Chapter 7The Normal Probability Distribution

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 obtainingsuccess in independent trials of a binomial experiment is
calculated by plugging the value of into the binomial formula as shown below:

Continuous Random variable

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

i.e.

Continuous Probability Distribution

We can only compute probability over an interval of values. Since and fora
continuous variable,

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

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.

for a uniform distribution

Example 1: A continuous random variable is uniformly distributed with.

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

Area of rectangle = Height x Width

1 = Height x(b - a)

Height =

= =

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

= * (30 - 20)

= * 10

= = 0.25

P (x< 15) = P (x≤ 15)= * (15 – 10)

= P (10≤ x ≤ 15)= * 5

= = 0.125

Objective B: Normal distribution – Bell-shaped Curve

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 years and
standard deviation 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.

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.

Use the 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 -score.

Probability is the area under the curve over an interval.

The total area under the normal curve is 1.

Under the standard normal distribution,

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

(b) What is the area to the left?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

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 -score for a given probability

Example 1: Draw the standard normal curve and the -scoresuch that the area to the left of the
-score is 0.0418.Use StatCrunch to find the-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

Example 2: Draw the standard normal curve and the -score such that the area to the right of the
-score is 0.18.Use StatCrunch to find the-score.

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

P(Z0.91536509) = 0.18

Example 3: Draw the standard normal curve and two -scores such that the middle area of the standard
normal curve is 0.70. Use StatCrunch to find the two -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

Compute and record the results

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 to -scores using.

Step 3: Use StatCrunch to find the desired area.

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

and a standard deviation .

(Note: this is not the standard normal curve becauseand.)

(a)

(b)

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

Example 2: Redo Example 1
 Use StatCrunch and random variable directly without converting to first.

(a)

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)

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

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 -score.

Step 3: Obtain from by the formula or .

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

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

Compute and record the results

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

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

Compute and record the results

≈77.96

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

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

Compute and record the results

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

Example 2: Redo Example 1
 Use StatCrunch to find directly without converting from to.

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.

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

Mean = 125, Std. Dev. = 24, and P ( ≤x ≤ ) = 0.95

Compute and record the results

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 less than 30.

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.

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.