Review the case study on page 359 of the textbook.

The data on the number of chocolate chips per bag for 42 bags of Chips Ahoy! cookies were obtained by the students in an introductory statistics class at the United States Air Force Academy in response to the Chips Ahoy! 1,000 Chips Challenge sponsored by Nabisco, the makers of Chips Ahoy! Use the data collected by the students to answer the following questions and to conduct the analyses required in each part.

A. Obtain and interpret a point estimate for the mean number of chocolate chips per bag for all bags of Chips Ahoy! Cookies. (Note: The sum of the data is 52,986.) B .Construct and interpret a normal probability plot, boxplot, and histogram of the data. C. Use the graphs in part (b) to identify outliers, if any. D.Is it reasonable to use the one-mean t-interval procedure to obtain a confidence interval for the mean number of chocolate chips per bag for all bags of Chips Ahoy! cookies? Explain your answer. E. Determine a 95% confidence interval for the mean number of chips per bag for all bags of Chips Ahoy! cookies, and interpret your result in words. (Note: = 1261.6; s = 117.6.)

A. ͞X =∑Xi/n = 52986/42 = 1261.57
BASED ON THE SAMPLE DATA ,WE ESTIMATE THAT MEAN CHIPS PER BAG TO APPROXIMATLY 1261.57 CHIPS PER BAG.

B. GRAPHS ARE ATTACHED
C. I NOTICED MAY BE 1545 OR 1546 ARE OUTLINERS, THAT WHAT IT LOOKED LIKE AT BOX PLOT.

D. I THINK YOU CAN USE ONE MEAN T-INTERVAL PROCEDURE EVEN IF THE STANDARDIZED DEVIATION NOT GIVEN BY USING THE FOLLOWING

STEP 1. FOR A CONFIDENCE LEVEL OF 1-α, USE TABLE IV TO FINDTα/2 WITH DF= N-1 WHERE N IS SAMPLE SIZE WE WANT 95 % CONFIDENCE INTERVAL SO α= 1-0.95 =0.05
FOR N = 42 DF = 42-1= 41
FROM TABLE IV Tα/2=T0.05/2= T0.025=2.020
STEP 2 THE CONFIDENCE INTERVAL FOR µ IS FROM
͞X –T(α/2) * S/√N TO ͞X + T(α/2)*S/√N
͞X = 1261.57
S= ͞X-µ/(N-1/√N) = 1261.57-1000...

...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
blu34978_ch06.qxd
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4:39 PM
Page 300
Confirming Pages
300
Chapter 6 The NormalDistribution
Statistics
Today
What Is Normal?
Medical researchers have determined so-called normal intervals for a person’s blood
pressure, cholesterol, triglycerides, and the...

...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....

...
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...

...
PGEG371: Data Analysis & Geostatistics
NormalDistributions
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 NormalDistribution to find information about a data population.
On successful completion of this exercise, you should be able to
Describe what aNormalDistribution is;
How the histogram for a whole population looks like;
How to find information about a population using probability distribution function (PDF), and cumulative distribution function (CDF);
How to calculate the best estimate of the mean and standard deviation, and the confidence interval.
2. Introduction
In this section lab, you will study PDF and CDF of a Normaldistribution. PDF is the Probability Distribution Function, and CDF is the Cumulative Distribution Function. A NormalDistribution is the simplest model that exists. Could you speculate why?
Normal (Gaussian) distribution has very important characteristics. Some of them are: (1) it has always the same shape,
(2) the mean values is in the centre of the distribution (any thoughts about the median and mode?),
(3) the standard deviation...

... 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 are doing in the statewide achievement test, and somebody gives you a list of all 600 of their scores, that’s useless. This everyday problem is even more obvious and staggering when you’re dealing, let’s say, with the population data for the nation....

...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)...

...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....

...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 probabilitydistributions, sampling distributions, confidence intervals and testing statistical hypotheses. The emphasis of the course will be on the proper use of statistical techniques and their implementation rather than on mathematical proofs. (Prerequisite: MATH110 formerly MA112).
Table of Contents
Course Scope |
Successful completion of this course will provide you with a working knowledge of the principles of statistics and enable you to solve problems involving simple probability, averages and variations, normal probability distributions, sampling distributions, confidence intervals, and the testing of statistical hypotheses. The course is designed for students majoring in business administration or management programs of study. The emphasis of the course will be on the proper use of statistical techniques and their implementation rather than on mathematical proofs. However, some mathematics is necessary in order...

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