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...

...
PGEG371: Data Analysis & Geostatistics
Normal Distributions
Laboratory Exercise # 3
1st and 5th February, 2015
Read through this instruction sheet then answer the ‘pre-Lab’ quiz BEFORE starting the exercises!
1. Aim
The purpose of this laboratory exercise is to use a Normal Distribution to find information about a data population.
On successful completion of this exercise, you should be able to
Describe what a Normal Distribution 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...

...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....

...(a) Suppose we take a random sample of size 100 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...

...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...

...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?...

...mean of a random variable weights each 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...

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