Great Deal! Get Instant $10 FREE in Account on First Order + 10% Cashback on Every Order Order Now

PowerPoint Presentation ITEC 210 DATA ANALYSIS FOR BUSINESS Normal Probability Distribution and Confidence Intervals Prof. Itir KARAESMEN AYDIN 1 Outline and Learning Outcomes In this presentation,...

1 answer below »
PowerPoint Presentation
ITEC 210 DATA ANALYSIS FOR BUSINESS
Normal Probability Distribution and Confidence Intervals
Prof. Itir KARAESMEN AYDIN
1
Outline and Learning Outcomes
In this presentation, you will learn
Normal distribution functions in Excel to compute probabilities.
Functions and tools in Excel to compute confidence intervals for a population mean.
Functions and tools in Excel to compute confidence intervals for a population proportion.
NOTE: This presentation does not show you *how* the work is done on Excel.
2
Topic:
Normal Distribution
and
Probability Calculations
3
Learning Objective
After completing these examples, you will be able to
Compute probabilities for a random variable that is standard normal distributed.
Compute probabilities for a random variable that has a general normal distribution.
Compute the value of the normal-distributed random variable co
esponding to a given cumulative probability.
4
Continuous Random Variables
A quantitative variable that takes any value on one or more intervals on the real line.
5
0
X
X
0
Continuous Random Variables
A quantitative variable that takes any value on one or more intervals on the real line.
We cannot compute the probability of the variable taking a precise value, but we can compute the probability of the variable lying within a given range.
6
0
X
X
0
What is the probability that X takes values in a given interval?
Normal Distribution
The density function f(x) determines the shape of the distribution for a continuous random variable.
The normal distribution is symmetric (skewness=0).
7
x
f(x)
Normal Distribution
Normal Probability Density Function
The mean and standard deviation are unique.
The distribution is symmetric around the mean.
Mean = Median = Mode for normal distribution.
 = mean
 = standard deviation
 = XXXXXXXXXX
e = XXXXXXXXXX
 
Normal Distribution
The probability that X takes values between X1 and X2 is the same as the area under the density function between X1 and X2 in the graph below:
9
x
x1
x2
f(x)
Normal Distribution
The probability that X takes values greater than X1 is the same as the area under the density function to the right of X1.
10
x
x1
f(x)
Normal Distribution
The probability that X takes values smaller than X1 is the same as the area under the density function to the left of X1.
11
x
x1
f(x)
Normal Distribution
The probability that X takes values smaller than X1 or greater than X2 is the same as the area under the density function to the left of X1 plus the area to right of X2.
12
x
x1
f(x)
x2
Mean and Standard Deviation
Mean can be positive, zero, or negative.
Mean () is the center value.
Standard deviation () affects the variation or spread around the mean.
13
x
=-10
=+10
=15
=25
Normal Distribution
P(- < X < + ) = 1 = total area under the density function
P(X ≤ ) = 0.5 = the area to the left of 
P(X ≥ ) = 0.5 = the area to the right of 
14
x
f(x)

The Empirical Rule
15
x
m–3s
m–s
m–2s
m+s
m+2s
m+3s
m
68.3%
95.4%
99.7%
Standard Normal Distribution
The normal distribution with =0 and =1 is called the standard normal distribution.
We use the letter Z for the random variable that is standard normal distributed
16
Z
f(x)
0
Excel Functions
NORMSDIST(cutoff)
NORM.S.DIST(cutoff,TRUE)
NORM.S.DIST(cutoff,1)
These functions calculate P(Z ≤ cutoff).
These functions always compute the “cumulative” probability: the function computes the “left tail” of the distribution, i.e., the area under the probability density curve to the left of a given a cutoff value.
The output is a numerical value between 0 and 1.
17
Excel Exercise #16
Cutoff value is 0.9. What is the probability that the observed values of the random variable Z will be less than or equal to 0.9?
P(Z ≤ 0.9) = NORM.S.DIST(0.9,TRUE)=0.8159
18
Z
0.9
f(x)

