Chapter 7The Normal Probability Distribution Chapter … is the area under the curve over an interval. The total area under the normal curve is 1. 3.5 2 1 1 2 3.5 Z Negative Z Positive

1

Chapter 7The Normal Probability Distribution

Chapter 7.1 Uniform and Normal Distribution

Objective A: Uniform Distribution

A1. Introduction

Recall: Discrete random variable probability distribution

Special case: Binomial distribution

Finding the probability of obtaining a success in n independent trials of a binomial experiment is

calculated by plugging the value of a into the binomial formula as shown below:

( ) (1 )a n a

n aP x a C p p

Continuous Random variable For a continued random variable the probability of observing one particular value is zero.

i.e. ( ) 0P x a

Continuous Probability Distribution

We can only compute probability over an interval of values. Since ( ) 0P x a and ( ) 0P x b fora

continuous variable,

( ) ( )P a x b P a x b

To find probabilities for continuous random variables, we use probability density functions.

2

ab

1

a b

Two common types of continuous random variable probability distribution:

1) Uniform distribution.

2) Normal distribution.

A2. Uniform distribution

Note: The area under a probability density function is 1.

Area of rectangle Height Width

1 Height ( )b a

1Height

( )b a

for a uniform distribution

Example 1: A continuous random variable x is uniformly distributed with10 50x .

(a) Draw a graph of the uniform density function.

Area of rectangle = Height x Width

1 = Height x(b - a)

Height = 1

(𝑏 − 𝑎)

= 1

(50−10) =

1

40

(b) What is (20 30)P x ? Area of rectangle = Height x Width

= 1

40 * (30 - 20)

= 1

40 * 10

= 1

4 = 0.25

10 50

40

1

2030

40

1

3

(c) What is ( 15)P x ? Area of rectangle = Height x Width

P (x< 15) = P (x≤ 15) = 1

40 * (15 – 10)

= P (10≤ x ≤ 15) = 1

40 * 5

= 1

40 = 0.125

Objective B: Normal distribution – Bell-shaped Curve

40

1

1015

4

430330 530 630 730X

1 1

2 2

Example 1: Graph of a normal curve is given. Use the graph to identify the value of and .

Example 2: The lives of refrigerator are normally distributed with mean 14 years and

standard deviation 2.5 years.

(a) Draw a normal curve and the parameters labeled.

(b) Shade the region that represents the proportion of refrigerator that lasts

for more than 17 years.

530

100

5.115.6 14 5.16 199 5.21X

5

(c) Suppose the area under the normal curve to the right 17x is 0.1151 .

Provide twointerpretations of this result.

Notation: P (x≥17) = 0.1151

The area under the normal curve for any interval of values of the random variable x

represents either:

− The proportions of the population with the characteristic described by the interval of

values.

11.51% of all refrigerators are kept for at least 17 years.

−the probability that a randomly selected individual from the population will have the

characteristic described by the interval of values.

The probability that a randomly selected refrigerator will be kept for at least 17 years is

11.51%.

Chapter 7.2 Applications of the Normal Distribution

Objective A: Area under the Standard Normal Distribution

The standard normal distribution

– Bell shaped curve

– =0 and =1

The random variable for the standard normal distribution is Z .

Use the 𝑍 table (Table V) to find the area under the standard normal distribution. Each value in

the body of the table is a cumulative area from the left up to a specific Z -score.

Probability is the area under the curve over an interval.

The total area under the normal curve is 1.

1 1 22 Z5.3 5.3

ZNegative ZPositive

1

0 0

0

Z

Z

6

Under the standard normal distribution,

(a) What is the area to the right 0 ? 0.5

(b) What is the area to the left 0 ?0.5

Example 1: Draw the standard normal curve with the appropriate shaded area and use StatCrunch

to determine the shaded area.

Open StatCrunch → select Stat → Calculators → Normal →select Standard→select ≤ → Input

desired value for X → compute → record results

(a) Find the shaded area that lies to the left of -1.38.

(b) Find the shaded area that lies to the right of 0.56.

Similar steps as in part (a) except you want to select ≥ and input value, compute and record

results

P (Z> 0.56) = 0.28773972

Z

0838.0)38.1( ZP

56.00

Z

0.0838

-1.38

7

5.0Area 5.0Area 5.0Area

(c) Find the shaded area that lies in between 1.85 and 2.47.

Open StatCrunch → select Stat → Calculators → Normal →select Between → Input desired

values for X range → compute → record results

P (1.85 ≤ Z ≤ 2.47) = 0.02540112

Objective B: Finding the 𝒁-score for a given probability

Example 1: Draw the standard normal curve and the z -score such that the area to the left of the

z -score is 0.0418. Use StatCrunch to find the z -score.

Open StatCrunch→Select Stat→ Calculator→ Normal → Standard → Input the value for

P (x ≤ ) = 0.0418

Compute and record the results

P (Z <-1.73) = 0.0418

0 85.1 47.2Z

0

0418.0

?Z

85.1 47.2

8

Example 2: Draw the standard normal curve and the 𝑍-score such that the area to the right of the

𝑍-score is 0.18.Use StatCrunch to find the 𝑍-score.

Similar to example 1, input P (x ≥________) = 0.18, compute

P (Z > 0.91536509) = 0.18

Example 3: Draw the standard normal curve and two 𝑍-scores such that the middle area of the standard

normal curve is 0.70. Use StatCrunch to find the two 𝑍-scores.

If the middle area is 0.70, the total tailed areas is 0.30 (1-0.70) and the left tailed area is 0.15

(0.30/2). We will use StatCrunch to find the z –score for the lower bound then use the

symmetric concept to find the z –score for the upper bound.

