1. 5 Normal Probability Distributions
Elementary Statistics
Larson Farber

2. Properties of a Normal Distributionx
The mean, median, and mode are equal
Bell shaped and is symmetric about the mean
The total area that lies under the curve is one or 100%

3. Properties of a Normal Distribution
Inflection point
Inflection point x
As the curve extends farther and farther away from the
mean, it gets closer and closer to the x-axis but never touches it.
The points at which the curvature changes are called inflection points. The graph curves downward between the inflection points and curves upward past the inflection points to the left and to the right.

4. Means and Standard Deviations
Curves with different means, same standard deviation 10 11 12 13 14 15 16 17 18 19 20
Curves with different means, different standard deviations 9 10 11 12 13 14 15 16 17 18 19 20 21 22

5. Empirical Rule
About 68% of the area lies within 1 standard deviation of the mean
68%
About 95% of the area lies within 2 standard deviations
About 99.7% of the area lies within 3 standard deviations of the mean

6. Determining Intervalsx
3.3
3.6
3.9
4.2
4.5
4.8
5.1
An instruction manual claims that the assembly time for a product is normally distributed with a mean of 4.2 hours
and standard deviation 0.3 hour. Determine the
interval in which 95% of the assembly times fall.
95% of the data will fall within 2 standard deviations of the mean.
4.2 – 2 (0.3) = 3.6 and (0.3) = 4.8.
95% of the assembly times will be between 3.6 and 4.8 hrs.

7. The Standard
Normal Distribution

8. The Standard Score
The standard score, or z-score, represents the number of standard deviations a random variable x falls from the mean.
The test scores for a civil service exam are normally distributed with a mean of 152 and a standard deviation of 7. Find the standard z-score for a person with a score of:
(a) (b) (c) 152
(a)
(b)
(c)

9. The Standard Score
The standard score, or z-score, represents the number of standard deviations a random variable x falls from the mean.
The test scores for a civil service exam are normally distributed with a mean of 152 and a standard deviation of 7. Find the standard z-score for a person with a score of:
(a) (b) (c) 152
(a)
(b)
(c)

10. The Standard Normal Distribution
The standard normal distribution has a mean of 0 and a standard deviation of 1.
Using z-scores any normal distribution can be transformed into the standard normal distribution. z –4 –3 –2 –1 1 2 3 4

11. Cumulative Areas
The
total
area
under
the curve
is one. z –3 –2 –1 1 2 3
The cumulative area for z = 0 is , indicating that the probability of getting a z value of 0 or less is .5

12. Cumulative Areas - Finding
Find the cumulative area for a z-score of –1.25.
0.1056 z –3 –2 –1 1 2 3
Table in Appendix
Table on fold-out card
Excel
Internet calculator

13. Using the table … Solution: Find -0.2 in the left hand column.
Find the cumulative area that corresponds to a z-score of
Solution:
Find -0.2 in the left hand column.
Move across the row to the column under 0.04

14. Using the table … The area to the left of z = -0.24 is 0.4052.
Find the cumulative area that corresponds to a z-score of
Solution:
Find -0.2 in the left hand column.
Move across the row to the column under 0.04
The area to the left of z = is

15. Using Excel
standard deviation
mean
x value of interest

16. Online: Rossman-Chance

17. Online: SurfStat

18. The probability that z is at most –1.25 is 0.1056.
Cumulative Areas
Find the cumulative area for a z-score of –1.25.
0.1056 z –3 –2 –1 1 2 3
Read down the z column on the left to z = –1.25 and across to the column under .05. The value in the cell is , the cumulative area.
The probability that z is at most –1.25 is

19. “Less than” z Find P(z < –1.45). –3 –2 –1 1 2 3
To find the probability that z is less than a given value, read the cumulative area in the table corresponding to that z-score.
Find P(z < –1.45).
P (z < –1.45) = z –3 –2 –1 1 2 3
Read down the z-column to –1.4 and across to .05. The cumulative area is

