chapter six normal curves and sampling probability distributions

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


The curve approaches the horizontal axis but never touches or crosses it.

μ


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%⇒


Percentages of data that lies between given values€

μ


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


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


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?


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



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



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



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



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



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?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?


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?


μ =0.97 + 0.44

2

μ =1.41

2μ = 0.705



b. What is the standard deviation of the grades?


σ =0.97 − 0.44

12

σ =0.53

3.4641σ = 0.1530



c. What is the probability of getting a grade between 65% and 75%?


P 0.65 ≤x≤0.75( )0.75−0.650.97−0.44

0.100.53

0.1887



d. What is the probability of getting a grade 80% or higher?


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?



every 40 days.

a. What is the mean number of days between accidents?


μ =40


every 40 days.

b. What is the standard deviation of the number of days between

accidents?


σ =40


every 40 days.

c. What is the probability of having another

accident after 30 to 60 days?


P 30 ≤x≤60( )

e−3040 −e

−6040

e−0.75 −e−1.5

0.4724 −0.22310.2492


every 40 days.

d. What is the probability of having another

accident after more than 60 days?


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

chapter six normal curves and sampling probability distributions

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