Probability Distribution Memo
To: Howard Gray, CEO;
Jean Dubois, VP Mechanical Watch Division;
Uma Gardner, VP Production;
Amanda Hamilton, VP Marketing

After identifying the business problem of falling sales and an increase in rejections by the Swiss Official Chronometer Control, conducting a study for research will prove to identify a solution. Researchers performed a study of a sample population of 500 people. The study reveals 60% of the watches purchased are certified and the average rating of the SOCC certification as proof of quality is 3.9 therefore; consumers tend to believe the watches with SOCC certifications are of better quality. The study also reveals of the 39% planning on purchasing another watch within the next six months, 33% are willing to pay between $1,001.00 and $1,500.00 more for an Aquine watch. With that in mind, researchers do not believe advertising has anything to do with declining sales. Further research into the certification process is necessary because the study shows consumers rely on the certification as proof of quality and the rejection rate is increasing.

Reviewing the certification process reveals with a sample size of 120, the test with the highest rejection rates are Horizontal and Vertical Difference, Maximum Variation Rates, and Mean Variation Rates. The data for the three tests is evenly distributed therefore; “applying the Central Limit Theorem, researchers can conclude that the behavior sample represents the behavior of the population” (section 2). As a result of the tests, upgrading the Timing and Poising machines as well as purchasing Customized Movement Holders to aid in securing the base plates of the movement to a flat surface is the suggestion to resolve the business problem.

At this time, the study does not reveal a need for a Deep Water Simulation Unit or an Ultrasonic Cleaning Tank. Neither of these purchases will decrease the rejections by SOCC or increase sales.

...NORMAL DISTRIBUTION
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 normal distribution 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 normal distribution 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...

...
Normal Distribution
Normal distribution 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 normal distribution 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 normal distribution and the formula for normal distribution, which was found by...

...random variable is
A) generated by a random number table.
B) the variable for which an algebraic equation is solved.
C) a numerical measure of a probability experiment.. Ans = C
D) a qualitative attribute of a population.
4) Given the table of probabilities for the random variable x, does this form a probabilitydistribution? Answer yes or no.
x 5 10 15 25
P(x) 0.1 –0.1 0.3 0.8 Ans = No
5) True or False: The expected value of a discrete random variable may be negative Ans = True
6) The table of probabilities of the random variable x is given as:
x 0 1 2 5
P(x) 0.5 0.2 0.2 0.1
Find the mean, µ and standard deviation, σ of x. Round answers to one decimal place. Ans = µ = 1.1, σ = 1.5
7) If p is the probability of success of a binomial experiment then the probability of failure is
A) 1 B) –p C) 1–p D) p + 0.5 Ans = C
8) A binomial experiment has 6 trials with the probability of success on any trial = p = 0.5. Find the probability of exactly 2 successes in the 6 trials. (Use the binomial probabilitydistribution function.) Ans = 0.2344
9) Assume that male and female births are equally likely and the birth of any child does not affect the probability of the gender of any other...

...from a discrete distribution in this manner: A green die and a red die are thrown simultaneously 100 times and let Xi denote the sum of the spots on the two dice on the ith throw, i = 1, 2,...100. Find the probability that the sample mean number of spots on the two dice is less than 7.5.
n = 100
µ = 7 µ[pic] = 7
σ = 2.41 σ[pic] = 2.41 /[pic]
|X |2 |3 |4 |5 |
|P(X=x) |0.3333 |0.3333 |0.3333 | 1.00 |
|P.X |199.9800 |99.9900 |0.0000 | 299.97 |
|P².X | 119,988.00 | 29,997.00 | - | 149,985.00 |
σ² = 149,985.00 - 299.97²
= 60,003
σ = [pic]
= 244.9551
P([pic]≥320) = 1 - P ( Z < [pic] - µ[pic] )
σ[pic]
= 1 - P ( Z < 320-299.97 )
244.9551/[pic]
= 1 - P ( Z < 20.03 )
14.1425
= 1 - P ( Z < 1.42)
= 1 - 0.9222
= 0.0778 is the probability that at least 320 will attend.
1 - NORMDIST(320,299.97,244.9551/SQRT(300),1) = 0.0783
(c) A circuit contains three resistors wired in series. Each is rated at 6 ohms. Suppose, however, that the true resistance of each one is a normally distributed random variable with a mean of 6 ohms and a standard...

