COMP 211 DATA AND SYSTEM MODELING (PROB/STAT) Spring 2012 Assignment #2 Due: Monday, 5pm, 4/16/2012 Total points: 200 (each question 20 points) Please submit a softcopy (in PDF format) of your assignment to WebCT before the deadline. Late penalty: within 24 hours after the deadline: ‐20%; after 24 hours: 0 point. Question 1: [20 points] A film-coating process produces films whose thickness are normally distributed with a mean of 110 microns and a standard deviation of 10 microns. For a certain application, the minimum acceptable thickness is 90 microns. (a) What proportion of films will be too thin? (b) To what value should the mean be set so that only 1% of the films will be too thin? (c) If the mean remains at 110, what must the standard deviation be so that only 1% of the films will be too thin? Question 2: [20 points] If a resistor with resistance R ohms carries a current of I amperes, the potential difference across the resistor, in volts, is given by V=IR. Suppose that I is lognormal with parameters μI =1 and σI2 = 0.2, R is lognormal with parameters μR =4 and σR2 = 0.1, and that I and R are independent. (a) Show that V is lognormally distributed , and compute the parametersμV and σV2 (Hint: ln V = ln I + ln R) (b) Find P(V < 200) (c) Find P(150≦V≦300) (d) Find the mean of V (e) Find the median of V (f) Find the standard deviation of V Question 3: [20 points] The number of traffic accidents at a certain intersection is thought to be well modeled by a Poisson process with a mean of 3 accidents per year. (a) Find the mean waiting time between accidents. (b) Find the standard deviation of the waiting times between accidents. (c) Find the probability that more than one year elapses between accidents. (d) Find the probability that less than one month elapses between accidents. (e) If no accidents have occurred within the last six months, what is the probability that an accident will occur within the next year? Question 4: [20 points] If T is a continuous random...

...
NormalDistributionNormaldistribution is a statistics, which have been widely applied of all mathematical concepts, among large number of statisticians. Abraham de Moivre, an 18th century statistician and consultant to gamblers, noticed that as the number of events (N) increased, the distribution approached, forming a very smooth curve.
He insisted that a new discovery of a mathematical expression for this curve could lead to an easier way to find solutions to probabilities of, “60 or more heads out of 100 coin flips.” Along with this idea, Abraham de Moivre came up with a model that has a drawn curve through the midpoints on the top of each bar in a histogram of normally distributed data, which is called, “Normal Curve.”
One of the first applications of the normaldistribution was used in astronomical observations, where they found errors of measurement. In the seventeenth century, Galileo concluded the outcomes, with relation to the measurement of distances from the star. He proposed that small errors are more likely to occur than large errors, random errors are symmetric to the final errors, and his observations usually gather around the true values. Galileo’s theory of the errors were discovered to be the characteristics of normaldistribution and the formula for...

...is 60 points. Make sure your answers are as complete as possible and show your work/argument. In particular, when there are calculations involved, you should show how you come up with your answers with necessary tables, if applicable. Answers that come straight from program software packages will not be accepted. The quiz is due by midnight, Sunday, April 22, at 11:59 pm.
IMPORTANT: You are requested to include a brief note at the beginning of your submitted quiz, confirming that your work is your own. The note should say, "I have completed this assignment myself, working independently and not consulting anyone." Your submitted quiz will be accepted only if you have included this statement.
I have completed this assignment myself, working independently and not consulting anyone.
1. John made an experiment by tossing three fair coins. (Fair coin has the same probability for a tail and head ½).
(a) (3 points) List the sample space for this experiment. (All possible outcomes)
HHH, TTT, HTH, THT, TTH, HHT, HTT, THH = 8
(b)(2 point) What is a probability of three tails?
P(TTT) = 1/8 = 0.125
(c) (2 points) What is a probability of exactly two tails?
P(2 Tails) = 3/8 = 0.375
(d) (2 points) What is a probability of at least one tail?
P(1 Tail) = 7/8 = 0.875
2. Among UMUC students 70% own a car, 50% own a bike and 40% own both.
(a) (4 points) Draw a Venn...

