Question

a.

A population consists of the following five values: 17, 17, 24, 18, and 18.

	List all samples of size 3, and compute the mean of each sample. (Round your Mean values to 2 decimal places.)

Sample	Values	Sum	Mean
1	(Click to select)17,24,1817,18,1817,17,2417,17,18
2	(Click to select)17,17,2417,24,1817,17,1817,18,18
3	(Click to select)17,17,1817,17,2417,18,1817,24,18
4	(Click to select)17,17,1817,18,1817,17,2417,24,18
5	(Click to select)17,24,1817,17,2417,17,1817,18,18
6	(Click to select)17,24,1817,17,2417,17,1817,18,18
7	(Click to select)17,18,1817,24,1824,18,1817,17,24
8	(Click to select)17,17,2417,18,1824,18,1817,24,18
9	(Click to select)17,24,1817,17,2424,18,1817,18,18
10	(Click to select)17,18,1817,17,2417,24,1824,18,18

        

(b)	Compute the mean of the distribution of sample means and the population mean. (Round your answers to 2 decimal places.)

Mean of the distribution of the sample mean
Population mean

B.

A normal population has a mean of 80 and a standard deviation of 3. You select a sample of 44.

    

Compute the probability the sample mean is: (Round z values to 2 decimal places and final answers to 4 decimal places.)

   

(a)	Less than 79.

Probability

(b)	Between 79 and 81.

Probability

(c)	Between 81 and 82.

Probability

(d)	Greater than 82.

    

Probability

c.

At the downtown office of First National Bank, there are five tellers. Last week, the tellers made the following number of errors each: 3, 7, 2, 3, and 1.

    

(a)	How many different samples of 2 tellers are possible without replacement?

Different samples

    

(b)	List all possible samples of 2 observations each from left to right without replacement and compute the mean of each sample. (Round your Mean value answers to 1 decimal place.)

        

Sample	Values	Sum	Mean
1	(Click to select)7,37,23,23,7
2	(Click to select)3,37,13,27,3
3	(Click to select)3,37,37,13,1
4	(Click to select)7,33,17,22,3
5	(Click to select)3,22,17,37,2
6	(Click to select)2,17,37,12,3
7	(Click to select)7,37,12,32,1
8	(Click to select)3,12,37,32,1
9	(Click to select)2,33,12,17,3
10	(Click to select)7,33,12,32,1
		Total

     

(c)	Compute the mean of the sample means and compare it to the population mean. (Round your answers to 1 decimal places.)

The mean of the sample means is
The population mean is

        

Both means are

(Click to select)Not equalEqual

Answer 1

a)

values	sum	mean
17,17	34	17
17,24	41	20.5
17,18	35	17.5
17,18	35	17.5
17,24	41	20.5
17,18	35	17.5
17,18	35	17.5
24,18	42	21
24,18	42	21
18,18	36	18

Mean of the distribution of the sample mean =18.8

population mean =18.8

b)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	80
std deviation   =σ=	3.0000
sample size       =n=	44
std error=σx̅=σ/√n=	0.4523

a)

probability =

P(X<79)

=

P(Z<-2.21)=

0.0136

b)

probability =

P(79<X<81)

=

P(-2.21<Z<2.21)=

0.9864-0.0136=

0.9728

c)

probability =

P(X>82)

=

P(Z>4.42)=

1-P(Z<4.42)=

1-1=

0.0000

c)

different samples of 2 tellers are possible without replacement =10

b)

values	sum	mean
3,7	10	5
3,2	5	2.5
3,3	6	3
3,1	4	2
7,2	9	4.5
7,3	10	5
7,1	8	4
2,3	5	2.5
2,1	3	1.5
3,1	4	2

c)

The mean of the sample means is =3.2

The population mean is =3.2

Both means are equal