1. Random Variables and Probability Distributions
Chapter 16
Random Variables and Probability Distributions

2. Random Variable -
A grocery store manager might be interested in the number of broken eggs in each carton (dozen of eggs). OR
An environmental scientist might be interested in the amount of ozone in an air sample.
A numerical variable whose value depends on the outcome of a chance experiment
Associates a numerical value with each outcome of a chance experiment
Two types of random variables
Discrete
Continuous
Since these values change and are subject to some uncertainty, these are examples of random variables.

3. Two Types of Random Variables:
Discrete – its set of possible values is a collection of isolated points along a number line
Continuous - its set of possible values includes an entire interval on a number line
In this chapter, we will look at different distributions of discrete and continuous random variables.
This is typically a “count” of something
This is typically a “measure” of something

4. Identify the following variables as discrete or continuous
The number of broken eggs in each carton
The amount of ozone in samples of air
The weight of a pineapple
The amount of time a customer spends in a store
The number of gas pumps in use
Discrete
Continuous
Continuous
Continuous
Discrete

5. Probability Distributions for Discrete Random Variables
Probability distribution is a model that describes the long-run behavior of a variable.

6. What do you notice about the sum of these probabilities?
In a Wolf City (a fictional place), regulations prohibit no more than five dogs or cats per household.
Let x = the number of dogs and cats in a randomly selected household in Wolf City x
P(x)
This is called a discrete probability distribution. It can also be displayed in a histogram with the probability on the vertical axis.
What do you notice about the sum of these probabilities?
Is this variable discrete or continuous?
What are the possible values for x?
The Department of Animal Control has collected data over the course of several years. They have estimated the long-run probabilities for the values of x.
Number of Pets
Probability

7. Discrete Probability Distribution
Gives the probabilities associated with each possible x value
Each probability is the long-run relative frequency of occurrence of the corresponding x-value when the chance experiment is performed a very large number of times
Usually displayed in a table, but can be displayed with a histogram or formula

8. Properties of Discrete Probability Distributions
1) For every possible x value,
0 < P(x) < 1.
2) For all values of x,
S P(x) = 1.

9. Just add the probabilities for 0, 1, and 2
Dogs and Cats Revisited . . .
Let x = the number of dogs or cats per household in Wolf City x
P(x)
What is the probability that a randomly selected household in Wolf City has at most 2 pets?
Just add the probabilities for 0, 1, and 2
What does this mean?
P(x < 2) =
= .78

10. Notice that this probability does NOT include 2!
Dogs and Cats Revisited . . .
Let x = the number of dogs or cats per household in Wolf City x
P(x)
What is the probability that a randomly selected household in Wolf City has less than 2 pets?
Notice that this probability does NOT include 2!
What does this mean?
P(x < 2) =
= .57

11. Dogs and Cats Revisited . . .
Let x = the number of dogs or cats per household in Wolf City x
P(x)
What is the probability that a randomly selected household in Wolf City has more than 1 but no more than 4 pets?
When calculating probabilities for discrete random variables, you MUST pay close attention to whether certain values are included (< or >) or not included (< or >) in the calculation.
What does this mean?
P(1 < x < 4) =
= .40

12. Probability Distributions for Continuous Random Variables

13. Consider the random variable:
x = the weight (in pounds) of a full-term newborn child
Suppose that weight is reported to the nearest pound. The following probability histogram displays the distribution of weights.
Now suppose that weight is reported to the nearest 0.1 pound. This would be the probability histogram.
What type of variable is this?
If weight is measured with greater and greater accuracy, the histogram approaches a smooth curve.
What is the sum of the areas of all the rectangles?
The area of the rectangle centered over 7 pounds represents the probability 6.5 < x < 7.5
Notice that the rectangles are narrower and the histogram begins to have a smoother appearance.
The shaded area represents the probability 6 < x < 8.
This is an example of a density curve.

