Normal Distribution

1. Normal Distribution

2. Mean, Median, Mode at Line of Symmetry Properties of Normal Distribution • It is symmetrical; the mean, median, and mode are equal and fall at the line of symmetry for the curve. • It is shaped like a bell, peaking at the middle and sloping down toward the sides. • Approximately 68% of the data is within one standard deviation of the mean. • Approximately 95% of the data is within two standard deviations of the mean. • Approximately 99.7% of the data is within three standard deviations of the mean.

3. Properties of Normal Distribution • Approximately 68% of the data is within one standard deviation of the mean. • 68% of data is within 3 of the mean (975)… • 68% of data lies within the range 972 - 978

4. Properties of Normal Distribution • Approximately 95% of the data is within two standard deviations of the mean. • 95% of data is within 6 of the mean (975)… • 95% of data lies within the range 969 - 981

5. Example 1 As an engineer who designs roller coasters, Mary wants to develop a ride that 95% of the population can ride. The average adult in North America has a mass of 71.8 kg with a standard deviation of 13.6kg. • What range of masses should she be prepared to anticipate?

6. 99 44.6 71.8 Example 1 – Range of masses • In a normal distribution, 95% of the data is within 2 standard deviations of the mean • 71.8 - 2(13.6) = 44.6 • 71.8 + 2(13.6) = 99 • She should expect masses 44.6 - 90 kg to include 95% of the population

7. Example 2 – Area under a Normal Curve • If X~N(50,52), draw diagrams to represent the following values for a random variable, X: • X > 55 • 40 < X < 60 • X < 38

8. -1 0 1 Standardizing the Normal Distribution • X~N(,2) is any normal curve with mean  and standard deviation, . • Z~N(0,1) is the standard normal curve with  = 0 and  = 1. -   + 

9. -1 0 1 Standardizing the Normal Distribution • Z is the standard variable for any Normal Distribution • To convert from Normal distribution to Z score use the following formula:

10. Example 3 • The times people wait in line at a grocery store are normally distributed with a mean of 10 minutes and a standard deviation of 2.5 minutes. • What is the probability that a customer waits less than 5 minutes? • Let X = # of minutes waiting • = 10 •  = 2.5 • To find P(X<5), standardize the Normal Distribution and find the area under the curve.

11. Example 3 (continued) • P(X < 5) = P(Z < -2) = ??? • Using the Normal Distribution Curve to find the probability…. • P(Z< -2) = 0.0228 • The probability of waiting less than 5 minutes is 2.3%

12. Another Probability Problem • What is the probability of waiting 7 – 10.5 minutes? i.e. P(7<X<10.5) • What does the shaded area look like under the normal curve? • The Standard Normal Distribution tables calculate probabilities/area TO THE LEFT of a particular z-score. •  So we have to divide this into two parts…. • P(7<X<10.5) = P(X<10.5) – P(X<7)

13. If you need to find the area between two Z-scores,  you should look up the larger Z-score and record the area to the left of this number.  Then look up the smaller Z-score and again record the area to the left.  Finally subtract the smaller area from the larger area.

14. Example 4 • Michael is 190cm tall. In his high school, heights are normally distributed with a mean of 165cm and a standard deviation of 20cm. What is the probability that Michael’s best friend is taller than he is? • Let X = the height the student • Find P(X > 190)

15. Example 4 (continued) • Standardize • Calculate the Probability P(Z < 1.25) • P(Z < 1.25) = 0.8955 • Subtract this probability from 1 since we want P(Z > 1.25) …we are calculating an area to the RIGHT • P(Z > 1.25) = 1 – P(Z < 1.25) = 1 – 0.8944 = 0.1056