Question

The amount of pollutants that are found in waterways near large cities is normally distributed with mean 9.9 ppm and standard deviation 1.8 ppm. 39 randomly selected large cities are studied. Round all answers to 4 decimal places where possible.

Answer 1

a. 9.9, 1.8

b. 9.9, 0.2882

c. 0.5239

d. 0.6368

e. No

f.

Q1 = 9.7069

Q3 = 10.0931

IQR = 0.3862

Explanation:

a.

X ~ N (9.9, 1.8)

b.

x ~ N (9.9, 1.8/√39)

x ~ N (9.9, 0.2882)

c.

P(X > 9.8)

We calculate the probability as follows:

`\begin{gathered} P\left(X>9.8\right)=1-p\left((X-9.9)/(1.8)<(9.8-9.9)/(1.8)\right) \\ \\ P\left(X>9.8\right)=1-p(z<-0.06) \\ \\ P\left(X>9.8\right)=1-0.4761 \\ \\ P\left(X>9.8\right)=0.5239 \end{gathered}`

d.

p (x > 9.8)

We calculate the probability as follows:

`\begin{gathered} P\left(x>9.8\right)=1-p\left((X-9.9)/((1.8)/(√(39)))<(9.8-9.9)/((1.8)/(√(39)))\right) \\ \\ P\left(x>9.8\right)=1-p(z<-0.35) \\ \\ P\left(x>9.8\right)=1-0.3632 \\ \\ P\left(x>9.8\right)=0.6368 \end{gathered}`

e.

No, you don't need to make the assumption

f.

Q1 = 0.25

In this case the value of z = 0.25, so we look for the closest value in the normal table, like this:

Thanks to this, we make the following equation:

`\begin{gathered} -0.67=(x-9.9)/((1.8)/(√(35))) \\ \\ x-9.9=-0.19311 \\ \\ x=-0.1931+9.9 \\ \\ x=9.7069 \\ \\ Q_1=9.7069 \end{gathered}`

Q3 = 0.75

In this case the value of z = 0.75, so we look for the closest value in the normal table, like this:

Therefore:

`\begin{gathered} -0.67=(x-9.9)/((1.8)/(√(39))) \\ \\ x-9.9=0.1931 \\ \\ x=0.1931+9.9 \\ \\ x=10.0931 \\ \\ Q_3=10.0931-9.7069 \end{gathered}`

Therefore, the interquartile range would be:

`\begin{gathered} IQR=Q_3-Q_1 \\ \\ IQR=10.0931-9.7069 \\ \\ IQR=0.3862 \end{gathered}`