Answers: - Eastern Mediterranean University



2012 – 2013 SPRING SEMESTER - STAT 201 STUDY QUESTIONS FOR MID-TERM EXAMINATION.

Question 1: The following data gives the monthly expenses of families for a sample of households (in $);

400 410 200 170 370 510 415 325 160 115 190 215

300 240 260 330 150 600 145 135 350 230 485 540

a) How many families were questioned?

b) How many classes would you recommend?

c) What class interval would you suggest?

d) Organize the data into frequency distribution.

e) Draw histogram according to your answer in part c.

Answers:

a) Simply asking for what is n (sample size); n is sum of the families questioned. Thus, n = 24

b) 2 k > n Rule where n = 24 in this question. If k = 4 then 2 4 = 16 < 24 so number of classes must be more than 4.

If k = 5 then 2 5 = 32 > 24, so k = 5

c) i ≥ H – L / k from the data Highest Value = 600 and the Lowest Value = 115.

Thus i ≥ (600 – 115) / 5 = 97, so use 100.

d) Classes: Frequencies:

110 Up to 210 8

210 up to 310 5

310 up to 410 5

410 up to 510 3

510 up to 610 3

*Note that 115 must be included in the first class and 600 must be included in the last class, and in total you must have 5 classes and interval is 100 @ every class!!!! If these conditions satisfy then your frequency distribution is correct. Also, try if interval is 100 and if you start with 105 as lower limit of the first class. Then this distribution is correct? The answer is yes because, all conditions are also satisfied. In the exam, check the conditions.

** While you are organizing your frequency distribution and placing the frequencies make sure that 410 is on the 4th class, not on the 3rd and also 510 is part of last class not 4th!!!1

e) Also from histogram, you must be able to answer all questions a, b, c, d. Check if you can. Do not forget that on the

X-axis you will write down the classes, and on Y- axis you will write down frequencies. Get the distance between frequencies equally on Y-axis, (like increase the frequencies by 2 (2 4 6 8 10)) and then draw.

[pic]

*** Also look @ Page 35 and 39 to learn how to draw frequency polygon and cumulative frequency polygon.

Question 2. Following are the workers questioned among ABC Holding and following data frequency is

organized. Classes shows the hours of working and frequencies shows number of workers.

Hours of Working: No of Workers:

0 up to 3 8 a) How many people questioned @ ABC holding?

3 up to 6 11

6 up to 9 37 b) is 3 hours of working falls in the 1st class?

9 up to 12 20 c) Calculate the relative and cumulative frequencies.

12 up to 15 4 d) Calculate the mean and the standard deviation of this grouped data.

Answers:

a) In this question, you are asked what the sample size is, n is nothing but it is sum of all frequencies. n = 8+11+37+20+4 = 80

b) No it falls in 2nd class. Be careful!

c) Relative and Cumulative Frequencies are as follows

Classes Frequencies Relative Frequencies: Cumulative Frequencies:

0 up to 3 8 8/80 = 0.1 8

3 up to 6 11 11/80 = 0.1375 8+11 = 19

6 up to 9 37 37/80=0.4625 19+37=56

9 up to 12 20 20/80 = 0.25 56+20=76

12 up to 15 4 3/80 = 0.05 76+4=80

∑n = 80 ∑ = 1.00

d) [pic] if it is grouped data.

Classes Frequencies Mid Points (m) fm fm2 = fm*m

0 up to 3 8 1.5 8*1.5 = 12 12*1.5 =18

3 up to 6 11 4.5 11*4.5 =49.5 49.5*4.5 = 222.75

6 up to 9 37 7.5 37*7.5 = 277.5 277.5*7.5 = 2081.25

9 up to 12 20 10.5 20*10.5 =210 210*10.5 = 2205

12 up to 15 4 13.5 4*13.5 = 54 54*13.5 = 729

∑fm = 603 ∑fm2 = 5256

[pic] is the mean.

s = [pic] = 3.00 from s = [pic]

Question 3: Consider the annual incomes of five vice presidents of TMV industries are; $125000, $128000, $122000, $133000 and $140000 and consider that this is population.

a) What is range?

b) What is the arithmetic mean?

c) What is the population variance? Standard deviation?

d) What is the median?

Answers:

a. $18,000, found by $140,000 – $122,000; where R = H – L

b. $129,600, found by $648,000/5; from µ = [pic] and so; 125000+128000+122000+133000+140000 / 5

c. Variance = 40,240,000, found by 201,200,000/5 Standard Deviation = $6,343.50

σ 2=[pic]is variance of population for population data. (Notice that you use the population formulae)

σ = [pic]is standard deviation formula for population data.

x values: µ (x-µ)2

125000 129600 21160000

128000 129600 2560000

122000 129600 57760000

133000 129600 11560000

140000 129600 108160000 then add all these!

∑(x-µ)2 = 201, 200, 000

Thus σ 2 = 201200000/5 = 40,240,000

Thus σ = [pic]= 6,343.50

d) Median is determined after ordering the data from the smallest to the largest. The median is 128.

