Understanding Basic Statistics Solutions 1

 These are my work from taking 'Statistics' class at Hudson County Community College. (Mar 2015-May 2015)

• Book : Understanding Basic Statistics 
• ISBN-13:9780495831488, ISBN-10:0495831484
• For  students, please use these as resources, do not  plagiarize.
• If you have a question, please leave a comment.
• Graphs and Formulas may not shown. 

Solutions part 1.

http://jieunkimresume.blogspot.com/2015/05/understanding-basic-statistics.html

Solutions part 2.

http://jieunkimresume.blogspot.com/2015/05/understanding-basic-statistics_21.html

Page 45, #7, Jieun Kim Unit1

 (a) The class width = 25

Min : 236, Max: 360

The Class width, Calculation: (360-236)/5=24.8≈25

Increase the value to 25.

(b) Frequency Table

Class Limits
Lower-Upper	Class Boundaries	Midpoints	Frequency	Relative Frequency
236-260	235.5-260.5	248	4	0.070
261-285	260.5-285.5	273	9	0.158
286-310	285.5-310.5	298	25	0.438
311-335	310.5-335.5	323	16	0.281
336-360	335.5-360.5	348	3	0.053

(c) Histogram

Class Boundaries	Frequency
235.5-260.5	4
260.5-285.5	9
285.5-310.5	25
310.5-335.5	16
335.5-360.5	3

(d)Relative-Frequency Histogram

X : Finish Times(hours)

Y: Relative Frequency

Class Boundaries	Relative Frequency
235.5-260.5	0.070
260.5-285.5	0.158
285.5-310.5	0.438
310.5-335.5	0.281
335.5-360.5	0.053

(e) Mound-shaped symmetrical histogram

46p, #8.

(a) The class width = 11

min=45

max=109

(109-45)/6=64/6=10/6≈11

Increase the value to 11.

(b) Frequency Table

Class Limits
Lower-Upper	Class Boundaries	Midpoints	Frequency	Relative Frequency
45-55	44.5-55.5	50	3	0.043 (3/70)
56-66	55.5-66.5	61	7	0.010 (7/70)
67-77	66.5-77.5	72	22	0.314 (22/70)
78-88	77.5-88.5	83	26	0.371 (26/70)
89-99	88.5-99.5	94	9	0.129 (9/70)
100-110	99.5-110.5	105	3	0.043 (3/70)

(c) Histogram

X : Glucose blood levels(mg/100ml), Y : Frequency

Class Boundaries	Frequency
44.5-55.5	3
55.5-66.5	7
66.5-77.5	22
77.5-88.5	26
88.5-99.5	9
99.5-110.5	3

(d) Relative-Frequency Histogram

X : Glucose blood levels(mg/100ml), Y : Relative Frequency

Class Boundaries	Relative Frequency
44.5-55.5	0.043
55.5-66.5	0.010
66.5-77.5	0.314
77.5-88.5	0.371
88.5-99.5	0.129
99.5-110.5	0.043

(e) Mound-shaped symmetrical histogram

#6 (page 54)

(a) Pareto Chart

- Pareto chart is displayed by the heights of vertical bars, which are arranged in order from highest to lowest. So, the highest number of spearheads is 33, Shannon and the lowest number of spearheads is 8, Blackwater.

 (b) Circle Graph

Bann 19/89≈0.21, 21%,  21% x 360˚= 75.6˚

Blackwater 8/89≈0.09, 9%, 9% x 360˚=32.4˚

Erne 15/89≈0.17, 17%, 17% x 360˚=61.2˚

Shannon 33/89≈0.37, 37%, 37% x 360˚=133.2˚

Barrow 14/89≈0.16, 16%, 16% x 360˚=57.6˚

#12 (page 56)

 #1 (page 60)

(a) Make a stem-and-leaf display.

<Leaves not ordered>

4	4 represents 44years old.
Stem	Leaves
4	4
5	8	2	7	8
6	8	6	6	8	1
7	2	3	0	5	2	3	7	6
8	6	9	4	7	6	5	4	4
9	7	1	1	2	0	3

<Final display with leaves ordered>

4	4 represents 44 years old.
Stem	Leaves
4	4
5	2	7	8	8
6	1	6	6	8	8
7	0	2	2	3	3	5	6	7
8	4	4	4	5	6	6	7	9
9	0	1	1	2	3	7

(b) Their average age is 76years old(sum of their age/32=76) and 69%(22/32) of cowboys' age range is between 70 and 97. So, these data support the quote. Because if the cowboys are drunken, gambling lot, quick to draw and fire their pistols, they die earlier.   

