Question

The amount of pollutants that are found in waterways near large cities is normally distributed with mean 9 ppm and standard deviation 1.5 ppm. 38 randomly selected large cities are studied. Round all answers to 4 decimal places where possible.

1. What is the distribution of XX? XX ~ N(,)
2. What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)
3. What is the probability that one randomly selected city's waterway will have more than 9.6 ppm pollutants?
4. For the 37 cities, find the probability that the average amount of pollutants is more than 9.6 ppm.
5. For part d), is the assumption that the distribution is normal necessary? YesNo
6. Find the IQR for the average of 37 cities.
Q1 = ppm
Q3 = ppm
IQR: ppm

Answer 1

Explanation:

Hello!

There are two values of n in the text, I'll use the one that appears in all the questions.

The variable of interest is

X: pollutants found in waterways near large cities. (ppm)

This variable has a normal distribution with parameters μ= 9ppm and σ= 1.5ppm

1) X~N(μ;σ²)

X~N(9;2.25)

2) The distribution of the sample mean is X~N(μ;σ²/n)

σ²/n= 2.25/37= 0.06

X~N(9;0.06)

3) P(X>9.6)

To calculate this probability you have to use the standard normal distribution. Using the population parameters, you can calculate the corresponding Z value:

Z= (X-μ)/σ= (9.6-9)/1.5= 0.4

P(Z>0.4)= 1-P(Z≤0.4)= 1 - 0.65542= 0.34458

The probability of selecting a city at random and finding 9.6ppm pollutants.

4) In this item, instead of calculating the probability of one value of the variable you have to calculate the probability of the sample average taking a determined value. Because of this, you have to work using the distribution of the sample mean, instead of the distribution of the variable.

P(X[bar]>9.6)

Z= (X[bar]-μ)/(σ/√n)= (9.6-9)/√0.06= 2.45

P(Z>2.45)= 1 - P(Z≤2.45)= 1 - 0.99286= 0.00714

5) The assumption of a normal distribution is not necessary for item 4. Since the sample size is large enough (greater than 30) you can apply the central limit theorem and approximate the distribution of the sample mean to normal, regarding the distribution of the original variable.

6)

In this case, you have to work starting with the standard normal distribution and then "translate" the Z values into values of the average amount of pollutants.

The first quartile divides the bottom 25% of the distribution from the top 75%, symbolically:

P(Z≤z₁)= 0.25

z₁= -0.674

z₁= (X[bar]-μ)/(σ/√n)

z₁*(√n/σ)=X[bar]-μ

X[bar]=z₁*(√n/σ)+μ

X[bar]=(-0.674)*(√37/1.5)+9= 6.27ppm

The third quartile divides the bottom 75% of the distribution from the top 25%, symbolically:

P(Z≤z₂)= 0.75

z₂= 0.674

z₂= (X[bar]-μ)/(σ/√n)

z₂*(√n/σ)=X[bar]-μ

X[bar]=z₂*(√n/σ)+μ

X[bar]=(0.674)*(√37/1.5)+9= 11.7.3ppm

IQR= Q₃-Q₁= 11.73-6.27= 5.46ppm

I hope this helps!