STAT assignment - worksheet For introduction to statics PDF

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Addis Ababa Unive Univerrsity Colle College ge of Na Natural tural and Computational Scienc Sciences es Depar Department tment of Sta Statistics tistics Probability & Sta f Engineers (Stat 2171) W Statistics tistics for or Wor or orksheet ksheet I

GROUP MEMBERS NAME 1. 2. 3. 4. 5. 6.

MIKIYAS MOHAMMED MULUKEN WALLE YASIN MOHAMMED GELETA MATHEWAS DEGSEW ABEBAW TAMIRAT LEMESSA

ID UGR/3095/12 UGR/9931/12 UGR/4544/12 UGR/4712/12 UGR/1341/12 UGR/5507/12

SUBMITTED TO:- M.s SELOME BEKELE SUBMISSION DATE:- MAY 4,2021

1,Distinguish the following statistical terms with examples: a) Descriptive vs inferential statistics  Descriptive statistics is a type of statistics that consists of the collection, organization, summarization, presentation of data.  Example:- The average age of students in our dorm is 20.  Inferential statistics other type of statistics that consists of generalizing from sample to populations, performing estimations and hypothesis testing, determining relationships among variables and making predictions.  Example :- The result of a recent sample survey indicated that the fertility rate of Ethiopians decreased from 6.4 to 4.6 children per women in the last two and half decades. b) Sample vs population •

Population: a totality of things, objects, peoples, etc. about which information is being collected.  Example:-Ethiopian people population • Sample: It is a part of population selected in statistical manner to study the population.  Example:-selecting 500 students for research from AAU c) Parameter vs statistic • Parameter :- It is statistical value which refers to the population characteristics or it is a result obtained from the population.  Example:- census of Ethiopian population. • Statistic: - It is statistical value which refers to the sample characteristics or it is a result obtained from the sample.  Example:- mortality rate of a given country d) Census vs sample survey 

Census: a complete enumeration of the population.  Example : Population and housing census.  Sample survey:- is a method of collecting data from or about the members of a population so that inferences about the entire population can be obtained from a subset or sample of the population members.  Example;-taking questionnaires from 343 students for research from AAU students e) Quantitative vs qualitative variables • Qualitative: - are variables that can be placed into distinct categories or classified in accordance with an attribute that cannot be measured or expressed in numbers.  Example: - sex, religion, intelligence, honesty. • Quantitative :- are variables that can be measured, ordered or ranked .

 Example :- age, height, weight, income, width, pressure. f) Continuous vs discrete variables • discrete :- is one which takes only whole number values. It is usually obtained by counting .there is a gap between consecutive values , it varies only by finite jumps.  Example:- number of students the class, number of chair in the class, number of house along in the street. • Continuous;- is one which takes all real values between two given real values. Between consecutive values we can assume an infinite number of values.  Example;- weight of students, length of road, temperature of room g) Nominal vs ordinal scales of measurements Both of them are for qualitative data type but,  Nominal:- level of measurement classifies data into non-overlapping and exhausting categories in which no order or ranking can be imposed on the data. No arithmetic and relational operation can be applied. Example:-



 • Classifying students according to their field of study;  Categorizing survey subjects as male and female;  Religion (Christianity, Islam, Judaism, etc.)  Marital status (single, married, divorced, widowed, separated) Ordinal :- level of measurement classifies data into categories that can be ranked; however, precise differences between the ranks do not exist. Arithmetic operations are not applicable but relational operations are applicable. Examples:  Rating: Excellent, Very good, Good, Fair, Poor  Letter grades: A,B,C,D,F  Status: Large, Medium, Small  Judging (1st place, 2nd place, 3rd place)

h) Interval vs ratio scales of measurements Both are for quantitative data type but • Interval;- level of measurement ranks data, and precise differences between units of measure do exist; however, there is no meaningful zero. All arithmetic operations except division and multiplication are applicable. Relational operations are also possible. • Examples:  Temperature in C (0C does not mean no heat at all).