14. Probability Distributions for Continuous Variables
Is specified by a curve called a density curve.
The function that describes this curve is denoted by f(x) and is called the density function.
The probability of observing a value in a particular interval is the area under the curve and above the given interval.

15. Properties of continuous probability distributions
1. f(x) > 0 (the curve cannot dip below the horizontal axis)
2. The total area under the density curve equals one.

16. Let x denote the amount of gravel sold (in tons) during a randomly selected week at a particular sales facility. Suppose that the density curve has a height f(x) above the value x, where
The density curve is shown in the figure:

17. P(x < ½) = 1 – ½(0.5)(1) = .75 Gravel problem continued . . .
What is the probability that at most ½ ton of gravel is sold during a randomly selected week?
P(x < ½) =
1 – ½(0.5)(1) = .75
This area can be found by use the formula for the area of a trapezoid:
OR, more easily, by finding the area of the triangle,
and subtracting that area from 1.
The probability would be the shaded area under the curve and above the interval from 0 to 0.5.

18. P(x = ½) = Gravel problem continued . . .
What is the probability that exactly ½ ton of gravel is sold during a randomly selected week?
P(x = ½) =
How do we find the area of a line segment?
The probability would be the area under the curve and above 0.5.
Since a line segment has NO area, then the probability that exactly ½ ton is sold equals 0.

19. Does the probability change whether the ½ is included or not?
Gravel problem continued . . .
What is the probability that less than ½ ton of gravel is sold during a randomly selected week?
P(x < ½) =
P(x < ½)
= 1 – ½(0.5)(1) = .75
Hmmm This is different than discrete probability distributions where it does change the probability whether a value is included or not!
Does the probability change whether the ½ is included or not?

20. The following is the graph of f(x), the density curve:
Suppose x is a continuous random variable defined as the amount of time (in minutes) taken by a clerk to process a certain type of application form. Suppose x has a probability distribution with density function:
The following is the graph of f(x), the density curve:
Time (in minutes)
Density

21. P(x > 5.5) = .5(.5) = .25 Application Problem Continued . . .
What is the probability that it takes more than 5.5 minutes to process the application form?
P(x > 5.5) =
.5(.5) = .25
When the density is constant over an interval (resulting in a horizontal density curve), the probability distribution is called a uniform distribution.
Find the probability by calculating the area of the shaded region (base × height).
Time (in minutes)
Density

22. Other Density Curves
Some density curves resemble the one below. Integral calculus is used to find the area under the these curves.
Don’t worry – we will use tables (with the values already calculated). We can also use calculators or statistical software to find the area.

23. This will be useful later in this chapter!
The probability that a continuous random variable x lies between a lower limit a and an upper limit b is
P(a < x < b) = (cumulative area to the left of b) – (cumulative area to the left of a)
P(a < x < b) = P(x < b) – P(x < a)
This will be useful later in this chapter!

24. Means and Standard Deviations of Probability Distributions
The mean value of a random variable x, denoted by mx, describes where the probability distribution of x is centered.
The standard deviation of a random variable x, denoted by sx, describes variability in the probability distribution

25. Mean and Variance for Discrete Probability Distributions
Mean is sometimes referred to as the expected value (denoted E(x)).
Variance is calculated using
Standard deviation is the square root of the variance.

26. mx = 1.51 pets Dogs and Cats Revisited . . .
Let x = the number of dogs and cats in a randomly selected household in Wolf City x
P(x)
What is the mean number of pets per household in Wolf City?
xP(x)
xP(x)
First multiply each x-value times its corresponding probability.
Next find the sum of these values.
mx = 1.51 pets

27. Dogs and Cats Revisited . . .
Let x = the number of dogs or cats per household in Wolf City x
P(x)
What is the standard deviation of the number of pets per household in Wolf City?
This is the variance – take the square root of this value.
Next multiply by the corresponding probability. Then add these values.
First find the deviation of each x-value from the mean. Then square these deviations.
sx2 = (0-1.51)2(.26) + (1-1.51)2(.31) + (2-1.51)2(.21) + (3-1.51)2(.13) + (4-1.51)2(.06) + (5-1.51)2(.03) =
sx = pets

28. Mean and Variance for Continuous Random Variables
For continuous probability distributions, mx and sx can be defined and computed using methods from calculus.
The mean value mx locates the center of the continuous distribution.
The standard deviation, sx, measures the extent to which the continuous distribution spreads out around mx.