Excel Exercise #16 (cont’d)
Cutoff value is -0.2. What is the probability that the observed values of the random variable Z will be higher than or equal to -0.2?
P(Z ≥ -0.2) = 1 - P(Z ≤ -0.2) = 1- NORM.S.DIST(-0.2,TRUE)
                    = XXXXXXXXXX
                    = 0.5793.
19
-0.2
f(x)
Z
Excel Exercise #16 (cont’d)
P(-0.2 ≤ Z ≤ 0.9)
= NORM.S.DIST(0.9,TRUE) - NORM.S.DIST(-0.2,TRUE)
= 0.8159 – 0.4207 = 0.3952
20
0.9
f(x)

-0.2

Z
Excel Functions (cont’d)
NORMSINV(probability)
NORM.S.INV(probability)
These functions return the value of C such that
P(Z ≤ C) = probability.
The probability value is known. C is the unknown.

21
Excel Exercise #16 (cont’d)
Z is standard normal distributed.
The probability of observing values below C is 0.20.
Find C.
22
f(x)
C=?

Z
C = NORM.S.INV(0.2)
= XXXXXXXXXX
Excel Exercise #16 (cont’d)
Z is standard normal distributed.
The probability of observing values above C is 0.75.
Find C.
23
f(x)
C=?

Z
C = NORM.S.INV(1-0.75)
= NORM.S.INV(0.25)
= XXXXXXXXXX
Excel Exercise #17
Use Excel to compute the following for a standard normal distributed random variable Z.
P(Z ≤ -0.3)=?
P(Z ≤ 0.3)=?
P(Z ≥ -0.3)=?
P(Z ≥ 0.3)=?
P(-0.3 ≤ Z ≤ 0.3)=?
P(0.3 ≤ Z ≤ 1.0)=?
Find C such that P(Z ≤ C) = 0.88.
Find C such that P(Z ≤ C) = 0.3.
Find C for the 10% upper tail: P(Z>C)=0.1.
24
#17 – Excel functions
25
General Normal Distribution
A random variable has the “standard” normal distribution only when its mean is 0 and its standard deviation is 1.
Excel can calculate probabilities or cutoff values for random variables that are more “general” with means different from 0 and/or standard deviations different from 1.
26
Excel Functions for
General Normal Distribution
NORMDIST(cutoff,,, TRUE)
NORM.DIST(cutoff, ,,TRUE)
NORM.DIST(cutoff, ,,1)
These functions calculate P(X ≤ cutoff) for a random variable X that is Normal distributed with a known mean value  and a known standard deviation value .
The output is a numerical value between 0 and 1.
27
Excel Functions for
General Normal Distribution
NORMINV(probability, ,)
NORM.INV(probability, ,)
These functions return the value of C such that
P(X ≤ C) = probability
for a random variable X that is normal distributed with a known mean value  and a known standard deviation value .
The probability value is known. C is the unknown.
28
Excel Exercise #18
Use Excel to compute the following for a normal-distributed random variable X with =10 and =25.
P(X ≤ 15)=?
P(X ≤ -5)=?
P(X ≥ -10)=?
P(X ≥ 10)=?
P(-5 ≤ X ≤ 10)=?
P(10 ≤ X ≤ 15)=?
Find C such that P(X ≤ C)=0.88.
Find C such that P(X ≤ C)=0.3.
Find C for the 10% upper tail: P(X>C)=0.1.
29
#18 – Excel functions
30
Excel Exercise #19
An electronics retailer has only 20 webcams left in stock.
The retailer places a new order with their supplier.
The shipment a
ives from the supplier in 10 days.
The demand for the webcams is known to be normal distributed. The mean demand for webcams in the next 10 days is expected to be 18 and the standard deviation is 3.
31
#19 (cont’d)
a) What is the probability that the retailer will sell out the webcam inventory within the next 10 days?
Let X be the random demand.
The retailer sells out when demand ≥ stock level.
P(X ≥ Stock level)
    = P(X ≥ 20)
    = 1 - NORM.DIST(20,18,3,TRUE)
    = 1 – 0.7475 = 0.2525.    
32
#19 (cont’d)
) What is the probability that the retailer will have 6 or more units remaining in stock when the next shipment a
ives?
The retailer has 6 or more units remaining in stock,
if (stock level – demand) ≥ 6.
P(stock level – demand ≥ 6)
P(demand ≤ stock level-6)
    = P(X ≤ 20-6) = P(X ≤ 14)
    = NORM.DIST(14,18,3,TRUE) = 0.0912
    