•

 IQ (IQ test do not measure people who have no intelligence).  SAT scores. Ratio :- level of measurement possesses all the characteristics of interval measurement, and there exists a true zero. All arithmetic operations and Relational operations are also possible.  Examples: Weight , Height , Age , Time, Salary

2. Classify the following sentences as belonging to the area of descriptive statistics or inferential statistics. a) As a result of recent cutbacks by oil-producing nations, we can expect the price of gasoline to double in the next year.  inferential b) At least 5% of all killings reported last year in city X were due to tourists.  descriptive c) Of all patients who received this particular type of drug at a clinic Y, 75% later developed significant side effect.  descriptive d) Adane concludes that his chance of passing the first year this academic year is at least 80% based on the statistics that 75% of the freshmen passed last year. 

Inferential

3. Classify each of the following first as qualitative and quantitative and second as nominal, ordinal, interval and ratio measure. a) Monthly income of persons.  Quantitative and ratio b) Socio-economic status of a family when classified as low, middle and upper class.  Qualitative and ordinal c) Temperature inside 10 refrigerators.  Quantitative and intreval d) Classification of marital status as single, married, divorced and widowed.  Qualitative and nominal e) Times for swimmers to complete a 50-meter race  Quantitative and ratio f) Months of the year Meskerm, Tikimit, …  Qualitative and ordinal g) Regions numbers of Ethiopia (1, 2, 3 etc.)  Qualitative and nominal h)The number of students in a college  Quantitative and ratio i)the net wages of a group of workers  Quantitative and ratio

j)The expansion of a rod of metal when heated  Quantitative ad ratio k) the height of the men in the same town  Quantitative and ratio l. For 16 persons arrested for driving while intoxicated you record whether they live in urban, suburban, or rural areas. 

Qualitative and nominal

7. Suppose data collected for heights (in cms) 0f 390 cows were tabulated in a frequency distribution and the following results were obtained. frequency

6

25

Class mark

11 2

117

48

72

116

60

3 8

2 2

3

a.)Construct a complete frequency distribution with class limits, class boundaries, frequencies, and the less than type cumulative frequencies. Class limit Class boundaries Class mark Frequency LCF 110 – 114 115 – 119 120 – 124 125 – 129 130 – 134 135 – 139 140 – 144 145 – 149 150 – 154

109.5 – 114.5 114.5 – 119.5 119.5 – 124.5 124.5 – 129.5 129.5 – 134.5 134.5 – 139.5 139.5 – 144.5 144.5 – 149.5 149.5 – 154.5

112 117 122 127 132 137 142 147 152

6 25 48 72 116 60 38 22 3

b. Determine a height above which 50% of the cows found?  50%=median so using median formula we can get the median n −cfb 2 md = Lm + w f n/2=halfway point 390/2 = 195 The median class 195th

( )

6 31 79 151 267 327 365 387 390

390 −151 2 MD=129.5+ 5 116 = 129.5 + 1.896 MD=131.4 Above 50% of cow height is greater than 131cm

(

)

c. Below which height do we get 25% of the cows?  25% 25%=quartile 1orQ1 Q1 is found at n/4=390/4=97.5th This means 97th and 98th in this case they are the same.

(

)

kn −cfb 4 Q k =Lm+ w f

(

)

1(390) −79 4 Q k =124.5+ 5 72 = 124.5 + 1.285 = 125.8 25% of cow is found below 125.8cm . d. Compute the mean, median and modal height of the distribution. Mean ∑ fixi ´x = ∑ fi 51215 ¿ 390 =131.1cm is the mean Median

( )

n −cfb 2 w md = Lm + f n/2=halfway point 390/2 = 195 The median class 195th 390 −151 2 MD=129.5+ 5 116

(

)

= 129.5 + 1.896 MD=131.4 Mode The modal class is 129.5 – 134.5 d1 M=Lm+ w d 1+ d 2 44 M =129.5+ 5 44 + 56 ¿ 129.5 + 2.2 = 131.7 is the mode

(

)

(

)

e. Draw a histogram, frequency polygon and less than type Ogive for the above data histogram

frequency 140 120 100

frequency

80 60 40 20 0

10

5 9.

