Answer:
[tex]P(X<22)=P(\frac{X-\mu}{\sigma}<\frac{22-\mu}{\sigma})=P(Z<\frac{22-28}{2})=P(z<-3)=0.00135[/tex]
[tex]P(X>34)=P(\frac{X-\mu}{\sigma}>\frac{34-\mu}{\sigma})=P(Z>\frac{34-28}{2})=P(z>3)=0.00135[/tex]
And if we add the two values we got 0.00135+0.00135 = 0.0027 and in % would be 0.27% and rounded would be 0.3 %
d. 0.3%
D
Step-by-step explanation:
Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(28,2)[/tex]
Where [tex]\mu=28[/tex] and [tex]\sigma=2[/tex]
We are interested on this probability
[tex]P(X<22 \cup X>34)[/tex]
We can find the individual probabilities. We can use the z score formula given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X<22)=P(\frac{X-\mu}{\sigma}<\frac{22-\mu}{\sigma})=P(Z<\frac{22-28}{2})=P(z<-3)=0.00135[/tex]
[tex]P(X>34)=P(\frac{X-\mu}{\sigma}>\frac{34-\mu}{\sigma})=P(Z>\frac{34-28}{2})=P(z>3)=0.00135[/tex]
And if we add the two values we got 0.00135+0.00135 = 0.0027 and in % would be 0.27% and rounded would be 0.3 %
d. 0.3%
D
