Download Normal Distribution Lecture Notes - NIU Math Department

Normal Distribution Lecture Notes
Professor Richard Blecksmith
[email protected]
Dept. of Mathematical Sciences
Northern Illinois University
Math 101 Website: http://math.niu.edu/∼richard/Math101
Section 2 Website: http://math.niu.edu/∼richard/Math101/fall06
1. Normal Distribution Curve
34%
34%
2.5%
13.5%
µ−σ
µ − 2σ
µ
2.5%
13.5%
µ+σ
µ + 2σ
In a normal distribution
• Fact 1.
• Fact 2.
– 50%
– 50%
Center = mean = median
The data lies equally distributed on each side of the center.
of the data lies to the left of µ and
of the data lies to the right of µ.
2. The 68 – 95 – 99 Rule
• Fact 3.
– 68% of the data lies within 1 standard deviation of the mean
– 95% of the data lies within 2 standard deviations of the mean
– 99% of the data lies within 3 standard deviations of the mean
1
2
3. Standardizing Data
Given normally distributed data, with mean µ and starndard deviation σ.
If x is a data point, we wish to know:
• how many standard deviations is x to the right (or left) of the center?
• That is, x = µ + z · σ. Solve for z.
µ+z·σ =x
z·σ =x−µ
z = (x − µ)/σ
4. The z–Rule
Original
Data Value
x
Standardized
Data Value
z = (x − µ)/σ
• A negative value of z represents a data point to the left of the center
• A positive value of z represents a data point to the right of center
5. Example from Text (page 51)
The lifetime of 20,000 flashlight batteries are normally distributed, with a
mean of µ = 370 days and a standard deviation of σ = 30 days.
1. What percentage of the batteries are expected to last more than 340
days?
Solution: z = (x − µ)/σ
= (340 − 370)/30 = −1.00
• Look up z = 1 in the chart.
• (The negative means that this value occurs one standard deviation to
the left of the center µ.)
• The corresponding P value is 34.1%.
3
6. Draw the picture
34.1
µ − 1.00σ
50
µ
The answer is 34.1 + 50 = 84.1%.
7. Question 2
2. How many batteries can be expected to last less than 325 days?
Solution: Work with percentages.
• z = (x − µ)/σ = (325 − 370)/30 = −1.50
• Look up z = 1.5 in the chart.
• The corresponding P value is 43.3%.
8. Draw the picture
43.3
µ − 1.50σ
µ
• Fifty percent of the data lies to the left of the center.
• Since 43.3% lies between µ − 1.50σ and the center µ,
• the percentage to the left of µ − 1.50σ is 50.0 − 43.3 = 6.7%
The final answer is: 6.7 percent of 20,000 = .067 × 20, 000 = 1340
9. SAT Example
• In 2001 a total of 1,276,320 college-bound students took the SAT exam.
4
• The mean and standard deviation of the test scores was µ = 506 and
σ = 111.
• 68% of the students fall within 1 standard deviation of the mean,
• that is in the range µ−σ = 506−111 = 395 to µ+σ = 506+111 = 617.
• 95% of the students fall within 2 standard deviations of the mean, that
is in the range µ − 2σ = 506 − 222 = 284 to µ + 2σ = 506 + 222 = 728.
• Where is the cutoff between the first and second Quartile?
10. SAT Example Cont’d
• We want P = 25%.
• The (3-digit) chart shows the z-value corresponding to P = .25 is z =
.675.
• This means that 25% of the data occurs before you get within .675
standard deviations of µ (on the left).
• Another 25% lies between µ − .675σ and µ itself.
• So the first quartile occurs at
• Q1 = µ − .675σ = 506 − (.675)111 = 431
• It turns out Q1 was exactly 430.
• The third quartile occurs at
• Q1 = µ + .675σ = 506 + (.675)111 = 581
11. Draw the Picture
2001 SAT Scores
25%
µ − 0.675σ
Q1 = 431
25%
25%
µ
506
25%
µ + 0.675σ
Q3 = 581