Chapter Six
Normal Curves and Sampling Probability
Distributions
Chapter 6Section 1
Graphs of Normal Probability Distributions
Properties of TheNormal Distribution
The curve is bell-shaped with the highest point over the mean, μ.
μ
Properties of The Normal Distribution
The curve is symmetrical about a vertical line through μ.
μ
Properties of The Normal Distribution
The curve approaches the horizontal axis but never touches or crosses it.
μ
Properties of The Normal Distribution
The transition points between cupping upward and downward occur
above μ + σ and μ – σ .€
μ
€
μ+1σ
€
μ−1σ
Inflection Point ⇒ ⇐ Inflection Point
The Empirical Rule
Approximately 68.2% of the data values lie within one standard deviation of the
mean. €
μ
€
μ+1σ
€
μ−1σ
€
⇐ 68.2%⇒
The Empirical RuleThe Empirical Rule
Approximately 95.4% of the data values lie within two standard
deviations of the mean.€
μ
€
μ+ 2σ
€
μ−2σ
€
⇐ 95.4%⇒
Almost all (approximately 99.7%) of the data values will be within three standard
deviations of the mean. €
μ
€
μ+ 3σ
€
μ−3σ
€
⇐ 99.7%⇒
The Empirical RuleThe Empirical Rule
Percentages of data that lies between given values€
μ
The Empirical RuleThe Empirical Rule
μ−3σ μ + 3σμ−2σ μ + 2σμ−σ μ +σ
0.34100.34100.13600.1360 0.02150.0215 0.00150.0015
Application of the Empirical Rule
Each of the variables in the left hand column of the table has a
normal probability distribution with the given mean (μ) and standard deviation (σ ). Use the empirical rule to complete the table.
Variable68.2% fall between
95.4% falls between
99.7% fall between
Height of adult females
65” 2.5”
Contents of a box of cereal
20 oz. 0.2 oz
Life span of a battery
1000 hours 50 hours
Diameter of an engine part
3” 0.05”
62.5−67.5
19.8−20.2
60−70 57.5−72.5
19.4 −20.6
2.85−3.15
850−1150900−1100950−1050
2.9−3.12.95−3.05
19.6−20.4
μ σ
Application of the Empirical Rule
The life of a particular type of light bulb is normally distributed with a mean of 1100 hours and a standard deviation of 100 hours.
Question: What is the probability that a light bulb of this type will last between 1000 and 1200 hours?
Answer: Approximately 0.6820
P 1000 ≤x≤1200( )
P 1100−100 ≤x≤1100 +100( )
P μ−1σ ≤x≤μ +1σ( )0.682
Application of the Empirical Rule
The life of a particular type of light bulb is normally distributed with a mean of 1100 hours and a standard deviation of 75 hours.
Question: What is the probability that a light bulb of this type will last between 950 and 1325 hours?
Answer: Approximately 0.9755
P 950 ≤x≤1325( )
P 1100 −150 ≤x≤1100 + 225( )
P 1100 −2 75( ) ≤x≤1100 + 3 75( )( )
P μ −2σ ≤x≤μ + 3σ( )0.1360 + .3410 + .3410 + .1360 + .0215
0.9755
Application of the Empirical Rule
Answer: Approximately 0.3410
P 100 ≤x≤115( )
P 100 ≤x≤100 +15( )
P 100 ≤x≤100 +1 15( )( )
P μ ≤x≤μ +1σ( )0.3410
I.Q. is normally distributed with μ =100 and σ=15. Fill in the values that correpsond to the standard deviation marks on the number line and find the probability that a person picked at random out of the general population has an I.Q. in the general interval. a. Between 100 and 115
Application of the Empirical Rule
Answer: Approximately 0.8180
P 85 ≤x≤130( )
P 100−15 ≤x≤100 + 30( )
P 100−1 15( ) ≤x≤100 +2 15( )( )
P μ−1σ ≤x≤μ +2σ( )0.3410 + 0.3410 + 0.1360
0.8180
I.Q. is normally distributed with μ =100 and σ=15. Fill in the values that correpsond to the standard deviation marks on the number line and find the probability that a person picked at random out of the general population has an I.Q. in the general interval. b. Between 85 and 130
Application of the Empirical Rule
Answer: Approximately 0.0215
P 130 ≤x≤145( )
P 100 + 30 ≤x≤100 + 45( )
P 100 + 2 15( ) ≤x≤100 + 3 15( )( )
P μ +2σ ≤x≤μ + 3σ( )0.0215
I.Q. is normally distributed with μ =100 and σ=15. Fill in the values that correpsond to the standard deviation marks on the number line and find the probability that a person picked at random out of the general population has an I.Q. in the general interval. c. Between 130 and 145
Application of the Empirical Rule
Answer: Approximately 0.0230
P x ≥130( )
P x≥100 + 30( )
P x≥100 +2 15( )( )
P x≥μ +2σ( )0.0215 + 0.0015
0.0230
I.Q. is normally distributed with μ =100 and σ=15. Fill in the values that correpsond to the standard deviation marks on the number line and find the probability that a person picked at random out of the general population has an I.Q. in the general interval. d. Over 130
Application of the Empirical Rule
Answer: Approximately 0.0015
P x ≤55( )
P x≥100−45( )
P x≥100−3 15( )( )
P x≥μ−3σ( )0.0015
I.Q. is normally distributed with μ =100 and σ=15. Fill in the values that correpsond to the standard deviation marks on the number line and find the probability that a person picked at random out of the general population has an I.Q. in the general interval. e. Under 55
Control Chart
A statistical tool to track data over a period of equally spaced time intervals or in some sequential
order.
Statistical Control
A random variable is in statistical control if it can be described by the same probability distribution when it is
observed at successive points in time.