Jieun Kim Unit2-1

#5 (page 81)

2,3,4,5,5

(a) compute the mode, median, and mean.

<Mode>

mode=5. Because the mode of a data set is the value that occurs most frequently. The value,5 occurs the most. So, the value 5 is the mode.

<Median>

median=4.

First, put those numbers in order. 2,3,4,5,5.

These are five number, so the middle number is the third number, 4.

<Mean>

mean=3.8

mean= sum of all entries/number of entries=(2+3+4+5+5)/5=19/5=3.8

(b) If the numbers represented codes for the colors of T-shirts ordered from a catalog, which average(s) would make sense?

Mode makes sense for this case. Because, Two of Code 5 T-shirts are sold. The code 5 T-shirt is popular so far.

(c) If the numbers represented one-way mileages for trails to different lakes, which average(s) would make sense?

Mean makes sense for this case. Because people will compare the distance of the five ways and the mean of distances.

Median makes sense too. Because the median is the central value of an ordered distribution and some people just choose the middle number of the five ways. 

(d) Suppose the numbers represent survey responses from 1 to 5, with 1 = disagree strongly, 2 = disagree, 3=agree, 4 = agree strongly, and 5= agree very strongly. Which averages make sense?  

Mean makes sense for this case. Because it needs mean for sum of all entries to get the mean score of survey responses.

#9 (page 82)

(a)

Mean=3.27

Mean=sum of all entries/number of entrie=36/11≈3.27

sum=36=2+3+1+1+3+4+6+9+3+1+3

Median=3

Numbers in order : 1,1,1,2,3,3,3,3,4,6,9

These are 11 numbers, 6th number is the median. So, number 3 is the median. 

Mode=3

Number 3 occurs most frequently, so number 3 is the mode.

(b)

Mean=4.21

Mean=sum of all entries/number of entries=59/14≈4.21

sum=8+1+1+0+6+7+2+14+3+0+1+13+2+1=59

Median=2

Numbers in order : 0,0,1,1,1,1,2,2,3,6,7,8,13,14

These are 14 numbers, 7th and 8th numbers are the middle pair numbers.

(2+2)/2=2

Mode=1

Number 1 occurs most frequently, so number 1 is the mode.

(c) compare the results of parts (a) and (b).

At the upper canyon,  11 accidents happened and resulted 36 injured people for a 5-year period. The mean is 3.27, the median is 3, and the mode is 3. So, each accident resulted usually 3(rounded) injured.

At the lower canyon,  14 accidents happened and resulted 59 injured people for a 5-year period. The mean is 4.21, the median is 2 and the mode is 1. But at lower canyon, each accident resulted various number of injured people. Sometimes, there was accident, but no one injured. Or there was accident and it was quite serious so it resulted 13 injured or 14 injured.  So, I think the mean can be the best average for the lower canyon accidents. In this case, the lower canyon has one more injured than the upper canyon for each accident. But, if compare median or mode to the upper canyon, it is different. So, the lower canyon' data needs trimming.

(d)

14 x 5%=0.7≈1

I eliminate one data value from the bottom of the list and one from the top. Then take the mean of the remaining 12 entries.

0,0,1,1,1,1,2,2,3,6,7,8,13,14

0	0	1	1	1	1	2	2	3	6	7	8	13	14
	0	1	1	1	1	2	2	3	6	7	8	13

sum of the 12 entries=45

mean=45/12=3.75

After I trimmed data of the lower canyon,  mean is 3.75 and the upper canyon's mean is 3.27. So difference is 0.27. So, the lower canyon has 0.27 more injured than the upper canyon.

#10(page 82)

(a) compute the mean, median, and mode of the ages.

First, I made these ages in order.

21	21	22	22	22	22	23	23	23	23
24	24	24	25	25	25	25	25	25	25
26	26	26	26	27	27	28	28	28	29
29	29	29	29	30	31	31	32	33	37

<mean>

mean=sum  of all entries/number of entries=1050/40=26.25

Mean is 26.25

<median>

For forty numbers, 20th and 21th are the pair of middle numbers. So, 25 and 26 are the pair of middle numbers. To find the value half-way between them, (25+26)/2=51/2=25.5

Median is 25.5

<mode>

Age	21	22	23	24	25	26	27	28	29	30	31	32	33	37
Frequency	2	4	4	3	7	4	2	3	5	1	2	1	1	1

The mode of a data set is the value that occurs most frequently. So the value 25 occurs most frequently.

Mode is 25.

(b) compare the average.

 Mode has the lowest average value because there are 7 foot ball players who are 25 years old. Median is 25.5. I think Mode and Median didn't cover all range of the football player's age. I think that the Mean which has the highest average value is the most accurate average. Because it included 37years old football player and it included all data as well as.  

