Question

the amount of pollutants that are found in waterways near large cities is normally distributed with mean 9.9 ppm and standard deviation 1.4 ppm. 10 randomly selected large cities are studied. round all answers to 4 decimal places where possible.
a. what is the distribution of x? x - n(, )
b. what is the distribution of \\( \overline { x } \\)? \\( \overline { x } \\) - n(, )
c. what is the probability that one randomly selected citys waterway will have more than 10 ppm pollutants?
d. for the 10 cities, find the probability that the average amount of pollutants is more than 10 ppm.
e. for part d), is the assumption that the distribution is normal necessary? \\( \bigcirc \\) no \\( \bigcirc \\) yes
f. find the iqr for the average of 10 cities.
q1 = ppm
q3 = ppm
iqr: ppm

Explanation:

Step1: Distribution of \(X\)

Given \(X\) (amount of pollutants) is normally distributed. Mean \(\mu = 9.9\), standard deviation \(\sigma=1.4\). So \(X\sim N(9.9,1.4^{2})\)

Step2: Distribution of \(\bar{X}\)

For a sample of size \(n = 10\), the mean of \(\bar{X}\) is \(\mu_{\bar{X}}=\mu = 9.9\), and the standard deviation \(\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{1.4}{\sqrt{10}}\approx0.4427\). So \(\bar{X}\sim N(9.9,0.4427^{2})\)

Step3: Probability \(P(X > 10)\)

First, calculate the \(z\) - score: \(z=\frac{x-\mu}{\sigma}=\frac{10 - 9.9}{1.4}=\frac{0.1}{1.4}\approx0.0714\)
\(P(X>10)=1 - P(X\leq10)=1-\Phi(0.0714)\)
Using the standard normal table or calculator, \(\Phi(0.0714)\approx0.5287\)
\(P(X > 10)=1 - 0.5287=0.4713\)

Step4: Probability \(P(\bar{X}>10)\)

Calculate the \(z\) - score for \(\bar{X}\): \(z=\frac{\bar{x}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}=\frac{10 - 9.9}{0.4427}\approx0.226\)
\(P(\bar{X}>10)=1 - P(\bar{X}\leq10)=1-\Phi(0.226)\)
Using the standard normal table or calculator, \(\Phi(0.226)\approx0.589\)
\(P(\bar{X}>10)=1 - 0.589 = 0.411\)

Step5: Necessity of normality assumption for part (d)

Since \(n = 10<30\), the central limit theorem (which says that for \(n\geq30\), \(\bar{X}\) is approximately normal regardless of the population distribution) does not apply. So the assumption that the population is normal is necessary.

Step6: Find \(Q1\) and \(Q3\) for \(\bar{X}\)

For \(Q1\) (25th percentile), \(z_{0.25}=- 0.6745\)
\(\bar{x}_{Q1}=\mu_{\bar{X}}+z_{0.25}\sigma_{\bar{X}}=9.9+( - 0.6745)\times0.4427\)
\(\bar{x}_{Q1}=9.9-0.6745\times0.4427\approx9.9 - 0.2987=9.6013\)
For \(Q3\) (75th percentile), \(z_{0.75}=0.6745\)
\(\bar{x}_{Q3}=\mu_{\bar{X}}+z_{0.75}\sigma_{\bar{X}}=9.9 + 0.6745\times0.4427\approx9.9+0.2987 = 10.1987\)
\(IQR=Q3 - Q1=10.1987 - 9.6013=0.5974\)

Answer:

a. \(X\sim N(9.9,1.4^{2})\)
b. \(\bar{X}\sim N(9.9,0.4427^{2})\)
c. \(0.4713\)
d. \(0.4110\)
e. Yes
f. \(Q1 = 9.6013\) ppm, \(Q3 = 10.1987\) ppm, \(IQR = 0.5974\) ppm