...NormalDistribution
It is important because of Central Limit Theorem (CTL), the CTL said that Sum up a lot of i.i.d random variables the shape of the distribution will looks like Normal.
Normal P.D.F
Now we want to find c
This integral has been proved that it cannot have close form solution. However, someone gives an idea that looks stupid but actually very brilliant by multiply two of them.
reminds the function of circle which we can replace them to polar coordinate
Thus
Mean
By symmetry if g(x) is odd function g-x=-g(x) then -abgxdx=0
Variance
Notation
CDF is standard Normal CDF
by symmetric
,CDF , , All the odd moment of standard normal are zero. However, even moment is not easy to calculate by integral
(Symmetry)
Then we say
Most of Statistics books will write the pdf then explain the mean and variance but it is not intuitive.
Standardization
Find PDF of
CDF:
The PDF is derivative of the CDF (using chain rule)
PDF:
Later we’ll show if independent
68-95-99.7% Rule
Because you can’t actually calculate the , somebody create a rule of thumb
The properties of variance
If you shift the variance by c, the mean also shift by c. Thus, the variance doesn’t change.
Remember to square. It is easy to validate if you think c is negative and don’t square it. The corresponding variance becomes negative...

...NORMALDISTRIBUTION
1. Find the
distribution:
a.
b.
c.
d.
e.
f.
following probabilities, the random variable Z has standard normal
P (0< Z < 1.43)
P (0.11 < Z < 1.98)
P (-0.39 < Z < 1.22)
P (Z < 0.92)
P (Z > -1.78)
P (Z < -2.08)
2. Determine the areas under the standard normal curve between –z and +z:
♦ z = 0.5
♦ z = 2.0
Find the two values of z in standard normaldistribution so that:
P(-z < Z < +z) = 0.84
3. At a university, the average height of 500 students of a course is 1.70 m; the standard
deviation is 0.05 m. Find the probability that the height of a randomly selected student is:
1. Below 1.75 m
2. Between 1.68 m and 1.78 m
3. Above 1.60 m
4. Below 1.65m
5. Above 1.8 m
4. Suppose that IQ index follows the normaldistribution with µ = 100 and the standard
deviation σ = 16. Miss. Chi has the IQ index of 120. Find the percentage of people who
have the IQ index below that of Miss. Chi.
5. The length of steel beams made by the Smokers City Steel Company is normally
distributed with µ = 25.1 feet and σ = 0.25 feet.
a. What is the probability that a steel beam will be less than 24.8 feet long?
b. What is the probability that a steel beam will be more than 25.25 feet
long?
c. What is the probability that a steel beam will be between 24.9 and 25.7
feet long?
d. What is the probability that a steel beam will be between 24.6 and 24.9
feet long?
e....

...require that we know whether we have a sample or a population. 2. The following numbers represent the weights in pounds of six 7year old children in Mrs. Jones' 2nd grade class. {25, 60, 51, 47, 49, 45} Find the mean; median; mode; range; quartiles; variance; standard deviation. Solution: mean = 46.166.... median = 48 mode does not exist range = 35 Q1 = 45 Q2 = median = 48 Q3 = 51 variance = 112.1396 standard deviation =10.59 3. If the variance is 846, what is the standard deviation? Solution: standard deviation = square root of variance = sqrt(846) = 29.086 4. If we have the following data
34, 38, 22, 21, 29, 37, 40, 41, 22, 20, 49, 47, 20, 31, 34, 66 Draw a stem and leaf. Discuss the shape of the distribution. Solution: 2 3 4 5 6 | | | | | 219200 48714 0197 6
This distribution is right skewed (positively skewed) because the “tail” extends to the right. 5. What type of relationship is shown by this scatter plot?
45 40 35 30 25 20 15 10 5 0 0 5 10 15 20
Solution: Weak positive linear correlation 6. What values can r take in linear regression? Select 4 values in this interval and describe how they would be interpreted. Solution: the values are between –1 and +1 inclusive. -1 means strong negative correlation +1 means strong positive correlation 0 means no correlation .5 means moderate positive correlation etc. 7. Does correlation imply causation? Solution: No.
8. What do we call the r value. Solution: The correlation coefficient....