...computers are infected. What is the probability that 3 randomly chosen client computers serviced by different servers (one per server) will all be infected?
The probability that Alice’s RSA signature on a document is forged is () What is the probability that out of 4 messages sent by Alice to Bob at least one is not forged?
Event A is selecting a “red” card from a standard deck at random. Suggest another event (Event B) that is compatible with Event A.
What is the probability of getting 6 tails in 10 trials of tossing a coin? Solve this problem by using :The approximation mentioned in Theorem 6
The Binomial Distribution
Then compare answers for a) and b) after you have solved the problem.
When transmitting messages from a point A to a point B, out of every 40 messages 6 need to be corrected by applying error correcting codes. What is the probability that in a batch of 200 messages sent from A to B, there will be between 38 and 42 messages that will have to be corrected? Please choose the appropriate method to approximate this quantity.
The probability of an event occurring in each of a series of independent trials is . Find the distribution function of the number of occurrences of in 9 trials. That is, provide a table with all possibilities for number of occurrences of in 9 trials and calculate each’s corresponding...

...(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 variable that is always positive (such as a waiting time), with probability density function f(t) and cumulative distribution function F(t), then the hazard function h(t) is defined to be the function
h(t) = f(t)/(1-F(t)) The hazard function is the rate of failure per unit time, expressed as a proportion of the items that have failed. (a) If T ~ Weibull(α,β), find h(t) (b) For what values of α is the hazard rate increasing with time? For what values of α is it decreasing? (c) If T has an exponential distribution, show that the hazard function is constant. Question 5: [20 points] Among the adults in a large city, 30% have a college degree. A simple random sample of 100 adults is chosen. What is the probability that more than 35 of them have a college degree? (Hint: Central Limit Theorem) Question 6: [20 points] (a) Discuss how the fundamental matrix can be used to analyze an absorbing Markov chain. (b) Consider the following Markov Chain with state S={1,2,3}.
Denote...

...outcome xi according to its probability, pi. The mean also of a random variable provides the long-run average of the variable, or the expected average outcome over many observations.The common symbol for the mean (also known as the expected value of X) is , formally defined by
Variance - The variance of a discrete random variable X measures the spread, or variability, of the distribution, and is defined by
The standard deviation is the square root of the variance.
Expectation - The expected value (or mean) of X, where X is a discrete random variable, is a weighted average of the possible values that X can take, each value being weighted according to the probability of that event occurring. The expected value of X is usually written as E(X) or m.
E(X) = S x P(X = x)
So the expected value is the sum of: [(each of the possible outcomes) × (the probability of the outcome occurring)].In more concrete terms, the expectation is what you would expect the outcome of an experiment to be on average.
2. Define the following;
a) Binomial Distribution - is the discrete probabilitydistribution of the number of successes in a sequence of n independent yes/no experiments, each of which yields success with probability p. Therewith the probability of an event is defined by its binomial distribution. A success/failure experiment is also...

...Probabilitydistribution
Definition with example:
The total set of all the probabilities of a random variable to attain all the possible values. Let me give an example. We toss a coin 3 times and try to find what the probability of obtaining head is? Here the event of getting head is known as the random variable. Now what are the possible values of the random variable, i.e. what is the possible number of times that head might occur? It is 0 (head never occurs), 1 (head occurs once out of 2 tosses), and 2 (head occurs both the times the coin is tossed). Hence the random variable is “getting head” and its values are 0, 1, 2. now probabilitydistribution is the probabilities of all these values. The probability of getting 0 heads is 0.25, the probability of getting 1 head is 0.5, and probability of getting 2 heads is 0.25.
There is a very important point over here. In the above example, the random variable had 3 values namely 0, 1, and 2. These are discrete values. It might happen in 1 certain example that 1 random variable assumes 1 continuous range of values between x to y. In that case also we can find the probabilitydistribution of the random variable. Soon we shall see that there are three types of probabilitydistributions. Two of them deal with discrete values of the...