Question

A random variable is normally distributed. It has a mean of 245 and a standard deviation of 21.

Answer 1

Answer:

a) $\bar X \sim N(\mu=245, \frac{21}{\sqrt{10}}=6.64)$

So then the distribution is also normal but with another deviation. And the reason is because"

$E(\bar X) = \mu , Var(\bar X)=\frac{\sigma^2}{n}, Sd(\bar X)= \frac{\sigma}{\sqrt{n}}$

b) $E(\bar X) = \mu = 245$

$\sigma_{\bar X}= \frac{21}{\sqrt{10}}=6.64$

c) $P(\bar X >241)=P(Z>\frac{241-245}{\frac{21}{\sqrt{10}}}=-0.602)$

And using the complment rule, the normal standard deviation or excel we got:

$P(Z>-0.602)=1-P(Z$

d) Since the distribution for X is normal then the distribution for the sample mean $\bar X$ is given by:

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

And the reason is because"

$E(\bar X) = \mu , Var(\bar X)=\frac{\sigma^2}{n}, Sd(\bar X)= \frac{\sigma}{\sqrt{n}}$

e) $E(\bar X) = \mu = 245$

$\sigma_{\bar X}= \frac{21}{\sqrt{35}}=3.550$

f) $P(\bar X >241)=P(Z>\frac{241-245}{\frac{21}{\sqrt{35}}}=-1.127)$

And using the complment rule, the normal standard deviation or excel we got:

$P(Z>-1.127)=1-P(Z$

g) The probability for part f is lower since the deviation for the sample size of n =35 is lower compared to the deviation with a sample size of n =10.

Step-by-step explanation:

For this case we assume that we have a random variable X and the distribution of X is given by:

$X\sim N(\mu = 245, \sigma =21)$

Part a

Since the distribution for X is normal then the distribution for the sample mean $\bar X$ is given by:

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

And for this case n =10 represent the sample size we have:

$\bar X \sim N(\mu=245, \frac{21}{\sqrt{10}}=6.64)$

So then the distribution is also normal but with another deviation. And the reason is because"

$E(\bar X) = \mu , Var(\bar X)=\frac{\sigma^2}{n}, Sd(\bar X)= \frac{\sigma}{\sqrt{n}}$

Part b

$E(\bar X) = \mu = 245$

$\sigma_{\bar X}= \frac{21}{\sqrt{10}}=6.64$

Part c

For this case we can use the z score formula given by:

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

And we want this probability:

$P(\bar X >241)=P(Z>\frac{241-245}{\frac{21}{\sqrt{10}}}=-0.602)$

And using the complment rule, the normal standard deviation or excel we got:

$P(Z>-0.602)=1-P(Z$

Part d

Since the distribution for X is normal then the distribution for the sample mean $\bar X$ is given by:

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

And the reason is because"

$E(\bar X) = \mu , Var(\bar X)=\frac{\sigma^2}{n}, Sd(\bar X)= \frac{\sigma}{\sqrt{n}}$

Part e

$E(\bar X) = \mu = 245$

$\sigma_{\bar X}= \frac{21}{\sqrt{35}}=3.550$

Part f

For this case we can use the z score formula given by:

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

And we want this probability:

$P(\bar X >241)=P(Z>\frac{241-245}{\frac{21}{\sqrt{35}}}=-1.127)$

And using the complment rule, the normal standard deviation or excel we got:

$P(Z>-1.127)=1-P(Z$

Part g

The probability for part f is higher since the deviation for the sample size of n =35 is lower compared to the deviation with a sample size of n =10.