29. mx = 4650 pounds/inch2 sx = 200 pounds/inch2
A company receives concrete of a certain type from two different suppliers.
Let x = compression strength of a randomly selected batch from Supplier 1
y = compression strength of a randomly selected batch from Supplier 2
Suppose that
mx = 4650 pounds/inch2 sx = 200 pounds/inch2
my = 4500 pounds/inch2 sy = 275 pounds/inch2
The first supplier is preferred to the second both in terms of mean value and variability.
4500
4300
4700
4900 my mx

30. The mean and standard deviation of the monthly salaries are
What would happen to the mean and standard deviation if we had to deduct $100 from everyone’s salary because of business being bad?
Suppose Wolf City Grocery had a total of 14 employees. The following are the monthly salaries of all the employees.
The mean and standard deviation of the monthly salaries are
mx = $1700 and sx = $603.56
Suppose business is really good, so the manager gives everyone a $100 raise per month. The new mean and standard deviation would be
mx = $1800 and sx = $603.56
3500
1200
1900
1400
2100
1800
1300
1500
1700
2300
Let’s graph boxplots of these monthly salaries to see what happens to the distributions . . .
What happened to the standard deviations?
What happened to the means?
We see that the distribution just shifts to the right 100 units but the spread is the same.

31. Notice that both the mean and standard deviation increased by 1.2.
Wolf City Grocery Continued . . .
mx = $1700 and sx = $603.56
Suppose the manager gives everyone a 20% raise - the new mean and standard deviation would be
mx = $2040 and sx = $724.27
Let’s graph boxplots of these monthly salaries to see what happens to the distributions . . .
Notice that multiplying by a constant stretches the distribution, thus, changing the standard deviation.
Notice that both the mean and standard deviation increased by 1.2.

32. Mean and Standard Deviation of Linear functions
If x is a random variable with mean, mx, and standard deviation, sx, and a and b are numerical constants, and the random variable y is defined by
and

33. my = 50 + 1.8(318) = $622.40 What is the equation for y?
Consider the chance experiment in which a customer of a propane gas company is randomly selected. Let x be the number of gallons required to fill a propane tank. Suppose that the mean and standard deviation is 318 gallons and 42 gallons, respectively. The company is considering the pricing model of a service charge of $50 plus $1.80 per gallon Let y be the random variable of the amount billed.
What is the equation for y?
What are the mean and standard deviation for the amount billed?
y = x
my = (318) = $622.40
sy = 1.8(42) = $75.60

34. List all the possible sums (A + B). 2 3 4 5 6 7 3 4 5 6 7 8 ?
Move
1 s
Suppose we are going to play a game called Stat Land! Players spin the two spinners below and move the sum of the two numbers.
mA = mB = 3.5
sA = sB = 1.708
List all the possible sums (A + B). 1 2 3 4
Spinner A 5 6
Spinner B
Find the mean and standard deviation for these sums.
Not sure – let’s think about it and return in just a few minutes!
Here are the mean and standard deviation for each spinner.
Notice that the mean of the sums is the sum of the means!
How are the standard deviations related?
mA+B = 6
sA+B =2.041

35. List all the possible differences (B - A). ?
Move
1 s
Stat Land Continued . . .
Suppose one variation of the game had players move the difference of the spinners
mA = mB = 3.5
sA = sB = 1.708
List all the possible differences (B - A). 1 2 3 4
Spinner A 5 6
Spinner B
Find the mean and standard deviation for these differences.
How do we find the standard deviation for the sums or differences?
Notice that the mean of the differences is the difference of the means!
WOW – this is the same value as the standard deviation of the sums!
mB-A= 1
sB-A =2.041

36. Mean and Standard Deviations for Linear Combinations
If x1, x2, …, xn are random variables with means m1, m2, …, mn and variances s12, s22, …, sn2, respectively, and
y = a1x1 + a2x2 + … + anxn
then
This result is true ONLY if the x’s are independent.
This result is true regardless of whether the x’s are independent.