#12 (page 83)

Lab	Major test	Major test	Final exam
25%	22.5%	22.5%	30%
92	81	93	85

 =87.65=weighted mean.

Jieun Kim Unit 2-2

#5 (page 93)

(a) (i),(ii),(iii). Because (i) has consecutive numbers and The numbers' distance from the mean in (ii) is shorter than the numbers' distance from the mean in (iii). So, I think that standard deviation of (iii) will have the highest value.

(b) The difference in standard deviation between (i) and (ii) is greater than the difference in standard deviation between (ii) and (iii).

(i)Value	Mean	x-Mean	(ii)Value	Mean	X-Mean	(iii)Value	Mean	X-Mean
8	10	2	7	10	3	7	10	3
9	10	1	9	10	1	8	10	2
10	10	0	10	10	0	10	10	0
11	10	1	11	10	1	12	10	2
12	10	2	13	10	3	13	10	3

 Because the difference between 8 in (i) and the mean is 2. The difference between 7 in (ii) and the mean is 3. So, there is difference between these differences (3-2=1). But the difference between 7 in (ii) and the mean is 3. The difference between 7 in (iii) and the mean is also 4. So, there is no difference(3-3=0).  So, I expect that the difference in standard deviation between (i) and (ii) is greater than the difference in standard deviation between (ii) and (iii).

 #7 (page 93)

x:23,17,15,30,25

(a) range=15

Range=Largest value-Smallest value=30-15=15

(b) means 23+17+15+30+25=110

 means 23²+17²+15²+30²+25²=529+289+225+900+625=2568

(c)

The sample variance

 ==(2568-2420)/4=148/4=37

The sample standard deviation

=

(d) Defining formulas (sample statistic)

The sample variance

==148/(5-1)=148/4=37

Data Value x	Mean 110/5=22
23	22	1	1
17	22	-5	25
15	22	-7	49
30	22	8	64
25	22	3	9
Total 110			Total 148

The sample standard deviation

=

(e)

Data Value x	Mean==110/5=22
23	22	1	1
17	22	-5	25
15	22	-7	49
30	22	8	64
25	22	3	9
Total 110			Total 148

the population variance  ==148/5=29.6

the population standard deviation ==5.44

#6 (page 106)

Arranged data.

3 6 7 14 15 17 18 20 22 22

24 25 25 26 29 31 31 32 42 72

1. The lowest value of the data set = 3

2. Q1=15

3. The median=(22+24)/2=23

4. Q3=31

5. The highest value of the data set=72

Interquartile range=Q3-Q1=31-15=16

(a)

(I used website to make a chart) www.mathwarehouse.com

(b)

The data of nurses(question5) have a larger range that the clerical staff's data. These two chart's median are both 23. But the median of nurses' data is to the right, the distribution is negatively skewed. The median of staff's data is near the center of the box, the distribution is approximately symmetric.

The middle halves of nurses' data are 8 and 29. The middle halves of staff's data are 15 and 31. Their first middle halves are quite different. But the second middle halves are similar.

The interquartile of nurses' hour is 29-8=21. The interquartile of staff's hour is 31-15=16. The distance to the extreme value of nurses' data is 42-21=21 and the distance to the extreme value of staff's data is 72-16=56. So, the distance to the extreme value of staff's data is much further than the nurses' one. These charts also show that staff's box is further from the extreme value 72.

#7 (page 106)

(a)

1. The lowest value of the data set = 17

2. Q1=21.5

3. The median=24

4. Q3=27

5. The highest value of the data set=38

Interquartile range=Q3-Q1=27-21.5=5.5

(b) The Bachelor's degree percentage of Illinois is 26% and this rate falls into Quartile3. Because 26% is higher than the median and lower than Q3.  

(Section 5.1) #3 (page 164)

(a)  An event A that is certain to occur means that the probability of A is 1.

Let's say that there are 20 students studying Statistics and they all wear a school uniform.

In this case, the probability of wearing a school uniform is P(A)=20/20=1.

(b) An event B that is impossible means that the probability of B is 0.

There are 20 girls attending prom, no one wears a school uniform.

In this case, the probability of wearing a school uniform is P(B)=0/20=0.

#6 (page 165)

(a) -0.41 cannot be the probability of some event. Because a probability formula is

.

0≤Number of outcomes favorable to event≤1 is the only possible range.

(b)1.21 cannot be the probability of some event. The reason is same as answer from (a)

 0≤Number of outcomes favorable to event≤1 is the only possible range. But, the 1.21 is over 1.

