Download Math 227 Ch 7 notes KEY F16

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
(Recall symbols: for a sample mean 𝑥̅ , sample standard deviation 𝑠, population mean  , population standard deviation
)
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%
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(b) What is the area to the left   0 ? 0.50 or 50%
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 ≤ 2.47) =0.0254 or 2.54%
Objective B: Finding the 𝒁-score for a given probability
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Area  0.5
Example 1:
Area  0.5
Area  0.5
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. Choose ‘Between’ in Statcrunch.
z = - 1.04
and 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 90 th percentile?
First: μ = _125_wpm__ σ = __24_wpm_____
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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 speeds 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’ reading 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 may not have come from a population that was normally distributed because not all points lie
within the boundary. Therefore, there is no guarantee that the sample will be normally distributed.
(b)
Yes, the sample data may have 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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