37. A commuter airline flies small planes between San Luis Obispo and San Francisco. For small planes the baggage weight is a concern.
Suppose it is known that the variable x = weight (in pounds) of baggage checked by a randomly selected passenger has a mean and standard deviation of 42 and 16, respectively.
Consider a flight on which 10 passengers, all traveling alone, are flying.
The total weight of checked baggage, y, is
y = x1 + x2 + … + x10

38. mx = m1 + m2 + … + m10 = 42 + 42 + … + 42 = 420 pounds
Airline Problem Continued . . .
mx = 42 and sx = 16
The total weight of checked baggage, y, is
y = x1 + x2 + … + x10
What is the mean total weight of the checked baggage?
mx = m1 + m2 + … + m10
= … + 42
= 420 pounds

39. To find the standard deviation, take the square root of this value.
Airline Problem Continued . . .
mx = 42 and sx = 16
The total weight of checked baggage, y, is
y = x1 + x2 + … + x10
What is the standard deviation of the total weight of the checked baggage?
Since the 10 passengers are all traveling alone, it is reasonable to think that the 10 baggage weights are unrelated and therefore independent.
To find the standard deviation, take the square root of this value.
sx2 = sx12 + sx22 + … + sx102
= … + 162
= 2560 pounds
s = pounds

40. Special Distributions
Two Discrete Distributions:
Binomial and Geometric
One Continuous Distribution:
Normal Distributions

41. These questions can be answered using a binomial distribution.
Suppose we decide to record the gender of the next 25 newborns at a particular hospital.
What is the chance that at least 15 are female?
What is the chance that between 10 and 15 are female?
Out of the 25 newborns, how many can we expect to be female?
These questions can be answered using a binomial distribution.

42. Properties of a Binomial Experiment
There are a fixed number of trials
Each trial results in one of two mutually exclusive outcomes. (success/failure)
Outcomes of different trials are independent
The probability that a trial results in success is the same for all trials
The binomial random variable x is defined as
x = the number of successes observed when a binomial experiment is performed
We use n to denote the fixed number of trials.

43. Are these binomial distributions?
Toss a coin 10 times and count the number of heads
Yes
Deal 10 cards from a shuffled deck and count the number of red cards
No, probability does not remain constant
The number of tickets sold to children under 12 at a movie theater in a one hour period
No, no fixed number

44. Binomial Probability Formula:
Let
n = number of independent trials in a binomial experiment
p = constant probability that any trial results in a success
Where:
Appendix Table 9 can be used to find binomial probabilities.
Technology, such as calculators and statistical software, will also perform this calculation.

45. Is this a binomial experiment?
Instead of recording the gender of the next 25 newborns at a particular hospital, let’s record the gender of the next 5 newborns at this hospital.
Is this a binomial experiment?
Yes, if the births were not multiple births (twins, etc).
Define the random variable of interest.
x = the number of females born out of the next 5 births
What are the possible values of x? x
What is the probability of “success”?
What will the largest value of the binomial random value be?
Will a binomial random variable always include the value of 0?

46. Newborns Continued . . .
What is the probability that exactly 2 girls will be born out of the next 5 births?
What is the probability that less than 2 girls will be born out of the next 5 births?