1 –1

5 4. 4 11

.5

1 –1

5 9. 11

5 9.

2 –1

5 4. 4 12

.5

2 –1

5 9. 12

5 9.

3 –1

5 4. 4 13

class mark

Frequency polygon

.5

3 –1

5 9. 13

5 9.

4 –1

5 4. 4 14

.5

4 –1

5 9. 9 14

.5

5 –1

4.5

Series 1 140 120

frequency

100 80 Series 1 60 40 20 0

112

117

122

127

132

137

142

147

152

class boundary

Less type Ogive 450 400

less camultive frequency

350 300 250 200

Column1

150 100 50 0

6

25

48

72

116

60

38

22

3

frequency

11. Suppose the average salary of male employees is 520 Birr and that of females is 420 Birr. The mean salary of all employees is 500 Birr. Find the ratio of the number of male and female employees.

Nf is number of female

Sm salary of male

Nm is number if male Sm X´ m= Nm Sm 520= Nm Sm=520*Nm

Sf salary of female

Sf X´ f = Nf Sf 420= Nf Sf=420*Nf Sf + Sm Nf∗420+520∗Nm = X´ s= Nf +Nm Nf + Nm 500 (Nf + Nm )= Nf ∗420 + 520∗Nm 500∗Nf +500∗Nm=Nf ∗ 420 + 520∗Nm Nf ∗80 =20∗Nm Nm = 80 = 4 =¿ 4 :1 Nf 20 1 13. The price of a commodity increased by 5% from 1996 to 1997, by 8% from 1998 to 1999 and by 77% from 2000 to 2001. What was the average yearly price increase? G . M =√ x 1∗x 2∗… . x n G . M =√3 1.05∗1.08∗1.77 −1 G . M =1.26 −1= 0.26=26 % 14. Suppose you have given the following distribution. n

Class limits 0-9 10-19 20-29 30-39 40-49 50-59 60-69 Total

Frequency 4 16 F3 F4 F5 6 4 230

cf 4 20 20+ F3 F3+F4+20 F3+ F5+F4+20 F3+ F5+F4+26 F3 +F5+F4+30=230

If the median and mode of the distribution are given to be 33.5 and 34.0 respectively, then. a) Determine the missing frequencies n −cfb 2 w md = Lm + f By using the median value we can get the Lm boundary of the median that is 29.5 also the with is 10 115−(20+ F 3) 33.5=29.5 + 10 F4 From this equation we get 5F3+2F4=475………………….eq1

(

)

( )

M=Lm+

( d 1+d 1d 2 ) w

By using the mode value we can get the Lm and the class so Lm=29.5 d1=F4-F3 d2=F4-F5 3 10 ( F 4−FF 4−F 3+ F 4−F 5 )

34 =29.5 +

From this equation we get 11F3-2F4-9F5=0………………..eq 2 Using the total frequency we get F3+F4+F5=200………..eq3 By using the above 3 equation simultaneously we can get F3 = 55 F4 = 100 F5 = 45

b) Compute the arithmetic mean. ∑ fiCmi = 7635 =33. 2 AM = 230 ∑f c) Compute the value below which 25% of the observations lie.

(

)

kn −cfb 100 Pk =Lm + w f 230/4=57.5th item 230∗25 −20 100 10 Pk =19.5+ 55

(

)

Pk=26.32 25% of observation lies below 26.32 d) Compute the value above which 25% of the observations lie. This means 75% kn −cfb 100 Pk =Lm + w f 230*3/4=172.5th item

(

)

75∗230 −75 100 Pk =29.5+ 10 100

(

)