(c) 120% cannot be the probability of some event. 120% equals to 1.2 and it has the same reason.

1.2 is not between 0≤Number of outcomes favorable to event≤1.

(d) 0.56 can be the probability of an event. Because it is between 0≤Number of outcomes favorable to event≤1. For example, I threw a coin 100 times and I got a tail 56 times. In this case probability of getting tail is 0.56

 #10 (page 165)

(a) sample space : 1,2,3,4,5,6

These outcomes are equally likely. Because, the shape of dice is a cubic and each side has same length. So, the movement of rolling a dice doesn't affect a outcome.

(b)

P(getting 1)=1/6

P(getting 2)=1/6

P(getting 3)=1/6

P(getting 4)=1/6

P(getting 5)=1/6

P(getting 6)=1/6

 The sum of the probabilities of all simple events in a sample space is 1.

1/6+ 1/6+ 1/6+ 1/6+ 1/6+ 1/6=1

The sum of the probabilities of all simple events in a sample space should be 1. Because, in this case, all the possible outcomes are 6 of sample space. 6/6=1.

(c)

P(getting 1)=1/6

P(getting 2)=1/6

P(getting 3)=1/6

P(getting 4)=1/6

1/6+ 1/6+ 1/6+ 1/6=4/6=2/3≈0.67

(d) Getting 5 or 6 is mutually exclusive events.

P(getting 5)=1/6

P(getting 6)=1/6

1/6+ 1/6=2/6=1/3≈0.33

#13 (page 166)

(a) 58/127 ≈0.457

58 of 127 people enter the store.

(b) 25/58 ≈0.431

25 of 58 people bought something.

(c) 58/127 x 25/58=25/127≈0.197

P(A)=58/127 and P(B)=25/58 are independent, then P(A and B)= P(A)xP(B).

(d)1-(25/58)=33/58≈0.569

Because the complement of P(A) equals to 1-P(A).

 (Section 5.2) #5 (page 180)

(a)P(A and B)=P(A) x P(B)

(b)P(B I A)=P(A and B)/P(A), when P(A)≠0.

(c)P( IB)=P( and B)/P(B), when P(B)≠0.

(d)P(A or B)=P(A)+P(B)

(e)P(  or A)=P( )+P(A)

#8 (page 181)

*Total number of arches in park=288

(a) 111/288≈0.39

(b) 0.28≈81/288 = 30/288 + 33/288 + 18/288

(c) 0.82≈237/288 = 111/288 + 96/288 + 30/288

(d) 0.55≈159/288 = 96/288 + 30/288 + 33/288

(e) 0.0625=18/288

#11 (page 181)

(a) 5/36≈0.139

Possible outcomes of getting a sum of 6:  5

 (1,5) (2,4) (3,3) (4,2) (5,1)

Sample space : 36

 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

There is no overlap because the colors of dice are different.

(b)3/36=1/12≈0.083

 Possible outcomes of getting a sum of 4 : 3

(1,3) (2,2) (3,1)

Sample space : 36

 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

There is no overlap because the colors of dice are different.

(c)These are mutually exclusive. Because those outcomes are different and it is not overlapped each other outcomes.

Possible outcomes of getting a sum of 6 : 5 , (1,5) (2,4) (3,3) (4,2) (5,1)

Possible outcomes of getting a sum of 4 : 3 , (1,3) (2,2) (3,1)

#15 (page 182)

(a) The outcomes on the two cards are independent. Because, after drawing the first card, the first card was not replaced. So, it doesn't affect the second drawing.

(b)P(ace on 1st card and king on 2nd)=4/52 x 4/51 = 16/2652=4/663≈0.006033

(c) P(king on 1st card and ace on 2nd)=4/52 x 4/51 = 16/2652=4/663≈0.006033

(d) P(drawing an ace and a king in either order)=16/2652 + 16/2652= 32/2652≈0.012066

 #17 (page 182)

(a) 27/100 + 14/100 + 22/100= 63/100= 0.63

(b) 15/100 + 22/100 + 27/100 + 14/100 = 78/100=0.78

(c) 27/100 + 14/100 = 41/100 = 0.41

(d) 22/100 + 27/100 = 49/100 = 0.49

How would you respond? Because the class limits are different. '13 and order' means that it could be 13, 14, 15, 16, 17 and 18 years old.

 #5 (page 193)

(a) 8 sequences.

(b) 3 sequences. (HHT), (HTH), (THH).

(c) 3/8

#9 (page193)

4x3x2x1=24.

Because there are four choices for the first wire, three choices for the second, two choices for the third and only one for the fourth.

#11 (page 193)

4x3x3=36

36 of different lawn plots she need.

