Question

A set of data is normally distributed with a mean of 28 and a standard deviation of 2. What is the probability that a value from the data is less than 22 or greater than 34?
a.

9%

c.

4.5%

b.

0.6%

d.

0.3%



Please select the best answer from the choices provided


A

B

C

D

Answer 1

Answer:

$P(X$

$P(X>34)=P(\frac{X-\mu}{\sigma}>\frac{34-\mu}{\sigma})=P(Z>\frac{34-28}{2})=P(z>3)=0.00135$

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:

$X \sim N(28,2)$  

Where $\mu=28$ and $\sigma=2$

We are interested on this probability

$P(X34)$

We can find the individual probabilities. We can use the z score formula given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

$P(X>34)=P(\frac{X-\mu}{\sigma}>\frac{34-\mu}{\sigma})=P(Z>\frac{34-28}{2})=P(z>3)=0.00135$

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