33
#19 (cont’d)
c) If the retailer had ordered the webcams earlier (i.e. before the stock level dropped to 20 units), then the risk of a stockout would have been lower. What is the co
ect “reorder point” (the stock level at the time the retailer contacts the supplier for a new shipment) if the retailer wants the probability of a stock out until the next shipment a
ives to be at or below 5%?
The retailer has a stock out if demand ≥ stock level.
P(demand ≥ stock level) ≤ 0.05
 P(X ≥ cutoff) = 0.05
    = NORMINV(1-0.05,18,3) = 22.93
 23 units or more are needed to keep stock out chance below 5%.
    
34
Confidence Interval
for
The Population Proportion
35
Learning Objective
After completing these examples, you will be able to
Compute confidence intervals for a population proportion given sample proportions.
Compute confidence intervals for a population proportion given a sample data.
36
Sample vs. Population
It is hard to obtain information about the entire population.
We can take random, unbiased samples.
Each sample may have a different sample proportion.
How good is the information we obtain from a sample in representing the population? Is there a relationship between the sample information and the population information?
37
Sample vs. Population Proportion
p: population proportion
Example: p is the fraction of all Twitter accounts with zero tweets.
n: sample size
: sample proportion
Example: Given a sample of n Twitter accounts, is the fraction of Twitter accounts with zero tweets in this sample.
38
Confidence Interval for
Population Proportions
Use the following information
the sample proportion ,
the sample size (n),
a significance level of α which gives us 100(1-α)% confidence
to compute the confidence interval
[Lower limit of the CI, Upper limit of the CI].
We will be 100(1-α)% confident that the unknown population proportion is contained in this interval, i.e., we will compute the interval using a method that is successful in giving co
ect results 95% of the time.
39
Distribution of Sample Proportions
40
p
Each bu
le represents the value of one sample proportion.
1
0
0
1
There are 36 sample proportions represented in this picture.
The statistical distribution of
Answered Same Day Feb 15, 2021