I used the Multiplication rule of counting.

#13 (page194)

* For the questions 18, 25, and 27, I used the Counting rule for combinations.

#18 (page 194)

 #25 (page 194)

==

#27 (page 194)

(a)===

(b)

(c)

Read all of the highlighted information on pages 204-209 as well as complete the suggested Study Guide exercises #1 (page 204), #2 (page 206-207), and #3 (page 209).

Submit your solutions to problems

 #2 (page 210)

(a) Continuous variables. Because the speed is the measurement. 

(b) Discrete variables. Because the age is countable.

(c) Discrete variables. Because the number of book is countable.

(d) Continuous variable. Because the weight is countless number of values in a line interval.

(e) Discrete variables. Because the number of lightning strikes is countable.

#3 (page 210)

(a) The sum of probabilities of all the events in the sample space is 1.

x	0	1	2
P(x)	0.25	0.60	0.15

So, (a) is a valid probability.

(b) The sum of probabilities of all the events in the sample space is more than 1.

x	0	1	2
P(x)	0.25	0.60	0.20	.05

The sum of probabilities of all the events in the sample space must equal 1. So, (b) is not a valid probability.

#7 (page 211)

(a) It is a valid probability distribution. Because, the sum of the percentage is 100%. The sum of probabilities must equal 1 (100%).

(b)

(c) µ=32.3

Midpoint x	10	20	30	40	50	60
Percent of super shoppers	21%	14%	22%	15%	20%	8%
	2.1	2.8	6.6	6	10	4.8

µ=2.1+2.8+6.6+6+10+4.8=32.3

(d)  ≈16

  =-

=21+56+198+240+500+288-1043.29=1303-1043.29=259.71

 ==≈16

#11 (page 212)

(a)

P(win)= 

P(not win)==

(b)

	Win	Not win
Gain X	$35	$-15
Probability P(x)
	0.73	-14.69

Lisa's expected earning is $0.73

Lisa contributed $13.96 to the hiking club.

$0.73+ (-14.69)=-13.96

Read all of the highlighted information on pages 214-229 as well as complete the suggested Study Guide exercises #4 (page215), #5 (page 220-221), #6 (page 227-228), and #7 (page 229).

Submit your solutions to problems

(Section 6.2) #9 (page 223)

(a) = = 0.125

n=3, r=3, P=0.5

(b) = = 0.375

n=3, r=2, p=0.5

(c) P(r≥2)=P(3)+P(2)=0.125+0.375

n=3, r≥2, p=0.5

(d) = = 0.125

#12 (page 224)

(a) n=7, r=0, p=0.1

P(0)=0.478

(b) n=7, r≥1, p=0.1

P(r≥1)=P(1)+P(2)+P(3)+P(4)=0.372+0.124+0.023+0.003=0.522

(c) n=7, r≤2, p=0.1

P(r≤2)=P(0)+P(1)+P(2)=0.478+0.372+0.124=0.974

#13 (page 224)

(a) n=6, r=6, p=0.90

P(6)=0.531

(b) n=6, p=0.90, r=0

P(0)=0.000

(c) n=6, p=0.90, r≥4,  r=4,5,6

P(r≥4)=P(4)+P(5)+P(6)=0.098+0.354+0.531=0.983

(d) n=6, p=0.90 r≤3, r=0,1,2 or 3

P(r≤3)=P(0)+P(1)+P(2)+P(3)=0.000+0.000+0.001+0.015=0.016

(Section 6.3) #4 (page 231)

(a) P=0.50 is the distribution symmetric.

The expected value

=np=10x0.5=5

The distribution (n=10, p=0.5, 0≤r≤10) is centered over 5.

(b) The distribution for small values of p skewed right.

(c) The distribution for large values of p skewed left.

 #5 (page 231)

(a)

 Symmetric distribution

(b)

 Skewed right

(c)

 Skewed left

(d) Based on the histograms, both charts are binomial probability distributions. So, (b) is skewed to right and (c) is skewed to left.

(e)  If the probability of success is p=0.73, the distribution is skewed to left. Because p=0.73 is between p=0.70 and p=0.75 and n=5, p=0.70 is also skewed to left.