20. “Greater than”
To find the probability that z is greater than a given value, subtract the cumulative area in the table from 1.
Find P(z > –1.24).
0.1075
0.8925 z –3 –2 –1 1 2 3
The cumulative area (area to the left) is So the area to the right is 1 – =
P(z > –1.24) =

21. “Between”
To find the probability z is between two given values, find the cumulative areas for each and subtract the smaller area from the larger.
Find P(–1.25 < z < 1.17). z –3 –2 –1 1 2 3
1. P(z < 1.17) =
2. P(z < –1.25) =
3. P(–1.25 < z < 1.17) = – =

22. Summary
To find the probability that z is less
than a given value, read the
corresponding cumulative area. z -3 -2 -1 1 2 3
To find the probability is greater than a given value, subtract the cumulative area in the table from 1. -3 -2 -1 1 2 3 z
To find the probability z is between two given values, find the cumulative areas for each and subtract the smaller area from the larger. z -3 -2 -1 1 2 3

23. Finding Probabilities
Normal Distributions
Finding Probabilities

24. Probability and Normal Distributions
If a random variable x is normally distributed, you can find the probability that x will fall in a given interval by calculating the area under the normal curve for that interval.
μ = 500
σ = 100
600
μ =500 x
Remember that the total area under the curve is 1.0 (equal to 100%).

25. Probability and Normal Distributions
If a random variable x is normally distributed, you can find the probability that x will fall in a given interval by calculating the area under the normal curve for that interval.
μ = 500
σ = 100
600
μ =500 x
P(x < 600) = Area
Remember that the total area under the curve is 1.0 (equal to 100%).

26. Probability and Normal Distributions
600
μ =500
P(x < 600)
μ = σ = 100 x

27. Probability and Normal Distributions
Standard Normal Distribution
600
μ =500
P(x < 600)
μ = σ = 100 x 1
μ = 0
μ = 0 σ = 1 z
P(z < 1)
Same Area
P(x < 500) = P(z < 1)

28. Example
A survey indicates that people use their computers an average of 2.4 years before upgrading to a new machine. The standard deviation is 0.5 year. A computer owner is selected at random. Find the probability that he or she will use it for fewer than 2 years before upgrading. Assume that the variable x is normally distributed.

29. Solution P(x < 2) = P(z < -0.80) = 0.2119 Normal Distribution
Standard Normal Distribution
-0.80
μ = 0 σ = 1 z
P(z < -0.80) 2
2.4
P(x < 2)
μ = σ = 0.5 x
0.2119
P(x < 2) = P(z < -0.80) =

30. Example:
A survey indicates that for each trip to the supermarket, a shopper spends an average of 45 minutes with a standard deviation of 12 minutes in the store. The length of time spent in the store is normally distributed and is represented by the variable x. A shopper enters the store. Find the probability that the shopper will be in the store for between 24 and 54 minutes.

31. Solution: Finding Probabilities for Normal Distributions
-1.75 z
Standard Normal Distribution μ = 0 σ = 1
P(-1.75 < z < 0.75)
0.75 24 45
P(24 < x < 54) x
0.7734
0.0401 54
P(24 < x < 54) = P(-1.75 < z < 0.75)
= – =

32. Example:
Find the probability that the shopper will be in the store more than 39 minutes. (Recall μ = 45 minutes and σ = 12 minutes)

33. Solution: Finding Probabilities for Normal Distributions
Standard Normal Distribution μ = 0 σ = 1
P(z > -0.50) z
-0.50 39 45
P(x > 39) x
0.3085
P(x > 39) = P(z > -0.50) = 1– =

34. Section 5.4
Normal Distributions
Finding Values

35. From Areas to z-Scores z –1 1 2 3 4 0.9803 –4 –3 –2
Find the z-score corresponding to a cumulative area of
z = 2.06 corresponds
roughly to the
98th percentile.
0.9803 –4 –3 –2 –1 1 2 3 4 z
Locate in the area portion of the table. Read the values at the beginning of the corresponding row and at the top of the column. The z-score is 2.06.

36. Finding z-Scores from Areas
Find the z-score corresponding to the 90th percentile.
.90 z
The closest table area is The row heading is 1.2 and column heading is .08. This corresponds to z = 1.28.
A z-score of 1.28 corresponds to the 90th percentile.

