Download Chapter 7The Normal Probability Distribution Chapter 7.1 Uniform

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:
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 fora
continuous variable,
P(a  x  b)  P(a  x  b)
To find probabilities for continuous random variables, we use probability density functions.
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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
Note: The area under a probability density function is 1.
Area of rectangle  Height  Width
1  Height  (b  a)
Height 
1
for a uniform distribution
(b  a)
Example 1: A continuous random variable x is uniformly distributed with 10  x  50 .
(a) Draw a graph of the uniform density function.
1
40
10 50
Area of rectangle = Height x Width
1 = Height x(b - a)
1
Height = (𝑏 − 𝑎)
1
1
= (50−10) = 40
(b) What is P(20  x  30) ?
Area of rectangle = Height x Width
1
= 40 * (30 - 20)
1
40
1
= 40 * 10
1
= 4 = 0.25
20
30
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(c) What is P( x  15) ?
Area of rectangle = Height x Width
1
P (x< 15) = P (x≤ 15)
= 40 * (15 – 10)
= P (10≤ x ≤ 15)
=
1
40
= 40 = 0.125
1
40
*5
1
10
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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  and  .

  530
  2
  2
  1   1
  100
X
330 430 530 630 730
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.

6.5
9 11.5 14 16.5 19 21.5
X
(b) Shade the region that represents the proportion of refrigerator that lasts
for more than 17 years.
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(c) Suppose the area under the normal curve to the right x  17 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
 1
 0
 3.5
The random variable for the standard normal distribution is Z .
2
1
Negative Z
0
1
2
3.5 Z
Positive 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.

0
Z
Z
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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.
0.0838
Z
-1.38
P(Z  1.38)  0.0838
(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
Z
0 0.56
P (Z> 0.56) = 0.28773972
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(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
1.85
2.47
P (1.85 ≤ Z ≤ 2.47) = 0.02540112
Objective B: Finding the 𝒁-score for a given probability
Z
1.85 2.47
0
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.
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
0.0418
Z ?
0
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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
0.18
Z ?
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
Compute and record the results
P(-1.04< Z <1.04 ) = 0.70
0.15
70%
Z  1.04
0.15
Z  1.04 (Due to symmetry)
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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Assume that the random variable X is normally distributed with mean   50
Example 1:
and a standard deviation   7 .
(Note: this is not the standard normal curve because   0 and   1 .)
(a) P( X  58)

Z
X 

50 58
58  50

7

X

8
 1.14
7
≈0.8735
P(Z  1.14)  0.8735
0 1.14
Z
(b) P(45  X  63)
X  45
Z
X 

45  50
7
5

7

Z  0.71
X  63
Z
X 

63  50
7
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
7
Z  1.86

P (-0.71 ≤ Z ≤ 1.86) = 0.72970517 ≈ 0.7297
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Example 2:
Redo Example 1
 Use StatCrunch and random variable X directly without converting to Z first.
(a) P( X  58)
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) P(45  X  63)
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
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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
P( X  1650 )  0.0396822
0.0401
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 
or X    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 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
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X    Z 
X  125  1.2815516 (24)
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
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
X    Z 
X  125  (1.959964) (24)
X ≈ 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
X    Z 
X  125  (1.959964) (24)
X ≈ 172.04
The middle 95% reading speed are between 77.96 words per minute to 172.04 words per
minute.
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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.
Open StatCrunch → Select Stat → Calculator → Normal → Standard → Input the value for
Mean = 125, Std. Dev. = 24, and P (x ≤ _____) = 0.025 --> Compute
X = 77.96 words per minute
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Open StatCrunch → Select Stat → Calculator → Normal → Standard → Input the value for
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.
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
Z
Perfect normal curve. The curve is aligned with the dots.
0
 0
Almost a normal curve. The dots are within the
boundaries.
Z
Not a normal curve. Data is outside the boundaries.
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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.
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