

\\ z = \frac{-0.1}{1.3}

\\ z = -0.07692 \approx -0.08 (standard normal table has entries for two decimals places for z).

For \\ z = -0.08, is \\ P(z.

But, we are asked for \\ P(z>-0.08) \approx P(x>8.5).

\\ P(z-0.08) = 1

\\ P(z>-0.08) = 1 - P(z

\\ P(z>-0.08) = 0.5319

Thus, "the probability that one randomly selected city's waterway will have more than 8.5 ppm pollutants" is \\ P(z>-0.08) = 0.5319.

d. For the 38 cities, find the probability that the average amount of pollutants is more than 8.5 ppm.

Or \\ P(\overline{x} > 8.5)ppm?

This random variable follows a standardized random variable normally distributed, i.e. \\ Z \sim N(0, 1):

\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}} [2]

\\ z = \frac{\overline{8.5} - 8.6}{\frac{1.3}{\sqrt{38}}}

\\ z = \frac{-0.1}{0.21088}

\\ z = \frac{-0.1}{0.21088} \approx -0.47420 \approx -0.47

\\ P(z

Again, we are asked for \\ P(z>-0.47), then

\\ P(z>-0.47) = 1 - P(z

\\ P(z>-0.47) = 1 - 0.3192

\\ P(z>-0.47) = 0.6808

Then, the probability that the average amount of pollutants is more than 8.5 ppm for the 38 cities is \\ P(z>-0.47) = 0.6808.

e. For part d), is the assumption that the distribution is normal necessary?

For this question, we do not need to assume that the distribution from we take the sample is normal. We already know that the distribution for X is normally distributed. Moreover, the distribution for \\ \overline{X} is also normal because the sample was taken from a normal distribution. Additionally, the sample size is large enough to show a bell-shaped distribution.  

f. Find the IQR for the average of 38 cities.

We must find the first quartile (25th percentile), and the third quartile (75th percentile). For \\ P(z, \\ z \approx -0.68, then, using [2]:

\\ -0.68 = \frac{\overline{X} - 8.6}{\frac{1.3}{\sqrt{38}}}

\\ (-0.68 *0.21088) + 8.6 = \overline{X}

\\ \overline{x} =8.4566

\\ Q1 = 8.4566 ppm.

For Q3

\\ 0.68 = \frac{\overline{X} - 8.6}{\frac{1.3}{\sqrt{38}}}

\\ (0.68 *0.21088) + 8.6 = \overline{X}

\\ \overline{x} =8.7434

\\ Q3 = 8.7434 ppm.

\\ IQR = Q3-Q1 = 8.7434 - 8.4566 = 0.2868 ppm

Therefore, the IQR for the average of 38 cities is \\ IQR = 0.2868 ppm. \\ Q1 = 8.4566 ppm and \\ Q3 = 8.7434 ppm.

4 0
3 years ago
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