To Construct aControl Chart
• Draw a center horizontal line at μ.• Draw dashed lines (control limits) at
μ 2 σ and μ 3σ.• The values of μ and σ may be target values
or may be computed from past data when the process was in control.
• Plot the variable being measured using time on the horizontal axis.
Control Chart
μ+3σ
μ3σ
μ+2σ
μ2σ
μ
Out-Of-ControlWarning Signals
I. One point beyond the 3σ level.
II. A run of nine consecutive points on one side of the center line.
III. At least two of three consecutive points beyond the 2σ level on the same side of the center line.
Is the Process in Control?
μ+3σ
μ3σ
μ+2σ
μ2σ
μ
Is the Process in Control?
μ+3σ
μ3σ
μ+2σ
μ2σ
μ
Is the Process in Control?
μ+3σ
μ3σ
μ+2σ
μ2σ
μ
Is the Process in Control?
μ+3σ
μ3σ
μ+2σ
μ2σ
μ
Is the Process in Control?You are in charge of Quality Control for a manufacturing company that produces
parts for automobiles. A specific gear has been designed to have a diameter of
three inches. We have learned from that the standard deviation of the gear is
0.2 inches. The following ten measurements were taken from a random sample
of gears that came off the production line. Make a control chart on graph paper
for the measures given below. Does this indicate that the measures are in control?
Part 1 2 3 4 5 6 7 8 9 10
Diameter (inches)
2.9 2.6 3.1 3.5 2.8 2.9 3.4 3.2 2.7 3.3
a. Do any points fall beyond the LCL and UCL three standard deviation limits?
b. Is there a run of nine consecutive points on one side of the center line?
c. Is there an instance of two out of three points beyond the two standard
deviation limits on the same side of the center line?
Is the Process in Control?
μ+3σ
μ3σ
μ+2σ
μ2σ
μ
Is the Process in Control?a. Do any points fall beyond the LCL and UCL three standard
deviation limits? No points fall beyond the LCL and the UCL three standard
deviations limit.
a. Is there a run of nine consecutive points on one side of the center line? There is no run of nine consecutive points on one side of the
center line.
a. Is there an instance of two out of three points beyond the two standard deviation limits on the same side of the center line?There is no instance of two out of three points beyond the two standard deviation limits on the same side of the center line.
Uniform Probability Distributions
1. The equation is: y=1
β −α.
2. The mean is: μ=β +α2
.
3. The standard deviation is: σ=β −α12
.
4. P a≤x≤b( ) =b−aβ −α
.
βα a b
y=1
β −α
4. A professor noticed that the grades for his final examination fit a
Uniform Probability Distribution where the highest grade was a
97% and the lowest grade was a 44%.
a. What is the mean grade?
b. What is the standard deviation of the grades?
c. What is the probability of getting a grade between 65% and 75%?
d. What is the probability of getting a grade 80% or higher?
Uniform Probability Distributions
4. A professor noticed that the grades for his final examination fit a Uniform Probability
Distribution where the highest grade was a 97% and the lowest grade was a 44%.
a. What is the mean grade?
Uniform Probability Distributions
μ =0.97 + 0.44
2
μ =1.41
2μ = 0.705
4. A professor noticed that the grades for his final examination fit a Uniform Probability
Distribution where the highest grade was a 97% and the lowest grade was a 44%.
b. What is the standard deviation of the grades?
Uniform Probability Distributions
σ =0.97 − 0.44
12
σ =0.53
3.4641σ = 0.1530
4. A professor noticed that the grades for his final examination fit a Uniform Probability
Distribution where the highest grade was a 97% and the lowest grade was a 44%.
c. What is the probability of getting a grade between 65% and 75%?
Uniform Probability Distributions
P 0.65 ≤x≤0.75( )0.75−0.650.97−0.44
0.100.53
0.1887
4. A professor noticed that the grades for his final examination fit a Uniform Probability
Distribution where the highest grade was a 97% and the lowest grade was a 44%.
d. What is the probability of getting a grade 80% or higher?
Uniform Probability Distributions
P 0.80 ≤x≤0.97( )0.97−0.800.97−0.44
0.170.53
0.3208
Exponential Probability Distributions
1. The equation is: y=1β
e−
xβ .
2. The mean is: μ=β.3. The standard deviation is: σ=β.
4. P a≤x≤b( ) =e−
aβ −e
−bβ .
1
β
a b
y=1β
e−
xβ
The intersection in downtown Annville is experiencing an accident about
every 40 days.
a. What is the mean number of days between accidents?
b. What is the standard deviation of the number of days between
accidents?
c. What is the probability of having another accident after 30 to 60 days?
d. What is the probability of having another accident after more than
60 days?
Uniform Probability Distributions
The intersection in downtown Annville is experiencing an accident about
every 40 days.
a. What is the mean number of days between accidents?
Uniform Probability Distributions
μ =40
The intersection in downtown Annville is experiencing an accident about
every 40 days.
b. What is the standard deviation of the number of days between
accidents?
Uniform Probability Distributions
σ =40
The intersection in downtown Annville is experiencing an accident about
every 40 days.
c. What is the probability of having another
accident after 30 to 60 days?
Uniform Probability Distributions
P 30 ≤x≤60( )
e−3040 −e
−6040
e−0.75 −e−1.5
0.4724 −0.22310.2492
The intersection in downtown Annville is experiencing an accident about
every 40 days.
d. What is the probability of having another
accident after more than 60 days?
Uniform Probability Distributions
P x ≥60( )
e−6040 −e
−∞40
e−1.5 −e−∞
0.2231−00.2231
THE ENDOF
SECTION 1Homework Assignments
Pages:259 - 266 Exercises: #1 - 19, odd Exercises: #2 - 20, even
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