1. Chapter 5 Normal Probability Distribution
GOALS
List the characteristics of the normal probability distribution.
Define and calculate z values.
Determine the probability an observation will lie between two points using the standard normal distribution.

2. Characteristics Normal Probability Distribution
The normal curve is bell-shaped and has a single peak at the exact center of the distribution.
The arithmetic mean, median, and mode of the distribution are equal and located at the peak.
Thus half the area under the curve is above the mean and half is below it.

3. Characteristics Normal Probability Distribution
The normal probability distribution is symmetrical about its mean.
The normal probability distribution is asymptotic.
That is the curve gets closer and closer to the X-axis but never actually touches it

4. Characteristics Normal curve is symmetrical Theoretically, curve
extends to
infinity
Mean, median, and
mode are equal

5. The Standard Normal PD
The standard normal (or z) distribution is a normal distribution
Mean of 0; standard deviation of 1
A z-value is the distance between a selected value, designated X, and the population mean µ, divided by the population standard deviation, σ.

6. Standard Normal - Example
The bi-monthly starting salaries of recent MBA graduates follows the normal distribution with a mean of $2,000 and a standard deviation of $200.
z = (x-mu)/sd = (2200 – 2000)/200 = 2.00
Which “means” that a vale of 2200 is 2 standard deviations above the mean of 2000.

7. Standard Normal - Example
Mean $2000 / SD $200…
What is the z-value for a salary of $2,200?
What is the z-value of $1,700?
(x-mu)/sd = (1700 – 2200)/200 = -1.50
Which “means” that 1700 is 1.5 standard deviations below the mean of 2000
what do you think the z-value for the mean is?

8. Areas Under the Normal Curve
+/- …
1 Standard Deviation: About 68 %
2 Standard Deviations: About 95%
Practically all (99.7%) is within three
Applications
Estimates of population ranges/description
Translates to “probability”

9. Area - Example
The daily water usage per person in New Providence, New Jersey is normally distributed with a mean of 20 gallons and a standard deviation of 5 gallons.
About 68 percent of those living in New Providence will use how many gallons of water?
From the previous slide 68% translates to plus or minus 1 sd…
Therefore, you take the mean and range it plus/minus 1 sd… or
(20 – 5) to (20 + 5) or about 68% of the daily water usage will lie between 15 and 25 gallons.

10. Area - Example
What is the probability that a person from New Providence selected at random will use between 20 and 24 gallons per day?
Translate to “z value”
Use the book (or Excel) to determine the area under the curve… “the probability”

11. Z-Table

12. Area - Example
The area under a normal curve between a z-value of 0 and a z-value of 0.80 is
We conclude that percent of the residents use between 20 and 24 gallons of water per day.

13. Area - Example P(0<z<.8) =.2881 -4 -3 -2 -1 0 1 2 3 4
P(0<z<.8) =.2881
0 < x < 0.8

14. Area - Example
What percent of the population use between 18 and 26 gallons per day?

15. Area - Example Area associated with a z-value of – 0.40 is ____
Remember that the curve is symmetrical!
Area associated with a z-value of 1.20 is ______
Adding these areas, the result is _______
We conclude that…
/ / .5403
54.03 percent of the residents use between 18 and 26 gallons of water per day.

16. Area - Example
Professor Mann has determined that the scores in his statistics course are approximately normally distributed with a mean of 72 and a standard deviation of 5. He announces to the class that the top 15 percent of the scores will earn an A.
What is the lowest score a student can earn and still receive an A?

17. Area - Example
To begin let X be the score that separates an A from a B.
If 15 percent of the students score more than X, then 35 percent must score between the mean of 72 and X.
The z-value associated corresponding to 35 percent is about _______.
1.04

18. Area - Example
We let z equal 1.04 and solve the standard normal equation for X.
The result is the score that separates students that earned an A from those that earned a B.
Those with a score of 77.2 or more earn an A.
Using the equation… z = (X – mu)/sd…
1.04 = (X – 72)/5
X = (5) = = 77.2

19. Normal Approximation to the Binomial
The normal distribution (a continuous distribution) yields a good approximation of the binomial distribution (a discrete distribution) for large values of n.
Generally a good approximation to the binomial probability distribution when…
n*p and n*(1 - p) are both greater than 5.

20. The Normal Approximation continued
Recall for the binomial experiment:
There are only two mutually exclusive outcomes (success or failure) on each trial.
A binomial distribution results from counting the number of successes.
Each trial is independent.
The probability is fixed from trial to trial, and the number of trials n is also fixed.

21. Continuity Correction Factor
The value .5 subtracted or added, depending on the problem, to a selected value when a binomial probability distribution (a discrete probability distribution) is being approximated by a continuous probability distribution (the normal distribution).
Rules (pg 246)

22. Approximation - Example
A recent study by a marketing research firm showed that 15% of American households owned a video camera.
For a sample of 200 homes, how many of the homes would you expect to have video cameras?
Expect… mean of the binomial (mu) = n*p = (0.15)*(200) = 30 homes

23. Approximation - Example
What is the variance?
What is the standard deviation?
Variance = n*p*q = 30*(1-0.15) = 25.5
Standard deviation is srt(25.5) = homes

24. Approximation - Example
What is the probability that less than 40 homes in the sample have video cameras?
We use the correction factor… (pg 246)
So X is ________
The value of z is _______
39.5
Z = (39.5 – 30)/ = 1.88

25. Approximation - Example
From Appendix D the area between 0 and 1.88 on the z scale is
So the area to the left of 1.88 is =
The likelihood that less than 40 of the 200 homes have a video camera is about 97%.

26. P(z < 1.88) = =
z = 1.88