#6 (page 231)

(a)

 

(b) µ=np=8x0.01=0.08

(c) q=1-p=1-0.01=0.99

(d)0.28

=npq=8x0.01x0.99=0.0792

==0.28

#7 (page 235)

(a)

(b)

r≥6, n=10, p=0.55

P(6)+P(7)+P(8)+P(9)+P(10)=0.238+0.166+0.076+0.021+0.003=0.504

(c)

The expected number of claims=µ=np=10x0.55=5.5

The standard deviation ≈1.57

=npq=10x0.55x0.45=2.475

q=1-p=1-0.55=0.45

=≈1.57

#9 (page 235)

(a) n=16, p=0.5, r≥12

P(12)+P(13)+P(14)+P(15)+P(16)=0.028+0.009+0.002+0.000+0.000=0.039

(b) n=16, p=0.5, r≤7

P(7)+P(6)+P(5)+P(4)+P(3)+P(2)+P(1)+P(0)=0.175+0.122+0.067+0.028+0.009+0.002+0.000+0.000=0.403

(c) µ=np=16x0.5=8

#11 (page 235)

(a)

(b)

n=10, p=0.25, r≤1

P(0)+P(1)=0.056+0.188=0.244

n=10, p=0.75, r≥1

P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)

=0.000+0.003+0.016+0.058+0.146+0.250+0.282+0.188+0.056=0.999

(c) µ=np=10x0.75=7.5

(d) ≈1.37

=npq=10x0.75x0.25=1.875

=≈1.37

(Section 7.1) #5 (page 249)                                                                                   Jieun Kim 5-1

5.

(a) 49.85%

Because the normal curve is symmetric and has bell-shaped with a single peak. So, the percentage of the area under the normal curve is 49.85%, a half of total percentage.

(34%+13.5%+2.35%=49.85%)

(b) 68%

µ - σ = 34%, µ + σ = 34%, 34%+34%=68%

(c) 99.7%

µ - 3σ = 49.85%, µ + 3σ = 49.85%, 49.85% + 49.85% = 99.7%

#7 (page249)

µ=65, σ=2.5
µ - 3σ	µ - 2σ	µ - σ	µ	µ + σ	µ + 2σ	µ + 3σ
57.5	60	62.5	65	67.5	70	72.5

Between µ - 3σ and µ - 2σ : 2.35%

Between µ - 2σ and µ - σ : 13.5%

Between µ - σ and µ : 34%

Between µ  and µ + σ : 34%

Between µ + σ and µ + 2σ : 13.5%

Between µ + 2σ and µ + 3σ : 2. 35%

(a) 49.85 % (34%+13.5%+2.35%=49.85%)

(b) 49.85% (2.35%+13.5%+34%=49.85%)

(c) 68% (34%+34%=68%)

(d) 95% (13.5%+34%+34%+13.5%=95%)

#10 (page250)

µ=7.6, σ=0.4
µ - 3σ	µ - 2σ	µ - σ	µ	µ + σ	µ + 2σ	µ + 3σ
6.4	6.8	7.2	7.6	8	8.4	8.8

Between µ - 3σ and µ - 2σ : 2.35%

Between µ - 2σ and µ - σ : 13.5%

Between µ - σ and µ : 34%

Between µ  and µ + σ : 34%

Between µ + σ and µ + 2σ : 13.5%

Between µ + 2σ and µ + 3σ : 2. 35%

(a) 15.85% (13.5+2.35%=15.85%)

(b) 83.85% (2.35%+13.5%+34%+34%=83.85%)

(c) About 135 cups

850cups x 0.1585 = 134.7 ≈135

(Section 7.2) #7(page 259)

(a) Robert, Juan, Linda. Because their z score is above 0.

(b) Joel. Because z score 0 is the mean.

(c) Susan, Jan. Because their z score is below 0.

(d) µ=150, σ=20

Name	Robert	Juan	Susan	Joel	Jan	Linda
z	1.10	1.70	-2.00	0.00	-0.80	1.60
X=zσ+ µ	172=1.10x20+150	184=1.70x20+150	110=-2.00 x20+150	150=0.00 x20+150	134=-0.80 x20+150	182=1.60 x20+150
x	172	184	110	150	134	182

 

 #9 (page 259)

(a) -1 < z	(b) z <-2	(c) -2.67 < z < 2.33
4.5 < x (4.5-4.8)/0.3 < z -0.3/0.3 < z -1 < z	x < 4.2 z < (4.2-4.8)/0.3 z < -0.6/0.3 z <-2	4.0 < x < 5.5 (4.0-4.8)/0.3 < z < (5.5-4.8)/0.3 -0.8/0.3 < z < 0.7/0.3 -2.67 < z < 2.33

(d) x < 4.368	(e) 5.184 < x	(f) 4.125<x<4.5
z < -1.44 x < -1.44 x 0.3 + 4.8 x < 4.368	1.28 < z 1.28 x 0.3 + 4.8 < x 5.184 < x	-2.25<z<-1.00 -2.25x0.3+4.8<x<-1.00x0.3+4.8 4.125<x<4.5

