Question

According to one study, brain weights of men are normally distributed with a mean of 1.60 kg and a standard deviation of 0.15 kg. Use the data to answer questions (a) through (e). a. Determine the sampling distribution of the sample mean for samples of size 3. The mean of the sample mean ish- The standard deviation of the sample mean is :-D (Round to four decimal places as needed.) b. Determine the sampling distribution of the sample mean for samples of size 12. The mean of the sample mean is H. The standard deviation of the sample mean is on (Round to four decimal places as needed.) c. Construct a graph of the normal population distribution and the two sampling distributions for brain weights. Choose the correct graph below. OA OB. Oc. Q n = 12 n=3 Pop Q n = 3 2 = 12 n = 12 Pop 18 16 16 d. Determine the percentage of all samples of three men that have mean brain weights within 0.1 kg of the population mean brain weight of 1.60 kg. % Click to select your answeris)

Answer 1

a. n=3

The mean of sample mean

ui 1.6

The standard deviation of the sample mean is $\sigma_{\bar{x}}=0.0866$

b.n=12

The mean of sample mean

ui 1.6

The standard deviation of the sample mean is $\sigma_{\bar{x}}=0.0433$

c.The correct graph is B.

d. The percentage of all sample of three men that have mean brain weight within 0.1 kg of mean brain weight of 1.6 kg:75.18%

-----------------------------------------------------------------------------------------------------------------------------------------------------------------

The sample mean is same as $\mu$

For n=3, The standard deviation of the sample mean is

01 1.6 V3 = 0.0866 n

For n=3, The standard deviation of the sample mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{1.6}{\sqrt{12}}=0.0433$

d.

The percentage of all sample of three men that have mean brain weight within 0.1 kg of mean brain weight of 1.6 kg:

ie $P(1.5<\bar{X}<1.7)$ when n=3. We know that when n=3, we know that $\sigma_{\bar{x}}=0.0866$

When  $\bar{X}=1.5$, we have $Z=\frac{\bar{X}-1.6}{0.0866}=\frac{1.5-1.6}{0.0866}=-1.1547$

When $\bar{X}=1.7$ , we have $Z=\frac{\bar{X}-1.6}{0.0866}=\frac{1.7-1.6}{0.0866}=1.1547$

$P(1.5<\bar{X}<1.7)=P(-1.1547<Z<1.1547)=P(Z<1.1547)-P(Z<-1.1547)$

$P(1.5<\bar{X}<1.7)=0.8759-0.1241=0.7518$