The first step in building our simulation model was to confirm whether or not a normal distribution was in fact reasonable for the unknown demand of Egress’ ski jackets in the coming year. To do this we input Egress’ estimated demands into Best Fit. After analyzing the output (Exhibit 1 and 2) we see that the p-value for Chi Squared is .5637, this indicates that the assumption of a normal distribution of demand is indeed reasonable. Based on this output we determined that the mean of demand was 12,000 and the standard deviation was 3,497 (rounded up from 3496.8). Our next step was to build a simulation model based on the data provided to us by Egress using @RISK. Based upon our findings in Best Fit we used @RISK’s normal distribution function to generate random inputs of demand and the RISKSIMTABLE function to run the simulation multiple times using different production quantities. For each simulation run we did 1,000 iterations, and used the @RISK output function to analyze profit. At Egress’ request we ran our first simulation for a production quantity of 7,800 and production quantity at the mean demand of 12,000. From the output (Exhibit 3 and 4) we see that expected profit at Q = 7800 is $42,249 and the standard deviation is $52,026. At Q = 12,000 the expected profit is $42,296 with a standard deviation of $143,181. While the mean demand quantity of 12,000 yields a slightly better expected profit, it is a much riskier option. First, the standard deviation of Q = 12,000 is more than three times larger than that of Q = 7,800. Second, we can see in Exhibit 6 that there is a 16.4% chance that at Q = 12,000 Egress will lose $100,000 or more. In comparison when Q = 7,800 there is only a 3.3% chance that Egress will lose more than $100,000 (Exhibit 5). Finally, while at Q = 12,000 the company’s maximum profit is considerable larger than it is at Q = 7,800, the company stands to loose much more at Q = 12,000 as is reflected in the minimum profit...

...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....

...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...

...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...

...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 NormalDistribution
Statistics
Today
What Is Normal?
Medical researchers have determined so-called normal intervals for a person’s blood
pressure, cholesterol, triglycerides, and the...

...√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...

...
NormalDistribution
Student’s Name:
Instructorr:
Date ofSubmission:
The table 1 below shows a relationship between actual daily temperatures and precipitation in the month of January 2011. These data was adopted from a meterological station in the states of Alaska, in the United States. Normal distributed data aresymmetric with a single bell shaped peaks. Th maean of the data it significant in indication the point that the peak is likely to occur. In addition, standard deviation indicates the spread, which is usually referred to asthegirth of thebeell shapedcurve (Balakrishnan & Nevzorov, 2003).
Table 1: Relationship between actual daily temperatures and precipitation of January 2011
|Date |Actual Temperatures |Precipitation |
|Jan. 1 |30 |0 |
|Jan. 2 |25 |0 |
|Jan. 3 |31 |0 |
|Jan. 4 |33 |0 |
|Jan. 5 |29 |0 |
|Jan. 6 |36 |0.26 |
|Jan. 7 |36 |0 |
|Jan. 8 |37 |0.01 |
|Jan. 9 |32 |0.21 |
|Jan. 10 |28...