Question

Suppose a random sample of n = 25 observations is selected from a normal population, with mean 106 and standard deviation 12. a. Find the probability that x-bar exceeds 110. Show the mean and standard deviation you used.
b. Find the probability that the sample mean deviates from the population mean = 106 by no more than 4.
c. Sketch the probability distribution from part B, showing the area that represents the probability you found.

Answer 1

Answer:

a) On this case $\bar X \sim N(106,\frac{12}{\sqrt{25}})$

$P(\bar X>110)=P(\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}>\frac{110-\mu}{\frac{\sigma}{\sqrt{n}}})$

$=P(Z>\frac{110-106}{\frac{12}{\sqrt{25}}})=P(Z>1.667)$

And we can find this probability on this way:

$P(Z>1.667)=1-P(Z$

b) $P(106-4$

$=P(\frac{102-106}{\frac{12}{\sqrt{25}}}$

And we can find this probability on this way:

$P(-1.667$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

$P(-1.667$

c) Figure attached

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the variable of a population, and for this case we know the distribution for X is given by:

$X \sim N(106,12)$  

Where $\mu=106$ and $\sigma=12$

And let $\bar X$ represent the sample mean, the distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

Part a

On this case $\bar X \sim N(106,\frac{12}{\sqrt{25}})$

We are interested on this probability

$P(\bar X>110)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

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

$P(\bar X>110)=P(\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}>\frac{110-\mu}{\frac{\sigma}{\sqrt{n}}})$

$=P(Z>\frac{110-106}{\frac{12}{\sqrt{25}}})=P(Z>1.667)$

And we can find this probability on this way:

$P(Z>1.667)=1-P(Z$

Part b

For this case we want the probability that the sample mean lies between these two values:

$P(106-4$

$=P(\frac{102-106}{\frac{12}{\sqrt{25}}}$

And we can find this probability on this way:

$P(-1.667$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

$P(-1.667$

Part c

See the figure attached.