Solution

Sanjeev answered on Feb 15 2021
147 Votes
Problems #16, #17
Problem#16. Standard Normal Distribution
Answer the following questions using Excel functions. The random variable Z is standard normal distributed.
a) (2 points) P(Z ≤ 0.4) = ?
) (2 points) P(-0.75 ≤ Z ≤ 0.6) = ?
c) (2 points) Find the value of C such that P(Z<=C)=0.15.
d) (2 points) Find the value of C such that P(Z>=C)=0.08.
Problem#17. Standard Normal Distribution
Use Excel to compute the following for a standard normal distributed random variable Z.
a) (2 points) P(Z > 0.6)=?
) (2 points) P(-0.2 ≤ Z ≤ 0.2)=?
c) (2 points) P(0.2 ≤ Z ≤ 0.5)=?
d) (2 points) Find the value of C such that P(Z > C) = 0.75.
Data #16, #17
    There is no data. Use the information provided in the problem statement.
Problems #18, #19
Problem#18. General Normal Distribution
Use Excel to compute the following for a normal distributed random variable Z with =400 and =65.
a) (2 points) P(Z > 500)=?
) (2 points) P( 335 ≤ Z ≤ 465)=?
c) (2 points) Find the value of C such that P(X > C) = 0.1.
Problem #19. General Normal Distribution
An electronics retailer has only 100 webcams left in stock until the next shipment a
ives from a supplier in 5 days. The demand for the webcams is known to be normal distributed. The mean demand for webcams in the next 5 days is expected to be 80 and the standard deviation is 14.
Use Excel to compute the following.
a) (2 points) What is the probability that the retailer will sell out the webcam inventory within the next 5 days?
) (2 points) What is the probability that the retailer will have 10 or more units remaining in stock when the next shipment a
ives?
c) (2 points) If the retailer had ordered the webcams earlier (i.e. before the stock level dropped to 100 units), then the risk of a stockout would have been lower. What is the co
ect “reorder point” (the stock level at the time the retailer contacts the supplier for a new shipment) if the retailer wants the probability of a stock out until the next shipment a
ives to be at or below 1%?
Data #18, #19
    There is no data. Use the information provided in the problem statement.
Problem #20, #21, #22
Data #20, #21, #22
    There is no data. Use the information provided in the problem statement.
    Problem 20                Problem 21                    Problem 22
    a                a                    a
        n    100            n    250                n    200
        alpha    0.05            alpha    0.1                alpha    0.1
        sample proportion    0.75            sample proportion    0.76                sample proportion    0.3
        Z    1.9599639845            Z    1.644853627                Z    1.644853627
        margin of e
or    0.0848689301            margin of e
or    0.0444293203                margin of e
or    0.0532993492
        Lower limit of CI    0.6651310699            Lower limit of CI    0.7155706797                Lower limit of CI    0.2467006508
        Upper limit of CI    0.8348689301            Upper limit of CI    0.8044293203                Upper limit of CI    0.3532993492
        We are 95% confident that the unknown population mean is contained in this interval [0.665,0.835]                We are 99% confident that the unknown population mean is contained in this interval [0.716,804]                    We are 90% confident that the unknown population mean is contained in this interval [0.247,0.353]
    b                b                    
        n    300            n    250                n    200
        alpha    0.08            alpha    0.01                alpha    0.1
        sample proportion    0.9            sample proportion    0.14                sample proportion    0.73
        Z    1.7506860713            Z    2.5758293035                Z    1.644853627
        margin of e
or    0.0303227722            margin of e
or    0.0565275703                margin of e
or    0.0516363542
        Lower limit of CI    0.8696772278            Lower limit of CI    0.0834724297                Lower limit of CI    0.6783636458
        Upper limit of CI    0.9303227722            Upper limit of CI    0.1965275703                Upper limit of CI    0.7816363542
        We are 92% confident that the unknown population mean is contained in this interval [0.870,0.930]                We are 99% confident that the unknown population mean is contained in this interval [0.083,0.197]                    We are 90% confident that the unknown population mean is contained in this interval [0.678,0.782]
                                        c.    Sample size has a negative relation with the margin of e
or because as the sample size increases, the margin e
or reduces. If in the above question, the sample size would have been greater than 200, then margin of e
or would be lower than what we have calculated in the above question.
Problem #23
Problem#23. Confidence Interval for a Population Proportion
Use the retailer data provided in Problem#7. That includes a sample of 150 customer. Answer the following questions based on the data.
NOTE: Show all your work in Excel.
a) (4 points) Compute the 95% confidence interval for the population proportion of customers who are 40 years or older. Interpret the confidence interval.
) (4 points) Compute the 99% confidence interval for the population proportion of customers who spent $150 or more. Interpret the confidence interval.
c) (4 points) Compute the 90% confidence interval for the population proportion of customers who used their EStore Card for the transaction. Interpret the confidence interval.
Data #23
    Customer ID    Method of Payment    No. of Months Since Last Purchase    Discount Code was Emailed    Discount Code Used    Transaction value ($ sales...
SOLUTION.PDF

Answer To This Question Is Available To Download

Related Questions & Answers

More Questions »

Submit New Assignment

Copy and Paste Your Assignment Here