Question

Suppose x has a distribution with a mean of 75 and a standard deviation of 45. Random samples of size n = 81 are drawn.

Answer 1

Answer:

a)  $X \sim N(\mu=75,\sigma=45)$  

$\bar X \sim N(75,\frac{45}{\sqrt{81}})$

b) $z=\frac{90-75}{\frac{45}{\sqrt{81}}}=3$

c) $P(\bar X$

d) No. it would not be unusual because more than 5% of all such samples hav means less than 90.

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".  

(a) Describe the x distribution and compute the mean and standard deviation of the distribution.

Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:

$X \sim N(\mu=75,\sigma=45)$  

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

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

On this case  $\bar X \sim N(75,\frac{45}{\sqrt{81}})$

(b) Find the z value corresponding to $\bar X = 90$.

The z score on this case is given by this formula:

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

And if we replace we got:

$z=\frac{90-75}{\frac{45}{\sqrt{81}}}=3$

(c) Find $P(\bar X < 90)$.

For this case we can use a table or excel to find the probability required:

$P(\bar X$

(d) Would it be unusual for a random sample of size 81 from the x distribution to have a sample mean less than 90? Explain.

For this case the best conclusion is:

No. it would not be unusual because more than 5% of all such samples hav means less than 90.