47. What is the mean number of girls born in the next five births?
Newborns Continued . . .
Let’s construct the discrete probability distribution table for this binomial random variable:
What is the mean number of girls born in the next five births? x 1 2 3 4 5
p(x)
.03125
.15625
.3125
Notice that this is the same as multiplying
n × p
Since this is a discrete distribution, we could use:
mx = 0(.03125) + 1(.15625) + 2(.3125) + 3(.3125) + 4(.15625) + 5(.03125)
=2.5

48. Formulas for mean and standard deviation of a binomial distribution

49. Newborns Continued . . .
How many girls would you expect in the next five births at a particular hospital?
What is the standard deviation of the number of girls born in the next five births?

50. Remember, in binomial distributions, trials should be independent.
However, when we sample, we typically sample without replacement, which would mean that the trials are not independent. . .
In this case, the number of success observed would not be a binomial distribution but rather hypergeometric distribution.
But when the sample size, n, is small and the population size, N, is large, probabilities calculated using binomial distributions and hypergeometric distributions are VERY close!
When sampling without replacement if n is at most 5% of N, then the binomial distribution gives a good approximation to the probability distribution of x.
The calculation for probabilities in a hypergeometric distribution are even more tedious than the binomial formula!

51. How is this question different from a binomial distribution?
Newborns Revisited . . .
Suppose we were not interested in the number of females born out of the next five births, but
which birth would result in the first female being born?
How is this question different from a binomial distribution?

52. Properties of Geometric Distributions:
There are two mutually exclusive outcomes that result in a success or failure
Each trial is independent of the others
The probability of success is the same for all trials.
A geometric random variable x is defined as
x = the number of trials UNTIL the FIRST success is observed ( including the success). x
So what are the possible values of x
To infinity
How far will this go?
. . .

53. Probability Formula for the Geometric Distribution
Let
p = constant probability that any trial results in a success
Where x = 1, 2, 3, …

54. Suppose that 40% of students who drive to campus at your school or university carry jumper cables. Your car has a dead battery and you don’t have jumper cables, so you decide to stop students as they are headed to the parking lot and ask them whether they have a pair of jumper cables.
Let x = the number of students stopped before finding one with a pair of jumper cables
Is this a geometric distribution?
Yes

55. Jumper Cables Continued . . .
Let x = the number of students stopped before finding one with a pair of jumper cables
p = .4
What is the probability that third student stopped will be the first student to have jumper cables?
What is the probability that at most three student are stopped before finding one with jumper cables?
P(x = 3) =
(.6)2(.4) = .144
P(x < 3) = P(1) + P(2) + P(3) =
(.6)0(.4) + (.6)1(.4) + (.6)2(.4) = .784

56. How is this done mathematically?
Normal Distributions
Continuous probability distribution
Symmetrical bell-shaped (unimodal) density curve defined by m and s
Area under the curve equals 1
Probability of observing a value in a particular interval is calculated by finding the area under the curve
As s increases, the curve flattens & spreads out
As s decreases, the curve gets taller and thinner
How is this done mathematically?
To overcome the need for calculus, we rely on technology or on a table of areas for the standard normal distribution

57. A B 6 s s
Do these two normal curves have the same mean? If so, what is it?
Which normal curve has a standard deviation of 3?
Which normal curve has a standard deviation of 1?
YES B A

58. Notice that the normal curve is curving downwards from the center (mean) to points that are one standard deviation on either side of the mean. At those points, the normal curve begins to turn upward.

59. Standard Normal Distribution
Is a normal distribution with m = 0 and s = 1
It is customary to use the letter z to represent a variable whose distribution is described by the standard normal curve (or z curve).

60. Using the Table of Standard Normal (z) Curve Areas
For any number z*, from to 3.89 and rounded to two decimal places, the Appendix Table 2 gives the area under the z curve and to the left of z*.
P(z < z*) = P(z < z*)
Where
the letter z is used to represent a random variable whose distribution is the standard normal distribution.
To use the table:
Find the correct row and column (see the following example)
The number at the intersection of that row and column is the probability

61. Suppose we are interested in the probability that z. is less than -1
P(z < -1.62) =
In the table of areas:
Find the row labeled -1.6
Find the column labeled 0.02
Find the intersection of the row and column
.0526 … z*
.00
.01
.02
-1.7
.0446
.0436
.0427
.0418
-1.6
.0548
.0537
.0526
.0516
-1.5
.0668
.0655
.0643
.0618

