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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
300
240
260
330
150
600
145
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.
325
135
160
350
115
230
190
485
215
540
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:
110 Up to 210
210 up to 310
310 up to 410
410 up to 510
510 up to 610
Frequencies:
8
5
5
3
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.
Histogram
8
8
5
6
5
3
Frequencies 4
3
2
Values
0
110 up
to 210
210 up
to 310
310 up
to 410
410 up
to 510
Monthly Expenses
510 up
to 610
*** Also look @ Page 35 and 39 to learn how to draw frequency polygon and cumulative frequency polygon.
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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:
0 up to 3
3 up to 6
6 up to 9
9 up to 12
12 up to 15
No of Workers:
8
11
37
20
4
a) How many people questioned @ ABC holding?
b) is 3 hours of working falls in the 1st class?
c) Calculate the relative and cumulative frequencies.
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
0 up to 3
3 up to 6
6 up to 9
9 up to 12
12 up to 15
d) x 
 fm
n
0 up to 3
3 up to 6
6 up to 9
9 up to 12
12 up to 15
Frequencies
Mid Points (m)
8
11
37
20
4
1.5
4.5
7.5
10.5
13.5
Cumulative Frequencies:
8
8+11 = 19
19+37=56
56+20=76
76+4=80
fm
fm2 = fm*m
8*1.5 = 12
11*4.5 =49.5
37*7.5 = 277.5
20*10.5 =210
4*13.5 = 54
∑fm = 603
12*1.5 =18
49.5*4.5 = 222.75
277.5*7.5 = 2081.25
210*10.5 = 2205
54*13.5 = 729
∑fm2 = 5256
603
 7.5375 is the mean.
80
 fm
 fm 
2
6032
5256 
s=
Relative Frequencies:
8/80 = 0.1
11/80 = 0.1375
37/80=0.4625
20/80 = 0.25
3/80 = 0.05
∑ = 1.00
if it is grouped data.
Classes
x
Frequencies
8
11
37
20
4
∑n = 80
80
80  1
2
= 3.00 from s =
n 1
n
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?
2
Answers:
a.
$18,000, found by $140,000 – $122,000; where R = H – L
b.
$129,600, found by $648,000/5; from µ =
c.
σ 2=
 ( x  u)
x
N
and so; 125000+128000+122000+133000+140000 / 5
Variance = 40,240,000, found by 201,200,000/5 Standard Deviation = $6,343.50
2
is variance of population for population data. (Notice that you use the population formulae)
N
 x  u 
2
σ=
N
x values:
125000
128000
122000
133000
140000
is standard deviation formula for population data.
µ
129600
129600
129600
129600
129600
(x-µ)2
21160000
2560000
57760000
11560000
108160000 then add all these!
∑(x-µ)2 = 201, 200, 000
Thus σ 2 = 201200000/5 = 40,240,000
Thus σ =
201200000
= 6,343.50
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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.
X  13.85 from x 
MD =
XX
n
 x  10.6+12.6+14.8+18.2+12+14.8+12.2+15.6 / 8 = 13.85
n
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.
n 1 8 1
 4.5 th observation. Now rearrange the data either from lowest to highest or highest
median =
so
2
2
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
th
and 5 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!
(10.6  13.85) 2  ...  (15.6  13.85) 2
c. s 2 
 6.0086 from
8 1
s
2


x  x 
2
n 1
3
 x  x 
2
s = 2.4512 from s 
d.
n 1
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.
e.
sk=
3mean median 
3(13.85  13.7)
So,
= 0.183 (slightly positively skerwed)
s
2.4512
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;
x
w

w1 x1  w2 x2  w3 x3  ......wn xn
Thus,
w1  w2  ...  wn
x w = $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
n
Value @ the end
Value @ the beginninh
27
1.394
 1 from
0.605
-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 1 
1
1
 84%
====> 1 
2
k
2.5 2
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) x  1s about 68% Thus, 91.9  (1)4.67; between 87.23 and 96.57
b) x  2s about 95% Thus 91.9  (2)4.67; between 82.56 and 101.24
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Question 9: There are 200 workers @ ABC Company and the following information is gathered from those.
Marital Status:
Single
Married
Male
24
36
Female
56
84
If one worker is selected @ random from those 200 workers, what is the probability that this worker is;
a)
b)
c)
d)
e)
f)
g)
h)
Married
Male
Single given that he is male
Female given that she is married
Married and Male
Male and single
Single or male
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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