a. This is a GROUP assignment.
b. Each student must be a member of a group of 4 or 5 students, selected by lecturer. c. Solutions from each group must be submitted by 19 April 2013.

SPECIAL DISTRIBUTIONS

I. Concept of probability (3%)

1. Explain why the distribution B(n,p) can be approximated by Poisson distribution with parameter if n tends to infinity, p 0, and = np can be considered constant.

2. Show that – and + are the turning points in the graph of the p.d.f. of normal distribution with mean and standard deviation .

3. What is the relationship between exponential distribution and Poisson distribution?

II. Computation of probability (7%)

1. Let the random variable X follow a Binomial distribution with parameters n and p. We write X ~ B(n,p). * Write down all basic assumptions of Binomial distribution. * Knowing the p.m.f. of X, show that the mean and variance of X are = np, and 2 = np(1 – p), respectively.

2. A batch contains 40 bacteria cells and 12 of them are not capable of cellular replication. Suppose you examine 3 bacteria cells selected at random without replacement. What is the probability that at least one of the selected cells cannot replicate?

3. Redo problem No. 2 if the 3 bacteria cells are selected at random with replacement.

4. The number of customers who enter a bank in an hour follows a Poisson distribution. If P(X = 0) = 0.05, determine the mean and variance of the number of customers in an hour.

5. In a large corporate computer network, user log-ons to the system can be modeled as a Poisson process with a mean of 25 log-ons per hour. What is the probability that there are no log-ons in an interval of 6 minutes?

6. The time until recharge for a battery in a laptop computer under common conditions is normally distributed with a...

...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....

...What are the characteristics of a population for which a mean/median/mode would be appropriate? Inappropriate?
The analysis of data begins with descriptive statistics such as the mean, median, mode, range, standard deviation, variance, standard error of the mean, and confidence intervals. These statistics are used to summarize data and provide information about the sample from which the data were drawn and the accuracy with which the sample represents the population of interest. The mean, median, and mode are measurements of the “central tendency” of the data. The range, standard deviation, variance, standard error of the mean, and confidence intervals provide information about the “dispersion” or variability of the data about the measurements of central tendency.
MEASUREMENTS OF CENTRAL TENDENCY The appropriateness of using the mean, median, or mode in data analysis is dependent upon the nature of the data set and its distribution (normal vs non-normal). The mean (denoted by x) is calculated by dividing the sum of the individual data points (where Σ equals “sum of”) by the number of observations (denoted by n). It is the arithmetic average of the observations and is used to describe the center of a data set.
mean=x= One of the most basic purposes of statistics is simply to enable us to make sense of large numbers. For example, if you want to know how the students in your school...

...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....

...with μ = 110 grams and σ = 25 grams. A sample of 25 vitamins is to be selected. What is the probability that the sample mean will be less than 100 grams?
9)
The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with μ = 110 grams and σ = 25 grams. A sample of 25 vitamins is to be selected. What is the probability that the sample mean will be greater than 100 grams?
10)
The amount of pyridoxine (in grams) per multiple vitamin is normally distributed with μ = 110 grams and σ = 25 grams. A sample of 25 vitamins is to be selected. So, 95% of all sample means will be greater than how many grams?
TABLE 7-1
Times spent studying by students in the week before final exams follow a normaldistribution with standard deviation 8 hours. A random sample of 4 students was taken in order to estimate the mean study time for the population of all students.
11)
Referring to Table 7-1, what is the probability that the sample mean exceeds the population mean by more than 2 hours?
12)
Referring to Table 7-1, what is the probability that the sample mean is more than 3 hours below the population mean?
13)
Referring to Table 7-1, what is the probability that the sample mean differs from the population mean by less than 2 hours?
14)
Referring to Table 7-1, what is the probability that the sample mean differs from the population mean by more than 3 hours?...

...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)...

...6
The NormalDistribution
Objectives
Outline
After completing this chapter, you should be able to
1
2
3
Identify distributions as symmetric or skewed.
4
Find probabilities for a normally distributed
variable by transforming it into a standard
normal variable.
Introduction
6–1
NormalDistributions
Identify the properties of a normaldistribution.
Find the area under the standard normaldistribution, given various z values.
5
Find speciﬁc data values for given
percentages, using the standard normaldistribution.
6
6–3 The Central Limit Theorem
6–4 The Normal Approximation to the Binomial
Distribution
Use the central limit theorem to solve
problems involving sample means for large
samples.
7
6–2 Applications of the NormalDistribution
Use the normal approximation to compute
probabilities for a binomial variable.
Summary
6–1
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Confirming Pages
300
Chapter 6 The NormalDistributionStatistics
Today
What Is Normal?
Medical researchers have determined so-called normal intervals for a person’s blood
pressure, cholesterol,...

...
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...

...School: Science, Technology, Engineering, and MathCourse Number: Math 302Course Name: StatisticsCredit Hours: 3 Credit HoursLength of Course: 16 WeeksPrerequisite: Math 110, College Algebra or an equivalent course |
Table of Contents |
Instructor Information | Evaluation Procedures |
Course Description | Evaluation Criteria |
Course Scope | Course Outline |
Course Objectives | Policies |
Course Delivery Method | Academic Services |
Course Resources | Supplemental Materials |
Instructor Information |
Instructor: See the Syllabus section in the online classroom
Email: See the Syllabus section in the online classroom
Table of Contents
Course Description (Catalog) |
This is an interactive course designed to help students achieve a greater understanding of the statistical methods and models available to analyze and solve business management problems. The course is designed for students majoring in a business administration or management course of study. Successful completion of this course will provide students with a working knowledge of the principles of statistics, the ability to analyze and solve problems involving probability, and a working knowledge of averages and variations, normal probability distributions, sampling distributions, confidence intervals and testing statistical...

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