...NormalDistribution:- A continuous random variable X is a normaldistribution with the parameters mean and variance then the probability function can be written as
f(x) = - < x < , - < μ < , σ > 0.
When σ2 = 1, μ = 0 is called as standard normal.
Normaldistribution problems and solutions – Formulas:
X < μ = 0.5 – Z
X > μ = 0.5 + Z
X = μ = 0.5
where,
μ = mean
σ = standard deviation
X = normal random variable
NormalDistribution Problems and Solutions – Example Problems:
Example 1:
If X is a normal random variable with mean and standard deviation calculate the probability of P(X<50). When mean μ = 41 and standard deviation = 6.5
Solution:
Given
Mean μ = 41
Standard deviation σ = 6.5
Using the formula
Z =
Given value for X = 50
Z =
=
= 1.38
Z = 1.38
Using the Z table, we determine the Z value = 1.38
Z = 1.38 = 0.4162
If X is greater than μ then we use this formula
X > μ = 0.5 + Z
50 > 41 = 0.5 + 0.4162
P(X) = 0.5 + 0.4162
= 0.9162
Example 2:
If X is a normal random variable with mean and standard deviation calculate the probability of P(X< 37). When mean μ = 20 and standard deviation = 15
Solution:
Given
Mean μ = 20
Standard deviation σ = 15
Using the formula
Z =
Given value for X = 37
Z =
=
= 1.13
Z = 1.13
Using the Z...

...under a Standard Normal curve
a) to the right of z is 0.3632;
b) to the left of z is 0.1131;
c) between 0 and z, with z > 0, is 0.4838;
d) between -z and z, with z > 0, is 0.9500.
Ans : a) z = + 0.35 ( find 0.5- 0.3632 = 0.1368 in the normal table)
b) z = -1.21 ( find 0.5 – 0.1131 = 0.3869 in the normal table)
c ) the area between 0 to z is 0.4838, z = 2.14
d) the area to the right of +z = ( 1-0.95)/2 = 0.025, therefore z = 1.96
3. Given the Normally distributed variable X with mean 18 and standard deviation 2.5, find
a) P(X < 15);
b) the value of k such that P(X < k) = 0.2236;
c) the value of k such that P(X > k) = 0.1814;
d) P( 17 < X < 21).
Ans : X ~ N ( 18, 2.52)
a) P ( X < 15)
P ( Z < (15-18)/2.5) = P ( Z < -1.2) = 0.1151 ( 4 decimal places)
b) P ( X < k) = 0.2236
P ( Z < ( k – 18) / 2.5 ) = 0.2236
From normal table, 0.2236 = -0.76
(k-18)/2.5 = - 0.76, solve k = 16.1
c) P (X > k) = 0.1814
P ( Z > (k-18)/2.5 ) = 0.1814
From normal table, 0.1814 = 0.91
(k-18)/ 2.5 = 0.91, solve k = 20.275
d) P ( 17 < X < 21)
P ( (17 -18)/2.5 < Z < ( 21-18)/2.5)
P ( -0.4 < Z < 1.2) = 0.8849 – 0.3446 = 0.5403 ( 4 decimal places)...

...√9 = 3), we have
z19 = (19 – 25) / 3 = -2 and z31 = (31 - 25) / 3 = +2
From the area between z =±2 is 2(0.4772) = 0.9554
Therefore the probability that a measurement selected at random will be between 19 and 31 is about 0.95. This area (probability) is shown fir the X values and for the z values.
σ = 3 0.95 σ = 1 0.95
X
19 25 31 -2 0 +2
Normal curve showing Standard normal curve showing
area between 19 and 31 area between -2 and +2
Entry to a certain University is determined by a national test. The scores on this test are normally distributed with a mean of 500 and a standard deviation of 100.
Tom wants to be admitted to this university and he knows that he must score better than at least 70% of the students who took the test. Tom takes the test and scores 585. Will he be admitted to this university?
Solution:
Let x be the random variable that represents the scores. x is normally distributed with a mean of 500 and a standard deviation of 100. The total area under the normal curve represents the total number of students who took the test. If we multiply the values of the areas under the curve by 100, we obtain percentages.
For x = 585, z = (585 - 500) / 100 = 0.85
The proportion P of students who scored below 585 is given by
P = [area to the left of z = 0.85] = 0.8023 = 80.23%
Tom scored better than 80.23% of the students...

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