Question 4: The sample of 8 companies in the aerospace industry was surveyed as their return on investment last year. The results are; 10.6 12.6 14.8 18.2 12.00 14.8 12.2 15.6

a) Calculate the mean? Calculate the mean deviation (MD) as well.

b) Calculate the median? Mode?

c) Calculate the variance?

d) Calculate the standard deviation?

e) Determine the coefficient of skewness by using Pearson method.

Answers:

a. [pic]from [pic]( 10.6+12.6+14.8+18.2+12+14.8+12.2+15.6 / 8 = 13.85

MD = [pic] so

MD = │10.6-13.85│+ │12.6-13.85│+│14.8-13.85│+│18.2-13.85│+│12.00-13.85│+│14.8-13.85│+│12.2-13.85│+│15.6-13.85│=2.0

| |

8

b. median = [pic] so[pic]th observation. Now rearrange the data either from lowest to highest or highest to lowest; 10.6 12.00 12.2 12.6 14.8 14.8 15.6 18.2 and we found above median falls between 4th and 5th observation simply add these two and divide it by 2 ( 12.6+14.8 / 2 = 13.7 is our median value.

Mode is 14.8 because it appears most frequent; 2 times!

c. [pic]from [pic]

d. s = 2.4512 from [pic] remember s = ( s2)2 thus 6.0086 * 6.0086 = 2.4512 but you must show all your calculations. This is only to check your answer.[pic]

e. sk= [pic] So, [pic]= 0.183 (slightly positively skerwed)

Question 5: The Bookstall Inc is a specialty bookstore concentrating on used books sold via the internet. Paperbacks are $1.00 each, and hardcover books are $3.50. Of the 50 books sold last Tuesday morning, 40 were paperback and the rest were hardcover. What was the weighted mean price of a book?

Answer: Here you cannot use simple mean because, values and @ the same time weights are different from each others. Use weighed mean; [pic] Thus,

[pic]= $1.50 each found by (40*1.00) + (10*3.5) / 50 = 1.50 / each.

Question 6: In 1976 the nationwide average price of a gallon of unleaded gasoline @ self serve pump was 0.605 $. By 2003, the average price had increased to 1.394$. What was the weighted mean annual increase for the period?

Answer:

Here you have data for 1976 and for 2003. Between these years no data available. The question is asking for increase from 1976 to 2003 ANNUALLY. That is why; we are using geometric mean to calculate ANNUAL increase.

From 1976 to 2003 price of gasoline increased by 3.14% annually, found by [pic] from

[pic]-1 where n is no of years from 1976 to 2003.

Question 7: The mean income of a group of sample observations is 500$; the standard deviation is $40. According to Chebyshev`s theorem at least what % of the incomes will lie between $400 and $600?

Answer:

400_______500________600, so k = 100/40 = 2.5 ⌠k is standard deviation of the mean, distance between 400 and 500 is 100, same as distance between 500 and 600. Thus k is simply 100/4⌡

From [pic]====>[pic]

Question 8: Given that mean number of drinks sold per year is 91.9 and the standard deviation is 4.67 at the nearby Wendy’s for the last 141 days. Using Empirical Rule,

a) Sales will be between what values on 68% of the days?

b) Sales will be between what two values on 95% of the days?

Answers:

a) [pic]Thus, 91.9 ( (1)4.67; between 87.23 and 96.57

b) [pic] Thus 91.9 ( (2)4.67; between 82.56 and 101.24

Question 9: There are 200 workers @ ABC Company and the following information is gathered from those.

|Marital Status: |Male |Female |

|Single |24 |56 |

|Married |36 |84 |

If one worker is selected @ random from those 200 workers, what is the probability that this worker is;

a) Married

b) Male

c) Single given that he is male

d) Female given that she is married

e) Married and Male

f) Male and single

g) Single or male

h) Female or single

i) Are events single and male mutually exclusive?

j) Are events female and male mutually exclusive?

k) Are events married and male independent?

Answers:

a) P (Married) = 120 / 200 from 36+84 = 120 is total married people.

b) P (Male) = 60/200 from 24+36 = 60 is total males without looking their marital status.

c) P (S/M) = 24/60 from single but among males, that is why we divide it by 60!

d) P (F/Married) = 84/120, question is saying the worker is female – GIVEN; female and married, so choose the married females and divide it by total married people.

e) P (Married AND Male) = 36/200 question is saying the worker must satisfy both condition; male and married, so choose the married males and divide it by total number.

f) P (Male AND Single) = 24/200, same procedures u followed in part e.

g) P(S or Male) = P(S) +P (M) - P(S AND Male) = 80/200 + 60/200 – 24/200 = 116/200. Here simply add probability of being single with probability of being male, but question says only one of the condition must satisfy Single or Male. Thus you must subtract the probability of being single AND male.

h) P (F or S) = P(F) + P(S) – P(F and S) = 140/200+80/200 – 56/200 = 164/ 200.

i) No, they are not mutually exclusive because, it is possible to be a male and single @ the same time.

j) Yes, they are mutually exclusive because, it is not possible to be a female and male @ the same time.

k) They are independent because, P(Married and Male) = P(Married) * P(Male)

36/200 = 120/200 * 60/200 where 0.18 = 0.18

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