Let's solve the given problem step-by-step:
a. What is the distribution of [tex]\(X\)[/tex]?
The amount of pollutants in the waterways is normally distributed with a mean (μ) of 10 ppm and a standard deviation (σ) of 1.4 ppm.
So, we write:
[tex]\[ X \sim N(10, 1.4) \][/tex]
b. What is the distribution of [tex]\(\bar{x}\)[/tex]?
For the sample mean [tex]\(\bar{x}\)[/tex] of 35 randomly selected cities, the distribution also follows a normal distribution due to the Central Limit Theorem. The mean of [tex]\(\bar{x}\)[/tex] is the same as the population mean, and the standard deviation (standard error) is the population standard deviation divided by the square root of the sample size.
- Mean of [tex]\(\bar{x}\)[/tex], [tex]\(\mu_{\bar{x}} = 10\)[/tex]
- Standard error of [tex]\(\bar{x}\)[/tex], [tex]\(\sigma_{\bar{x}} = \frac{1.4}{\sqrt{35}} \approx 0.2366\)[/tex]
So, we write:
[tex]\[ \bar{x} \sim N(10, 0.2366) \][/tex]
c. What is the probability that one randomly selected city's waterway will have less than 10.3 ppm pollutants?
We need to find [tex]\(P(X < 10.3)\)[/tex]. This is the cumulative probability up to 10.3 ppm for a normal distribution with mean 10 ppm and standard deviation 1.4 ppm. The probability calculated is approximately:
[tex]\[ \approx 0.5848 \][/tex]
d. For the 35 cities, find the probability that the average amount of pollutants is less than 10.3 ppm.
We need to find [tex]\(P(\bar{x} < 10.3)\)[/tex] for [tex]\(\bar{x} \sim N(10, 0.2366)\)[/tex]. This is the cumulative probability up to 10.3 ppm for the sample mean distribution. The probability calculated is approximately:
[tex]\[ \approx 0.8976 \][/tex]
e. For part d), is the assumption that the distribution is normal necessary?
Yes, the assumption that the distribution is normal is necessary because we are using the properties of the normal distribution to find the probability.
f. Find the IQR for the average of 35 cities.
The Interquartile Range (IQR) is the difference between the third quartile ([tex]\(Q_3\)[/tex]) and the first quartile ([tex]\(Q_1\)[/tex]) of the distribution of the sample mean [tex]\(\bar{x}\)[/tex].
- First Quartile ([tex]\(Q_1\)[/tex]) for [tex]\(\bar{x} \sim N(10, 0.2366)\)[/tex]:
[tex]\[ Q_1 \approx 9.8404 \][/tex]
- Third Quartile ([tex]\(Q_3\)[/tex]) for [tex]\(\bar{x} \sim N(10, 0.2366)\)[/tex]:
[tex]\[ Q_3 \approx 10.1596 \][/tex]
Calculating the IQR:
[tex]\[ \text{IQR} = Q_3 - Q_1 \approx 0.3192 \][/tex]
So the final values are:
a. [tex]\(X \sim N(10, 1.4)\)[/tex]
b. [tex]\(\bar{x} \sim N(10, 0.2366)\)[/tex]
c. Probability for one city < 10.3 ppm:
[tex]\[ \approx 0.5848 \][/tex]
d. Probability for average of 35 cities < 10.3 ppm:
[tex]\[ \approx 0.8976 \][/tex]
e. Yes
f. [tex]\(Q_1 \approx 9.8404\)[/tex] ppm
[tex]\[ Q_3 \approx 10.1596 \][/tex] ppm
[tex]\[ \text{IQR} \approx 0.3192 \][/tex] ppm
