The document provides an overview of the standard normal distribution and how to calculate probabilities and find z-scores, confidence intervals, and proportions using the normal distribution in Minitab. Examples are presented on finding probabilities based on z-scores, determining if a sample value falls within a certain percentage, calculating confidence intervals for means and proportions, and interpreting what confidence intervals represent.

1. B. Heard
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download one copy for personal use.)

2.  Standard Normal Distribution
 The “standard” normal distribution is a normal
distribution with mean zero and where the standard
deviation (and variance) equals one.
 The Total Area under the curve is one (1) or 100%
(This is true for all normal distributions regardless of
the mean and standard deviation).

3.  Using Minitab for Normal Distribution
calculations.
 Use Calc >> Probability Distributions >>
Normal
 Examples Follow

4.  Example
 Theaverage fish in Happy Lake weighs 2
pounds with a standard deviation of 0.5
pounds. If Bob catches a fish that weighs 3.2
pounds. What could you say about the
catch?

5. Since this is the Cumulative
Distribution Function, it “fills”
from left to right.
Therefore, you could say his
catch was in the “Top 1 %”

6.  Example
 Theaverage fish in Happy Lake weighs 2
pounds with a standard deviation of 0.5
pounds. If Bob catches a fish that weighs
1.35 pounds. What could you say about the
catch?

7. Since this is the Cumulative
Distribution Function, it “fills”
from left to right.
Therefore, you could say his
catch was in the “Bottom 10 %”

8.  Other types of questions
 If you have a normal distribution with a mu =
100 and sigma = 15, what number corresponds
to a z = -2
-2 = (x – 100)/15
Multiply both sides by 15 to get
-30 = x – 100
Add 100 to each side to get
70 = x
So “70” is my answer, I just did a little Algebra.

9.  Another type of question
 Say we take 120 samples of size 81 each from a
distribution we know is normal. Calculate the
standard deviation of the sample means if we
know the population variance is 25.
 (Answer next chart)

10.  Answer
 The Central Limit Theorem tells us the variance
is the Population variance divided by the Sample
Size. We can just take the square root to get
the standard deviation.
Variance = 25/81 or 0.309
Standard Deviation = Square Root(25/81) = 5/9 = 0.556

11.  Findingz scores
 Example
 The area to the left of the “z” is 0.6262. What z
score corresponds to this area.
 Use Calc >> Probability Distributions >> Normal
 (Set Mean = 0 and Standard Deviation to 1 and
use “INVERSE Cumulative Probability”

12. Answer is 0.322
rounded to three
decimals. Remember
the distribution fills
from left to right.

13.  Another type of question
 In a normal distribution with mu = 40 and
sigma = 10 find P(32 < x < 44)
 Easy, but this takes a couple of steps.
 Using Calc >> Probability Distributions >> Normal
find the probabililties that x < 32 and x < 44
using the Cumulative Probability option.

14.  Continued
Get results for
Both 32 and
then 44.

15.  Answer
Subtract
0.655422 –
0.211855
To get
0.443567
Or 0.444 rounded
to
three decimals
P(32 < x < 44) = 0.444 based
on
the given mean and std
deviation.

16.  Confidence Intervals and Examples
 Charts follow

17.  Interpreting Confidence Intervals
 If you have a 90% confidence interval of
(15.5, 23.7) for a population mean, it simply
means “There is a 90% chance that the
population mean is contained in the interval
(15.5, 23.7)
 It’s really that simple.

18.  Finding Confidence Intervals
 A luxury car company wants to estimate the true
mean cost of its competitor’s automobiles. It
randomly samples 180 of its competitors sticker
prices. The mean cost is $65,000 with a standard
deviation of $3200. Find a 95% confidence
interval for the true mean cost of the
competitor’s automobiles. Write a statement
about the interval.

19.  It randomly samples 180 of its competitors
sticker prices. The mean cost is $65,000
with a standard deviation of $3200. Find a
95% confidence interval…
 Use Stat >> Basic Statistics >> 1 sample Z
 Make sure to click Options and set to 95%

21.  Click your OK buttons…
Confidence Interval is
(64533, 65467), which
means we can be 95%
confident the true mean
cost of the competitor’s
vehicles are between
those two values.

22.  FindConfidence Intervals of Proportions
 Example
 An student wants to estimate what proportion of
the student body eats on campus. The student
randomly samples 200 students and finds 120 eat
on campus. Using a 95% confidence
interval, estimate the true proportion of
students who eat on campus. Write a statement
about the confidence level and interval.

23.  Example Solution
 p hat = 120/200 = 0.60
 q hat = 1- 0.60 = 0.40
 n p hat = 200 * 0.60 = 120
 n q hat = 200 * 0.40 = 80
 Using E = Zc* Square Root ((p hat * q hat)/n)
= 1.96 * Square Root ((0.60*0.40)/200)
=0.0679
Now we subtract this from the mean for the left side of
the interval and add it to the mean for the right side.
(0.60 – 0.0679, 0.60 + 0.0679) = (0.5321, 0.6679)
So with 95% confidence, we can say the population
proportion of students who eat lunch on campus is
(0.5321, 0.6679) or between 53.21% and 66.79%.

24.  Link to charts will be posted at
 www.facebook.com/statcave
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NEXT SUNDAY NIGHT FOR A BONUS LECTURE
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