Download Chapter 7.1 Uniform and Normal Distribution

Chapter 7 The Normal Probability Distribution
First an example from Chapter 6:
A coin is tossed 10 times. What is the probability of getting 3 tails, 5 tails, and 9 tails? In other
words: P(3T), P(5T), P(9T)? Using Statcrunch the probability distribution is:
P(3T)= 0.117
P(5T)= 0.246
P(9T)= 0.010
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:
P( x  a)  n Ca p a (1  p ) n  a
Continuous Random variable
For a continued random variable the probability of observing one particular value is zero.
i.e. P( x  a )  0
Continuous Probability Distribution
We can only compute probability over an interval of values. Since P( x  a )  0 and P ( x  b)  0 for a
continuous variable,
P ( a  x  b)  P ( a  x  b)
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To find probabilities for continuous random variables, we use probability density functions.
Area under curve: 100% or 1
Area shaded: 50% or 0.5
Two common types of continuous random variable probability distribution:
1) Uniform distribution.
2) Normal distribution.
A2. Uniform distribution
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ba
a
b
Area of rectangle  Height  Width
1  Height  (b  a)
1
(𝑏−𝑎)
= Height (for a uniform distribution)
Example 1: A continuous random variable x is uniformly distributed with 10  x  50 .
(a) Draw a graph of the uniform density function.
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1
40
a
b
(b) What is P(20  x  30) ?
Diagram:
= Area under curve = length x width
= (30-20)(1/40) = 10(1/40) = 10/40 = ¼ = 0.25 or 25%
(c) What is P( x  15) ?
Diagram:
= Area under curve = length x width
= (15 – 10)(1/40) = 5(1/40) = 5/40 = 1/8 = 0.125 or 12.5%
Objective B: Normal distribution – Bell-shaped Curve
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Example 1: Graph of a normal curve is given.
Use the graph to identify the value of  (mean of a population) and  (standard deviation of a
population .

  2
  2
  1   1
 = 530,  =630-530 = 100
X
330 430 530 630 730
(symbol for a sample mean 𝑥̅ , sample standard deviation 𝑠)
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.
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(b) Shade the region that represents the proportion of refrigerator that lasts
for more than 17 years.
(c) Suppose the area under the normal curve to the right x  17 is 0.1151 .
Provide two interpretations of this result.


11.51% of all refrigerators last more than 17 years.
The probability that a randomly selected refrigerator will last more than 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 .
You can 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.

0
Z
Z
Under the standard normal distribution,
(a) What is the area to the right   0 ? 0.50 or 50%
(b) What is the area to the left   0 ? 0.50 or 50%
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Example 1:
Draw the standard normal curve with the appropriate shaded area and then use StatCrunch to
determine the shaded area.
(a) that lies to the left of -1.38.
STAT-CALCULATORS-NORMAL mean: 0 SD: 1 ; P( z ≤ - 1.38) – COMPUTE = 0.08379332
So area = P( z ≤ - 1.38) ≈0.0838 or 8.38%
(b) that lies to the right of 0.56. P( z ≥ 0.56) = 0.2877 or 28.77%
(c) that lies in between 1.85 and 2.47. P( 1.85 ≤ z ≤ 0.56) =0.0254 or 2.54%
Objective B: Finding the 𝒁-score for a given probability
Area  0.5
Area  0.5
Area  0.5
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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. (Find z score first.)
This time enter answer in Statcrunch to find the z score: P( z ≤ __ ) = 0.0418 – COMPUTE
z = - 1.7301695 ≈ - 1.73
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.
Find z score first: P( z ≥ ___) = 0.18 COMPUTE
z = 0.915 ≈ 0.92
Example 2:
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. ‘Between’ does not work here in
Statcrunch.
area to the left: P ( z ≤ ___) = 0.15 compute: z = - 1.04
area to the right: P ( z ≥ ___) = 0.15 compute: z = + 1.04
Objective C: Probability under a Normal Distribution
Step 1: Draw a normal curve and shade the desired area.
X 
Step 2: Convert the values X to Z -scores using Z 
.

Step 3: Use StatCrunch to find the desired area.
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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) P( X  58)
Method 1: Using Statcrunch enter mean 𝜇 = 50 and standard deviation = 7 ; P( X  58) = ____; compute
≈ 0.873 or 87.3 %
Method 2: Change to a z score using formula Z 
Using Statcrunch enter mean 𝜇 = 0; 𝜎 = 1; P( z ≤
X 

8
7
=
58−50
7
8
= 7 ≈ 1.14
) = _____ ; compute ≈ 0.873 or 87.3%
(b) P(45  X  63)
Method 1 Using the original values and ‘between’ option on Statcrunch.
P(45  X  63) ≈ 0.731 or 73.1 %
(use mean 50 and SD 7)
Method 2 Change to z scores
For 45: 𝑧45 =
45−50
7
≈ −0.71
For 63: 𝑧63 =
63−50
7
≈ 1.86
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Using Statcrunch enter mean 𝜇 = 0; 𝜎 = 1; between option
P(45  X  63) ≈ 0.731 or 73.1 %
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.
a. µ = _________
σ = ________
b. Interpretation on mean and standard deviation:
A randomly selected light bulb will typically last 1550 ± 57 hours on average, or between 1493 and
1607 hours.
c. Use StatCrunch to find what proportion of the light bulbs will last more than 1650 hours.
P( x >1650) = _____
≈ 0.0396822 ≈ 0.0397
d. Write a sentence interpreting what was found in context.
On average, 3.97 % of the light bulbs will last more than 1650 hours.
e. Is it unusual for a light bulb to last more than 1650 hours?
1650−1550

Changing to a z score: z =

Since 3.97% < 2.5%, then it is not unusual.
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≈ 1.8 < 2 SD Not unusual
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.
X 
Step 3: Obtain X from Z by the formula Z 
.

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 grader whose reading speed is at the 90th percentile?
μ = _______ σ = ________
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Method 1: using Statcrunch directly
P(x < ___) = 0.90
x ≈ 155.75724 ≈ 156 The reading speed of a 6th grader whose reading at the 90th percentile is 156 words per
minute.
Method 2 Using μ = 0, σ = 1
P(z < ____ ) = 0.90 using statcrunch
Z ≈ 1.28
Plug into the z score formula to solve for x : z =
𝑥−125
1.28 =
24
𝑥−125
24
24(1.28) = X- 125
30.72 = X – 125
X = 30.72 + 125 = 155.72 ≈ 156
(b) Determine the reading rates of the middle 95% percentile.
Method 1:
P( 𝑥1 ≤ ___) = 0.025 𝑥1 = 77.96 ≈ 78
Then you can do ‘between’: P(78 < x < ___) = 0.95
𝑥2 ≈ 172
th
95% of 6 grade students’ readings speeds are between 78 and 172 words per minute.
Method 2: Using μ = 0, σ = 1
P(𝑧1 <____) = 0.025 𝑧1 = -1.96
- 1.96 =
𝑥−125
24
solving for x, x = 78
Due to symmetry 𝑧2 = 1.96
-1.96 =
𝑥−125
24
solving for x, x = 172
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
has a bell-shaped curve or not.
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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.
Z score
Perfect normal curve. The curve is aligned with the dots.
x
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)
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The sample data did not come from a population that was normally distributed because not a points lie within
the boundary. Therefore, there is no guarantee that the sample is normally distributed.
(b)
Yes, the sample data did come from a population that was normally distributed since all the points are within
the boundary. Therefore, the sample is approximately normally distributed.
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