37. Finding z-Scores from Areas
Find the z-score with an area of .60 falling to its right.
.40
.60 z z
With .60 to the right, cumulative area is .40. The closest area is The row heading is 0.2 and column heading is .05. The z-score is 0.25.
A z-score of 0.25 has an area of .60 to its right. It also corresponds to the 40th percentile

38. Finding z-Scores from Areas
Find the z-score such that 45% of the area under the curve falls between –z and z.
.275
.275
.45 –z z
The area remaining in the tails is .55. Half this area is
in each tail, so since .55/2 = .275 is the cumulative area for the negative z value and = .725 is the cumulative area for the positive z. The closest table area is and the z-score is The positive z score is 0.60.

39. From z-Scores to Raw Scores
To find the data value, x when given a standard score, z:
The test scores for a civil service exam are normally distributed with a mean of 152 and a standard deviation of 7. Find the test score for a person with a standard z-score of:
(a) (b) – (c) 0

40. From z-Scores to Raw Scores
To find the data value, x when given a standard score, z:
The test scores for a civil service exam are normally distributed with a mean of 152 and a standard deviation of 7. Find the test score for a person with a standard z-score of:
(a) (b) – (c) 0
(a) x = (2.33)(7) =
(b) x = (–1.75)(7) =
(c) x = (0)(7) = 152

41. Finding Percentiles or Cut-off Values
Monthly utility bills in a certain city are normally distributed with a mean of $100 and a standard deviation of $12. What is the smallest utility bill that can be in the top 10% of the bills?
$ is the smallest
value for the top 10%.
90%
10% z
Find the cumulative area in the table that is closest to (the 90th percentile.) The area corresponds to a z-score of 1.28.
To find the corresponding x-value, use
x = (12) =

42. Sampling Distributions & The Central Limit Theorem

43. Sampling Distributions
A sampling distribution is the probability distribution of a sample statistic that is formed when samples of size n are repeatedly taken from a population. If the sample statistic is the sample mean, then the distribution is the sampling distribution of sample means.
Sample
Sample
Sample
Sample
Sample
Sample
The sampling distribution consists of the values of the sample means,

44. Sampling Distribution of x-bar
Population with μ, σ
Sample 5
Sample 3
Sample 1
Sample 4
Sample 2
The sampling distribution consists of the values of the sample means,

45. The Central Limit Theorem
If a sample n >= 30 is taken from a population with
any type distribution that has a mean =
and standard deviation = x
the sample means will have a normal distribution
and standard deviation

46. The Central Limit Theorem
If a sample of any size is taken from a population with a normal distribution with mean = and standard deviation = x
the distribution of means of sample size n, will be normal
with a mean
standard deviation

47. Application 69.2 mean Standard deviation
The mean height of American men (ages 20-29) is
inches. Random samples of 60 such men are selected. Find the mean and standard deviation (standard error) of the sampling distribution.
69.2
mean
Distribution of means of sample size 60,
will be normal.
Standard deviation

48. Interpreting the Central Limit Theorem
The mean height of American men (ages 20-29) is = 69.2”. If a random sample of 60 men in this age group is selected, what is the probability the mean height for the sample is greater than 70”? Assume the standard deviation is 2.9”.
Since n > 30 the sampling distribution of will be normal
mean
standard deviation
Find the z-score for a sample mean of 70:

49. Interpreting the Central Limit Theoremz
2.14
There is a probability that a sample of 60 men will have a mean height greater than 70”.

50. Application Central Limit Theorem
During a certain week the mean price of gasoline in California was $1.164 per gallon. What is the probability that the mean price for the sample of 38 gas stations in California is between $1.169 and $1.179? Assume the standard deviation = $0.049.
Since n > 30 the sampling distribution of will be normal
mean
standard deviation
Calculate the standard z-score for sample values of $1.169 and $1.179.

51. Application Central Limit Theorem
P( 0.63 < z < 1.90)
= – = z
.63
1.90
The probability is that the mean for the sample is between $1.169 and $1.179.