Very few of you attempted this fun problem! However, those few that did really hit the target! Congratulations to all of you that attempted this case study! The Life Insurance Case Study from Week 3 utilized the Z-score for Normal Distribution (Continuous Probability). (AKA Continuous Probability which is the Probability of AN event occurring). Let me walk you through this problem, step by step. The problem gives us 3 numbers The Mean/Average for this data set is 60% The data point of interest (X) is 64% (640 homes out of a sample of 1,000) - We are interested in understanding if At Least 64% of the homes (out of 1,000) purchased a Life Insurance Policy in the last year N = 1,000 Find the Standard Standard deviation of the Proportion. This is our 1st step because it is not provided = .0155 The Standard Deviation of the Sample Population is .0155 FInd the Z-score X = .64 U = .60 O = .0155 .64 - .60/.0155 = Z Z = 2.58 The 4th step is to find the Area of the Curve from Z = 0 to Z = 2.58 Go the Normal Distribution Table and find 2.5 and .08 at the Intersection we find = .4951 or 49.51% The data point of interest (X) .64 is aligned to 2.58 standard deviations away from the average of the data set (.60) In other words, If you look at a Bell Curve, from the middle of the Bell Curve (X) to Z = 2.58 the area of the curve that it covers is 49.51% (large area of the curve is covered) The question asks, what is the Probability of finding At Least 64% of the homes with a Life Insurance Policy purchase? At Least means anything greater than 64% If we are interested in everything Greater than 64% then we must find the area of the curve greater than 49.51%. If Each side of the bell curve is = to .5000, we must do the following .5000 - .4951 = .0049 OR .49% OR .5% In other words, the only remaining area is .5% (less than 1%) Which means that the Probability of finding At Least 64% of the homes with a Life Insurance Policy is .5% Probability. Is this a Low or High Probability?

Very few of you attempted this fun problem! However, those few that did really hit the target! Congratulations to all of you that attempted this case study! The Life Insurance Case Study from Week 3 utilized the Z-score for Normal Distribution (Continuous Probability). (AKA Continuous Probability which is the Probability of AN event occurring). Let me walk you through this problem, step by step. The problem gives us 3 numbers The Mean/Average for this data set is 60% The data point of interest (X) is 64% (640 homes out of a sample of 1,000) - We are interested in understanding if At Least 64% of the homes (out of 1,000) purchased a Life Insurance Policy in the last year N = 1,000 Find the Standard Standard deviation of the Proportion. This is our 1st step because it is not provided = .0155 The Standard Deviation of the Sample Population is .0155 FInd the Z-score X = .64 U = .60 O = .0155 .64 - .60/.0155 = Z Z = 2.58 The 4th step is to find the Area of the Curve from Z = 0 to Z = 2.58 Go the Normal Distribution Table and find 2.5 and .08 at the Intersection we find = .4951 or 49.51% The data point of interest (X) .64 is aligned to 2.58 standard deviations away from the average of the data set (.60) In other words, If you look at a Bell Curve, from the middle of the Bell Curve (X) to Z = 2.58 the area of the curve that it covers is 49.51% (large area of the curve is covered) The question asks, what is the Probability of finding At Least 64% of the homes with a Life Insurance Policy purchase? At Least means anything greater than 64% If we are interested in everything Greater than 64% then we must find the area of the curve greater than 49.51%. If Each side of the bell curve is = to .5000, we must do the following .5000 - .4951 = .0049 OR .49% OR .5% In other words, the only remaining area is .5% (less than 1%) Which means that the Probability of finding At Least 64% of the homes with a Life Insurance Policy is .5% Probability. Is this a Low or High Probability?

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Very few of you attempted this fun problem! However, those few that did really hit the target! Congratulations to all of you that attempted this case study!

The Life Insurance Case Study from Week 3 utilized the Z-score for Normal Distribution (Continuous Probability). (AKA Continuous Probability which is the Probability of AN event occurring).

Let me walk you through this problem, step by step.

The problem gives us 3 numbers

The Mean/Average for this data set is 60%
The data point of interest (X) is 64% (640 homes out of a sample of 1,000) - We are interested in understanding if At Least 64% of the homes (out of 1,000) purchased a Life Insurance Policy in the last year
N = 1,000

Find the Standard Standard deviation of the Proportion. This is our 1st step because it is not provided
= .0155
The Standard Deviation of the Sample Population is .0155
FInd the Z-score
X = .64
U = .60
O = .0155
.64 - .60/.0155 = Z
Z = 2.58

The 4th step is to find the Area of the Curve from Z = 0 to Z = 2.58
Go the Normal Distribution Table and find 2.5 and .08 at the Intersection we find = .4951 or 49.51%
The data point of interest (X) .64 is aligned to 2.58 standard deviations away from the average of the data set (.60)
In other words, If you look at a Bell Curve, from the middle of the Bell Curve (X) to Z = 2.58 the area of the curve that it covers is 49.51% (large area of the curve is covered)
The question asks, what is the Probability of finding At Least 64% of the homes with a Life Insurance Policy purchase? At Least means anything greater than 64%
If we are interested in everything Greater than 64% then we must find the area of the curve greater than 49.51%.
If Each side of the bell curve is = to .5000, we must do the following
.5000 - .4951 = .0049 OR .49% OR .5%
In other words, the only remaining area is .5% (less than 1%)
Which means that the Probability of finding At Least 64% of the homes with a Life Insurance Policy is .5% Probability.
Is this a Low or High Probability?

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