Pk=39.25 75% of observation lies below 39.25 15) In a certain test the pass mark is 30. The distribution of marks of passing candidates classified by sex is given below. Mark 30 – 34 35 – 39 40 – 44 45 – 49 50 – 54 55 – 59 Total Boys 5 10 15 30 5 5 70 Girls 15 30 30 20 4 1 90 Class 32 37 42 47 52 57 mark The overall (combined) mean mark for boys including the 20 failed was 39. While the overall (combined) mean mark for girls including the 10 failed was 37. a) Find the mean marks obtained by the 20 boys who failed in the test.

the value of mean for boy which passed is ∑ fibcmi = 5∗32 +10∗37 +15∗42 +30∗47 +5∗52+5∗57 =44.5 ´x b p= 70 ∑ fib Total mean is 39 ´X bf∗20+ ´X bp∗70 X´ t= 90 X´ bf ∗20 + 44.5∗70 39= 90 39∗90− 44.5∗70 =19.7 5 X´ bf = 20 b) Find the mean marks obtained by the 10 girls who failed in the test. the value of mean for girl which passed is ∑ figcmi = 15∗32 +20∗37 + 42∗30+ 47∗20 +4∗52 +1∗57 =40.9 4 ´x gp= 90 ∑ fig Total mean is 37 ´ gp∗90 ´X gf ∗10+ X X´ t= 100 37∗100 − 40.94∗90 X´ bf = =1.5 10 17. Two sections were given an examination on a certain course. For section 1, the average mark(score) was 72 with standard deviation of 6 and for section 2, the average mark (score) was 85 with

standard deviation of 7. If student A from section 1 scored 84 and student B from section 2 scored90, then who perform better relative to the group? Section 1

Section 2

μ 72

85

δ 6

7

X 84 90 Let us find standard score both and compare X−μ Z= δ X 1−72 Z 1= 6 6∗Z 1=84 −72 Z1 = 2 X−μ δ 90 −85 Z 2= 7 Z=

Z2=0.714 Therefore student A perform better than B relatively to the group 20. Four married couples have bought 8 seats in a row for a show. In how many different ways can they be seated A, If each couple set together C1

C2

C3

C4

2!

2!

2!

2!

= 4!2!2!2!2!= 4×3×2×2×2×2×2= 384 b. if all women set together M1

M2

M3

M4

4w

= 5! 4w =4! =5!4!= 5×4×3×2(4×3×2)=2880 c. if all women sit together to the right of all the men 4m

4w

4! 4! =4! 4! = 4×3×2(4×3×2) = 576 23. One urn contains three red balls, two white balls, and one blue ball. A second urn contains one red ball, two white balls and three blue balls. a) One ball is selected at random from each urn. i) Describe a sample space for this experiment ii) Find the probability that both balls will be of the same color. iii) Find the probability that at least one of the balls is red. b) The balls in the two urns are mixed together in a single urn, and then a sample of threeis drawn. Find the probability that all the three colors are represented, when: i) Sampling with replacement ii) Sampling without replacement

Urn1 3R 2W

Urn2 1R 2W

1B

3B a. One ball select from each random from urn

i.

a sample space for this experiment is

S.P={RR,RW,BB,WR,WW,WB,BR,BW,BB}

ii.

probability that both balls will be of the same color is

1 2 3 ∗3 ∗2 ∗1 5 1 1 1 6 6 6 ¿ = + + = + + 12 9 12 18 6 6 6 iii) the probability that at least one of the balls is red is

3 3 3 2 1 ∗1 ∗2 ∗3 ∗1 ∗1 6 6 6 6 6 7 1 1 1 1 1 + + + + ¿ = + + + + = 6 6 6 6 12 6 4 18 36 12 6 b) The balls in the two urns are mixed together in a single urn, and then a sample of three is drawn.

i)when sampling with replacement the probability is 4 ∗4 12 ∗4 2 ¿ 12 ∗3 != 9 12 ii) when sampling without replacement the probability is 4 ∗4 12 ∗4 16 11 ∗3 != ¿ 55 10

24. Consider four objects, say a, b, c and d. suppose that the order in which these objects are listed represents the outcome of an experiment. Let the events A and B be defined as follows: A = {a is in the first position}; B = {b is in the second position}. a. List all elements of the sample space.