(g) A female had an RBC count of 5.9 or higher would be considered unusually high because z score is greater than 3 and the percentage of z≥3.67 is out of 99.7% of area. So, an RBC count of 5.9 or higher is unusually high.

x≥5.9

z≥(5.9-4.8)/0.3

z≥3.67

 #13 (page260)

z=-1.32

Area to the left of -1.32 = 0.0934

#25 (page 260)

Between z=0.32 and z=1.92

Area of Between z=0.32 and z=1.92

0.9726-0.6255=0.3471

 #37(page 260)

P(z≥-1.20)=1.000-P(z≤-1.20)=1.000-0.1151=0.8849

 #43 (page 260)

P(0 ≤ z ≤ 1.62)=P(z≤ 1.62)-P(z≤0)=0.9474-0.5000=0.4474

 #47 ( page 260)

P(-0.45≤ z ≤ 2.73)=P(z≤2.73)-P(z≤-0.45)=0.9968-0.3264=0.6704

 (Section7.3) #11 (page 270)                                                                                    Jieun Kim 5-2

P(x≥30)=P(z≥)=P(z≥)=P(z≥2.94)=1-0.9984=0.0016

#15 (page271)

6%=0.0600

The closest area is 0.0594. This is to the left of z=-1.56.

#21 (page271)

82%=0.8200

1-0.8200=0.1800

The closest area is 0.1788. This is to the right of z=-0.92.

#25 (page271)

(a)

P(x>60)=P(z>)=P(z>-1)=1-0.1587=0.8413

(b)

P(x<110)=P(z<)=P(z<)=P(z<1)=1-0.8413=0.1587

(c)

P(60<x<110)=(a)-(b)=0.8413-0.1587=0.6826

Apply    to prove it.

1-0.6826=0.3174

0.3174/2=0.1587

(d)

P(x>140)=P(z>)=P(z>)=P(z>2.2)=1-0.9868=0.0132

 #27 (page 271)

(a)

P(x<3.0)=P(z<)=P(z<)=P(z<-2.33)=0.0099

(b)

P(x>7.0)=P(z>)=P(z>)=P(z>2.11)=1-0.9826=0.0174

(c)

P(3.0<x<7.0)=1-0.0099-0.0174=0.9727

(Section7.5) #9 (page 287)

(a)

=µ=15

=σ/=14/=14/7=2

P(15≤≤17)=P(≤z≤)=P(0≤z≤1)=0.8413-0.5000=0.3413

(b)

=µ=15

=σ/=14/=14/8=1.75

P(15≤≤17)=P(≤z≤)=P(0≤z≤1.14)=0.8729-0.5000=0.3729

(c)

z of (a) = < z of (b) =

Because dividing smaller number results larger number.

#11 (page 287)

(a)

P(x<74.5)=P(z<)=P(z<)=P(z<-0.63)=0.2643

(b)

P(z<)=P(z<)=P(z<)=P(z<-2.78)=0.0027

(c)

I will be suspicious of the loader because the probability of the weight of coal in one car was less than 74.5 tons is approximately 26%. But the probability of the weight of coal in 20 cars was less than 74.5 tons is only 0.27%. So, I won't be suspicious of the second case.

#13 (page 287)

(a)

P(x<40)=P(z<)=P(z<)=P(z<-1.80)=0.0359

(b) P(<40)=P(z<)=P(z<)=P(z<)=P(z<-2.54)=0.0055

(c) P(<40)=P(z<)=P(z<)=P(z<)=P(z<-3.11)=0.0009

(d) P(<40)=P(z<)=P(z<)=P(z<)=P(z<-4.03)=0.0000

(e) The probabilities decreased as n increased. If more samples has lower blood glucose than 40, that is very serious case and it barely happens based on the probabilities of (a), (b), (c) and (d). In addition, the opposite side of the normal curve also happens very rare, because the probability is also close to 0.0000.

(Section7.6) #5 (page 294)

(a)

µ=np=200 x 0.88= 176

σ===≈4.60

P(x≥50)=P(z≥)=P(z≥)=P(z≥-27.39)=1.0000

(b)

µ=np=200 x 0.09= 18

σ===≈4.05

P(x≥50)=P(z≥)=P(z≥)=P(z≥7.90)=0.0000

#7 (page 295)

(a)

P(x≥15)=P(z≥)=P(z≥)=P(z≥-2.25)=1-0.0122=0.9878

(b)

P(x≥30)=P(z≥)=P(z≥)=P(z≥0.72)=1-0.7642=0.2358

(c)

P(25<x<35)=P(z<)-P(z<)=P(z<-P(z<)=P(z<1.71)-P(z<-0.27)=0.9564-0.3936=0.5628

(d)