Open StatCrunch → Select Stat → Calculator → Normal → Standard → Input the value for

P (x ≤ _____) = 0.15


P(-1.04< Z <1.04 ) = 0.70

Objective C: Probability under a Normal Distribution

Step 1: Draw a normal curve and shade the desired area.

Step 2: Convert the values X to Z -scores usingX

Z

.

Step 3: Use StatCrunch to find the desired area.

18.0

?Z

%70

)symmetrytoDue(04.1Z04.1Z

0.15 0.15

9

Example 1: Assume that the random variable X is normally distributed with mean 50

and a standard deviation 7 .

(Note: this is not the standard normal curve because 0 and 1 .)

(a) ( 58)P X

(b) (45 63)P X

P (-0.71 ≤ Z ≤ 1.86) = 0.72970517 ≈ 0.7297

50 58 X

XZ

7

5058

14.17

8

8735.0)14.1( ZP

Z0 14.1

71.0Z

XZ

7

5045

45X

7

5

63X

XZ

7

5063

86.1Z7

13

≈0.8735

10

Example 2: Redo Example 1

Use StatCrunch and random variable X directly without converting to Z first.

(a) ( 58)P X

Open StatCrunch → Select Stat → Calculator → Normal → Standard → Input the values for

Mean, Std. Dev. and P (x ≤ 58) = _____ →Compute

P (x ≤ 58) = 0.87.45105

(b) (45 63)P X

Open StatCrunch → Select Stat → Calculator → Normal → Between → Input the values for

Mean, Std. Dev. and P (45 ≤ x ≤ 63) = _____ →Compute

P (45 ≤ x ≤ 63) = 0.73082932

11

Example 3: GE manufactures a decorative Crystal Clear 60-watt light bulb that it advertises will last

1,500 hours. Suppose that the lifetimes of the light bulbs are approximately normal distributed,

with a mean of 1,550 hours and a standard deviation of 57 hours, use StatCrunch to find what

proportion of the light bulbs will last more than 1650 hours?

Open StatCrunch → Select Stat → Calculator →

Normal → Standard → Input the values for Mean,

Std. Dev. and P (x ≥ 1650) = _____ →Compute

Objective D: Finding the Value of a Normal Random Variable

Step 1: Draw a normal curve and shade the desired area.

Step 2: Use StatCrunch to find the appropriate cutoff Z -score.

Step 3: Obtain X from Z by the formula X

Z

or ZX .

Example 1: The reading speed of 6th grade students is approximately normal (bell-shaped) with a mean

speed of 125 words per minute and a standard deviation of 24 words per minute.

(a) What is the reading speed of a 6th graderwhose reading speed is at the 90 percentile?


Mean = 0, Std. Dev. = 1, and P (x ≤ ) = 0.90


0401.0

0396822.0)1650( XP

12

ZX

)24(2815516.1125X

X ≈ 155.76

(b) Determine the reading rates of the middle 95 percentile.

95% in the middle means each tail is 5% divided by 2 = 2.5% = 0.025


Mean = 0, Std. Dev. = 1, and P (x ≤ ) = 0.025


ZX

)24()959964.1(125 X

X ≈ 77.96


Mean = 0, Std. Dev. = 1, and P (x ) = 0.025


ZX

)24()959964.1(125X

X ≈ 172.04

The middle 95% reading speed are between 77.96 words per minute to 172.04 words per

minute.

13

Example 2: Redo Example 1

Use StatCrunch to find X directly without converting from Z to X .

Open StatCrunch → Select Stat → Calculator → Normal → Standard → Input the values for

Mean = 125, Std. Dev. = 24, and P (x ___) = 0.90 → Compute

(a) What is the reading speed of a 6th grader whose reading speed is at the 90 percentile?

X ≈ 155.76

(b) Determine the reading rates of the middle 95% percentile.

If the middle area is 0.95, the total tailed areas is 0.05 and the left tailed area is 0.025 (0.05/2).

We will use StatCrunch to find the X –score for the lower bound then change the inequality sign

to find the X –score for the upper bound.


Mean = 125, Std. Dev. = 24, and P (x ≤ _____) = 0.025 --> Compute

X = 77.96 words per minute

14


Mean = 125, Std. Dev. = 24, and P (x _____) = 0.025 --> Compute

X = 172.04 words per minute

The middle 95% reading speed are between 77.96 words per minute to 172.04 words per

minute.

Chapter 7.3 Normality Plot

Recall: A set of raw data is given, how would we know the data has a normal distribution?

Use histogram or stem leaf plot.

Histogram is designed for a large set of data.

For a very small set of data it is not feasible to use histogram to determine whether the data

hasa bell-shaped curve or not.

We will use the normal probability plot to determine whether the data were obtained from

a normal distribution or not. If the data were obtained from a normal distribution, the data

distribution shape is guaranteed to be approximately bell-shaped for n is less than 30.

15

0

0

Z

Z

Perfect normal curve. The curve is aligned with the dots.

Almost a normal curve. The dots are within the

boundaries.

Not a normal curve. Data is outside the boundaries.

16

Example 1: Determine whether the normal probability plot indicates that the sample data

could have come from a population that is normally distributed.

(a)

Not a normal curve.

The sample data do not come from a normally distributed population.

There is no guarantee that this sample data set is normally distributed.

(b)

A normal curve.

The sample data come from a normally distribute population.

There is a guarantee that this sample data set is approximately normally distributed.

Documents

Chapter 7The Normal Probability Distribution Chapter … is the area under the curve over an interval. The total area under the normal curve is 1. 3.5 2 1 1 2 3.5 Z Negative Z Positive