The maximum contamination level of arsenic ion in a water system is 0.050 parts per million. If the arsenic is present as AsCl3,
1 answer:
Answer:
mass AsCl₃ in 8.2x10⁵ L = 99.2gAsCl₃
Step-by-step explanation:
Let's remember that 0.05 ppm (parts per million) would be write terms of mg/L. We have 0.05 mg/L, that is a concentration unit.
First of all we need to find the g/L of AsCl₃ present in water.
Now, we just need to multiply it by 8.2 x 10⁵ L to determine the mass of AsCl₃ present in that volume.
Finally we have:
.
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Answer:
1)
A) ~ N(8.5;0.108)
B) P( > 9)= 0.0552
C) P(X> 9)= 0.36317
D) IQR= 0.4422
2)
A) ~ N(30;2.5)
B) P( <30)= 0.50
C) P₉₅= 32.60
D) P( >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:
~ 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: ~ N(8.5;0.108)
B)
P( > 9)= 1 - P(
≤ 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=
Then using the Z table you'll find the probability of
P(Z≤1.51)= 0.93448
Then
1 - P( ≤ 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(≤
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 :
z₁= (₁-μ)/(σ/√n)
z₁*(σ/√n)= (₁-μ)
₁= z₁*(σ/√n)+μ
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(≤
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 :
z₃= (₃-μ)/(σ/√n)
z₃*(σ/√n)= (₃-μ)
₃= z₃*(σ/√n)+μ
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
~ 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
~ N(30;2.5)
B)
P( <30)
First you have to standardize the value and then look for the probability:
Z= (-μ)/(σ/√n)= (30-30)/1.58= 0
P(Z<0)= 0.50
Then
P( <30)= 0.50
Which is no surprise since 30 y the value of the mean of the distribution.
C)
P( ≤
₀)= 0.95
P( Z≤ z₀)= 0.95
z₀= 1.645
Now you have to reverse the standardization:
z₀= (₀-μ)/(σ/√n)
z₀*(σ/√n)= (₀-μ)
₀= z₀*(σ/√n)+μ
₀= (1.645*1.58)+30= 32.60
P₉₅= 32.60
D)
P( >36)= 1 - P(
≤36)= 1 - P(Z≤(36-30)/1.58)= 1 - P(Z≤3.79)= 1 - 1 = 0
E)
Q₃: P(≤
₃)= 0.75
⇒ P(Z≤z₃)= 0.75
z₃= 0.67
z₃= (₃-μ)/(σ/√n)
z₃*(σ/√n)= (₃-μ)
₃= z₃*(σ/√n)+μ
₃= (0.67*1.58)+30= 31.0586
Q₃: 31.0586