S.P={abcd,abdc,acbd,acdb,adbc,adcb,bacd,badc,bcad,bcda,bdca,bdac,cabd,cadb,cdab,cdba,cb da,cbad,dabc,dacb,dbac,dbca,dcab,dcba}

b. List all elements of the events AnB and AUB. AnB={abcd,abdc} AuB={ abcd,abdc,acbd,acdb,adbc,adcb,cbda,cbad,dbac,dbca} 25. A lot consists of 10 good articles, 4 with minor defects and 2 with major defects. i. One article is chosen at random. Find the probability that a. It has no defects, let S be the sample space, m be a lot with minor defect and M be a lot with major defect and g be a lot with no defects(i.e. good), then; g=10 m=4 M=2 Ta=16 This means probability of good. g 10 P(G)= = Ta 16 b. It has no major defects, P(M)c=1-p(M) =1- 2/16 7 ¿ 8 c) It is either good or has major defects. P(g or M)=P(g)+P(M) 10 2 12 3 P(g∨M )= + = = =0.7 5 16 16 16 4 ii. Two articles are chosen (without replacement), Find the probability that a,Both are good 10 ∗9 16 90 3 = = 240 8 15 b, Both have major defects 2 ∗1 16 1 = 120 15 c, At least one is good

6 10 10 ∗10 ∗9 ∗6 16 16 16 60 90 60 210 7 + + = = + + = 15 15 240 240 240 240 8 15 d, at most one is good 6 6 10 ∗10 ∗5 ∗6 16 16 16 150 5 + + = = 15 15 15 240 8 e, Exactly one is good 6 10 ∗10 ∗6 16 16 60 60 120 1 + = = + = 15 240 240 240 2 15 f. Neither has major defects to find this 1-p(Both have major defects)-P(exactly one is major defect) 14 2 ∗2 ∗14 1 28 91 1 16 16 1− = − )1− + −( 120 120 120 15 15 120 28. Four horses (A, B, C, and D) have raced many times. It is estimated that A wins 30 percent of the time, B 40 percent of the time, C 20 percent of the time, and D 10 percent of the time. The game allows only one of the horses to be a winner in any race. If these horses will compete in a race, a) what is the probability that A will win?

P(A)=0.3

b) what is the probability that A or B will win?

, P(A u B)=P(A)+P(B)=0.3+0.4=0.7

c) what is the probability that A or B but not both will win? (A u B ) n P(A n B)c=0.7-0=0.7

d) what is the probability that neither A nor B will win? P(A u B)c=1-0.7=0.3

e) what is the probability that A or B or C will win? P(A u B u C)=0.3+0.4+0.2=0.9 f) are A and B independent? Yes, they are. Because P(AnB) is zero 29. One bag contains 4 white balls and 3 black balls, and the second bag contains 3 white balls and 5 black balls. One ball is drawn at random from the second bag and placed in the first bag. What is the probability that a ball now drawn from the first bag is white? Ans P(W2)=3/8 P(B2)=5/8 One ball is randomly selected from box 2 and put inside 1 Using total theorem P(W1)=P(W2)*P(W1/W2)+P(B)*P(W1/B) =3/8*5/8+5/8*4/8 =15/64+20/64 =35/64 30. A factory has two machines M1 and M2 making 60% and 40% respectively of the total production. Machine M1 produces 3% defective items, and M2 produces 5% defective items. Find the probability that a given defective part came from M1. Let A be the set of all defective items P(M1)=0.6 P(M2)=0.4 P(A/M1)=0.03 P(A/M2)=0.05 By using total theorem P(A)=P(M1)*P(A/M1)+P(M2)*P(A/M2) = 0.6*0.03+0.4*0.05 =0.038 By using bayes theorem A P ∗P(M 1) M1 M1 = P A P( A)

( )

P

( )

( MA1 )= 0.6∗0.03 0.038

P

( MA1 )=¿

0.47...