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...

...decimal places)
2. Find the value of z if the area 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...

...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...

...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...

...
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...

...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...

...Pages
C H A P T E
R
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...

...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...

...
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...

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