P(x>40)= P(z>)=P(z>)= P(z>2.71)=1-0.9966=0.0034

 #9 (page 295)

n=66, p=0.80, q=1-0.80=0.20

µ=np=66x0.80=52.8

σ===≈3.25

(a)

P(x≥47)=P(z≥)=P(z≥)=P(z≥-1.78)=1-0.0375=0.9625

(b)

P(x≤58)=P(z≤)=P(z≤)=P(z≤1.6)=0.9452

n=66, p=0.20, q=1-0.20=0.80

µ=np=66x0.20=13.2

σ===≈3.25

(c)

P(x≥15)=P(z≥)=P(z≥)=P(z≥0.55)=1-0.7088=0.2912

(d)

P(x<10)=P(z<)=P(z<)P(z<-0.98)=0.1635

Jieun Kim 6-1

 (Section8.1) #11 (page 316)

(a)

Critical value of 80%=1.28

E=1.28=1.28x0.085≈0.11

-E<µ<+E

3.15-0.11<µ<3.15+0.11

3.04gram < µ < 3.26gram

(b) The distribution of bird's weight should be normal curve and it requires standard deviation(σ) and sample size.

(c) The level of confidence is 80% and with this, I can get an interval that includes these bird' s weight distribution.  It also means that there is 80% chance that the confidence interval is one of the intervals that contains the population average weight of birds.  

(d) E=1.28×=0.08

1.28x0.33x=0.08

=0.08=0.19

=

=

n=10000/361=27.7≈28

#13 (page 317)

(a)

Critical value of 99%=2.58

E==2.58=2.58x1.12≈2.88

-E<µ<+E

37.5-2.88<µ<37.5+2.88

34.62<µ<40.38

(b) It needs sample size which is larger than 30 and the standard deviation(σ)

(c) 99% of confidence level means, there is 99% chance that the confidence interval is one of the intervals that includes the population average blood plasma level.

(d) E=2.58×=2.50

=2.507.50=0.13=

=

=

n=10000/169=59.17≈60

#16 (page 317)

Critical value of 90%=1.645

(a)

E=1.645=1.645x=1.645x2521.61≈4148

-E<µ<+E

50340-4148<µ<50340+4148

46192<µ<54488

(b)

E=1.645=1.645=1.645x1606.56=2642.79≈2643

-E<µ<+E

50340-2643<µ<50340+2643

47697<µ<52983

(c)

E=1.645=1.645=1.645x719.82=1184.11≈1184

-E<µ<+E

50340-1184<µ<50340+1184

49156<µ<51524

(d)

As the standard deviation decrease, the margin of error also decrease.

σ :16920>10780>4830

E :4148>2643>1184

(e)

As the standard deviation decrease, the length of a 90% confidence interval also decrease.

The length of interval : 8296>5286>2368

(Section 8.2) #1 (page 326)

d.f.=n-1=18-1=17

=2.110

#3 (page326)

d.f.=n-1=22-1=21

=1.721

#11 (page 327)

(a)

sample mean=11450/9=1272.22≈1272

x	1189	1271	1267	1272	1268	1316	1275	1317	1275
	1413721	1615441	1605289	1617984	1607824	1731856	1625625	1734489	1625625

=14577854

==131102500

=====1363.75

s==36.92≈37

(b)

d.f.=n-1=9-1=8

E==1.860x37/=1.860x37/3=1.860x12.33=22.94≈23

1272-23<µ<1272+23

1249<µ<1295

#13 (page 327)

(a)

x	68	104	128	122	60	64
	4624	10816	16384	14884	3600	4096

=54404

==298116

=====943.8≈944

s=≈30.7

=546/6=91

(b)E==1.301=1.301x12.53=16.3

91-16.3<µ<91+16.3

74.7<µ<107.3

#15 (page 327)

x	9.3	8.8	10.1	8.9	9.4	9.8	10.0	9.9	11.2	12.1
	86.49	77.44	102.01	79.21	88.36	96.04	100	98.01	125.44	146.41

(a)

=999.41

==9900

===≈1.046

s=≈1.02

=99.5/10=9.95

(b)

E==4.781x=4.781x0.32=1.53

9.95-1.53<µ<9.95+1.53

8.42< µ<11.48

(c)

All the values are above 6mg/dl and 99% confidence interval is also above 6mg.dl. So, these people are not anymore issue with a calcium deficiency. 

Solutions part 1.

http://jieunkimresume.blogspot.com/2015/05/understanding-basic-statistics.html

Solutions part 2.

http://jieunkimresume.blogspot.com/2015/05/understanding-basic-statistics_21.html