Question

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

a.) If you take a sample of size 10, can you say what the shape of the distribution for the sample mean is? Why?
b.) For a sample of size 10, state the mean of the sample mean and the standard deviation of the sample mean.
c.) For a sample of size 10, find the probability that the sample mean is more than 241.
d.) If you take a sample of size 35, can you say what the shape of the distribution of the sample mean is? Why?
e.) For a sample of size 35, state the mean of the sample mean and the standard deviation of the sample mean.
f.) For a sample of size 35, find the probability that the sample mean is more than 241. g.) Compare your answers in part c and f. Why is one smaller than the other?

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:

$P(Z>-0.602)=1-P(Z<-0.602)= 1-0.274=0.726$

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:

$P(Z>-1.127)=1-P(Z<-1.127)= 1-0.130=0.870$

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:

$P(Z>-0.602)=1-P(Z<-0.602)= 1-0.274=0.726$

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:

$P(Z>-1.127)=1-P(Z<-1.127)= 1-0.130=0.870$

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.

Answer 2

Using the normal distribution and the central limit theorem, it is found that:

a) Since the underlying distribution is normal, the shape of the distribution for the sample mean is normal.

b) It has sample mean of 245 and standard deviation of 6.64.

c) 0.7257 = 72.57% probability that the sample mean is more than 241.

d) The sample size is greater than 30, hence, no matter the underlying distribution, the shape is normal.

e) It has sample mean of 245 and standard deviation of 3.35.

f) 0.883 = 88.3% probability that the sample mean is more than 241.

g) Due to different sample sizes, the standard errors will be different, hence, one probability is smaller than the other.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  
• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
• By the Central Limit Theorem, the sampling distribution of sample means of size n has mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$. Also, for normal variables, this distribution is always normal, while for skewed variables, it is normal for sample sizes above 30.

In this problem:

• Mean of 245, hence $\mu = 245$
• Standard deviation of 21, hence $\sigma = 21$.

Item a:

Since the underlying distribution is normal, the shape of the distribution for the sample mean is normal.

Item b:

Sample of size 10, hence $n = 10$ and:

$\mu = 245$

$s = \frac{21}{\sqrt{10}} = 6.64$

It has sample mean of 245 and standard deviation of 6.64.

Item c:

This probability is 1 subtracted by the p-value of Z when X = 241, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{241 - 245}{6.64}$

$Z = -0.6$

$Z = -0.6$ has a p-value of 0.2743.

1 - 0.2743 = 0.7257

0.7257 = 72.57% probability that the sample mean is more than 241.

Item d:

The sample size is greater than 30, hence, no matter the underlying distribution, the shape is normal.

Item e:

Sample of size 35, hence $n = 35$ and:

$\mu = 245$

$s = \frac{21}{\sqrt{35}} = 3.55$

It has sample mean of 245 and standard deviation of 3.35.

Item f:

$Z = \frac{X - \mu}{s}$

$Z = \frac{241 - 245}{3.35}$

$Z = -1.19$

$Z = -1.19$ has a p-value of 0.117.

1 - 0.117 = 0.883.

0.883 = 88.3% probability that the sample mean is more than 241.

Item g:

Due to different sample sizes, the standard errors will be different, hence, one probability is smaller than the other.

A similar problem is given at https://brainly.com/question/24663213