Question

The amount of pollutants that are found in waterways near large cities is normally distributed with mean 8.5 ppm and standard deviation 1.4 ppm. 18 randomly selected large cities are studied. Round all answers to two decimal places.
A. xBar~ N( ____) (____)
B. For the 18 cities, find the probability that the average amount of pollutants is more than 9 ppm.
C. What is the probability that one randomly selected city's waterway will have more than 9 ppm pollutants?
D. Find the IQR for the average of 18 cities.Q1 =
Q3 =
IQR:
2. X ~ N(30,10). Suppose that you form random samples with sample size 4 from this distribution. Let xBar be the random variable of averages. Let ΣX be the random variable of sums. Round all answers to two decimal places.
A. xBar~ N(___) (____)
B. P(xBar<30) =
C. Find the 95th percentile for the xBar distribution.
D. P(xBar > 36)=
E. Q3 for the xBar distribution =

Answer 1

Answer:

1)

A) $\frac{}{X}$ ~ N(8.5;0.108)

B) P($\frac{}{X}$ > 9)= 0.0552

C) P(X> 9)= 0.36317

D) IQR= 0.4422

2)

A) $\frac{}{X}$ ~ N(30;2.5)

B) P( $\frac{}{X}$<30)= 0.50

C) P₉₅= 32.60

D) P( $\frac{}{X}$>36)= 0

E) Q₃: 31.0586

Step-by-step explanation:

Hello!

1)

The variable of interest is

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

This variable has a normal distribution:

X~N(μ;σ²)

μ= 8.5 ppm

σ= 1.4 ppm

A sample of 18 large cities were studied.

A) The sample mean is also a random variable and it has the same distribution as the population of origin with exception that it's variance is affected by the sample size:

$\frac{}{X}$ ~ N(μ;σ²/n)

The population mean is the same as the mean of the variable

μ= 8.5 ppm

The standard deviation is

σ/√n= 1.4/√18= 0.329= 0.33 ⇒σ²/n= 0.33²= 0.108

So: $\frac{}{X}$ ~ N(8.5;0.108)

B)

P($\frac{}{X}$ > 9)= 1 - P($\frac{}{X}$ ≤ 9)

To calculate this probability you have to standardize the value of the sample mean and then use the Z-tables to reach the corresponding value of probability.

Z= $\frac{\frac{}{X} - Mu}{\frac{Sigma}{\sqrt{n} } } = \frac{9-8.5}{0.33}= 1.51$

Then using the Z table you'll find the probability of

P(Z≤1.51)= 0.93448

Then

1 - P($\frac{}{X}$ ≤ 9)= 1 - P(Z≤1.51)= 1 - 0.93448= 0.0552

C)

In this item, since only one city is chosen at random, instead of working with the distribution of the sample mean, you have to work with the distribution of the variable X:

P(X> 9)= 1 - P(X ≤ 9)

Z= (X-μ)/δ= (9-8.5)/1.44

Z= 0.347= 0.35

P(Z≤0.35)= 0.63683

Then

P(X> 9)= 1 - P(X ≤ 9)= 1 - P(Z≤0.35)= 1 - 0.63683= 0.36317

D)

The first quartile is the value of the distribution that separates the bottom 2% of the distribution from the top 75%, in this case it will be the value of the sample average that marks the bottom 25% symbolically:

Q₁: P($\frac{}{X}$≤$\frac{}{X}$₁)= 0.25

Which is equivalent to the first quartile of the standard normal distribution. So first you have to identify the first quartile for the Z dist:

P(Z≤z₁)= 0.25

Using the table you have to identify the value of Z that accumulates 0.25 of probability:

z₁= -0.67

Now you have to translate the value of Z to a value of $\frac{}{X}$:

z₁= ($\frac{}{X}$₁-μ)/(σ/√n)

z₁*(σ/√n)= ($\frac{}{X}$₁-μ)

$\frac{}{X}$₁= z₁*(σ/√n)+μ

$\frac{}{X}$₁= (-0.67*0.33)+8.5=  8.2789 ppm

The third quartile is the value that separates the bottom 75% of the distribution from the top 25%. For this distribution, it will be that value of the sample mean that accumulates 75%:

Q₃: P($\frac{}{X}$≤$\frac{}{X}$₃)= 0.75

⇒ P(Z≤z₃)= 0.75

Using the table you have to identify the value of Z that accumulates 0.75 of probability:

z₃= 0.67

Now you have to translate the value of Z to a value of $\frac{}{X}$:

z₃= ($\frac{}{X}$₃-μ)/(σ/√n)

z₃*(σ/√n)= ($\frac{}{X}$₃-μ)

$\frac{}{X}$₃= z₃*(σ/√n)+μ

$\frac{}{X}$₃= (0.67*0.33)+8.5=  8.7211 ppm

IQR= Q₃-Q₁= 8.7211-8.2789= 0.4422

2)

A)

X ~ N(30,10)

For n=4

$\frac{}{X}$ ~ N(μ;σ²/n)

Population mean μ= 30

Population variance σ²/n= 10/4= 2.5

Population standard deviation σ/√n= √2.5= 1.58

$\frac{}{X}$ ~ N(30;2.5)

B)

P( $\frac{}{X}$<30)

First you have to standardize the value and then look for the probability:

Z=  ($\frac{}{X}$-μ)/(σ/√n)= (30-30)/1.58= 0

P(Z<0)= 0.50

Then

P( $\frac{}{X}$<30)= 0.50

Which is no surprise since 30 y the value of the mean of the distribution.

C)

P( $\frac{}{X}$≤ $\frac{}{X}$₀)= 0.95

P( Z≤ z₀)= 0.95

z₀= 1.645

Now you have to reverse the standardization:

z₀= ($\frac{}{X}$₀-μ)/(σ/√n)

z₀*(σ/√n)= ($\frac{}{X}$₀-μ)

$\frac{}{X}$₀= z₀*(σ/√n)+μ

$\frac{}{X}$₀= (1.645*1.58)+30= 32.60

P₉₅= 32.60

D)

P( $\frac{}{X}$>36)= 1 - P( $\frac{}{X}$≤36)= 1 - P(Z≤(36-30)/1.58)= 1 - P(Z≤3.79)= 1 - 1 = 0

E)

Q₃: P($\frac{}{X}$≤$\frac{}{X}$₃)= 0.75

⇒ P(Z≤z₃)= 0.75

z₃= 0.67

z₃= ($\frac{}{X}$₃-μ)/(σ/√n)

z₃*(σ/√n)= ($\frac{}{X}$₃-μ)

$\frac{}{X}$₃= z₃*(σ/√n)+μ

$\frac{}{X}$₃= (0.67*1.58)+30= 31.0586

Q₃: 31.0586