62. Suppose we are interested in the probability that z* is less than 2.31.
P(z < 2.31) =
.9896 … z*
.00
.01
.02
2.2
.9861
.9864
.9868
.9871
2.3
.9893
.9896
.9898
.9901
2.4
.9918
.9920
.9922
.9925

63. Suppose we are interested in the probability that z. is greater than 2
P(z > 2.31) =
The Table of Areas gives the area to the LEFT of the z*.
To find the area to the right, subtract the value in the table from 1
= .0104 … z*
.00
.01
.02
2.2
.9861
.9864
.9868
.9871
2.3
.9893
.9896
.9898
.9901
2.4
.9918
.9920
.9922
.9925

64. Suppose we are interested in the finding the z* for the smallest 2%.
P(z < z*) = .02
To find z*:
Look for the area in the body of the Table. Follow the row and column back out to read the z-value.
Since doesn’t appear in the body of the Table, use the value closest to it.
z* = -2.08 z* … z*
.06
.07
.08
-2.1
.0162
.0158
.0154
-2.0
.0207
.0202
.0197
-1.9
.0262
.0256
.0250 …

65. Suppose we are interested in the finding the z* for the largest 5%.
P(z > z*) = .05
Since is exactly between and .9505, we can average the z* for each of these
.95
z* = 1.645 z*
Remember the Table of Areas gives the area to the LEFT of z*.
1 – (area to the right of z*)
Then look up this value in the body of the table. … z*
.03
.04
.05
1.5
.9382
.9398
.9406
1.6
.9495
.9505
.9515
1.7
.9591
.9599
.9608 …

66. Finding Probabilities for Other Normal Curves
To find the probabilities for other normal curves, standardize the relevant values and then use the table of z areas.
If x is a random variable whose behavior is described by a normal distribution with mean m and standard deviation s , then
P(x < b) = P(z < b*)
P(x > a) = P(z > a*)
P(a < x < b) = P(a* < z < b*)
Where z is a variable whose distribution is standard normal and

67. Data on the length of time to complete registration for classes using an on-line registration system suggest that the distribution of the variable
x = time to register
for students at a particular university can well be approximated by a normal distribution with mean m = 12 minutes and standard deviation s = 2 minutes.

68. Registration Problem Continued . . . x = time to register
m = 12 minutes and s = 2 minutes
What is the probability that it will take a randomly selected student less than 9 minutes to complete registration?
Standardized this value.
Look this value up in the table.
P(x < 9) =
.0668 9

69. Registration Problem Continued . . . x = time to register
m = 12 minutes and s = 2 minutes
What is the probability that it will take a randomly selected student more than 13 minutes to complete registration?
Standardized this value.
Look this value up in the table and subtract from 1.
P(x > 13) =
= .3085 13

70. Registration Problem Continued . . . x = time to register
m = 12 minutes and s = 2 minutes
What is the probability that it will take a randomly selected student between 7 and 15 minutes to complete registration?
Standardized these values.
Look these values up in the table and subtract
(value for a*) – (value for b*)
P(7 < x < 15) =
= .9270 7 15

71. Registration Problem Continued . . . x = time to register
m = 12 minutes and s = 2 minutes
Because some students do not log off properly, the university would like to log off students automatically after some time has elapsed. It is decided to select this time so that only 1% of students will be automatically logged off while still trying to register.
What time should the automatic log off be set at?
Look up the area to the left of a* in the table.
Use the formula for standardizing to find x.
P(x > a*) = .01
a* = 16.66
.99
.01 a*

72. Ways to Assess Normality
What should happen if our data set is normally distributed?
Some of the most frequently used statistical methods are valid only when x1, x2, …, xn has come from a population distribution that at least is approximately normal. One way to see whether an assumption of population normality is plausible is to construct a normal probability plot of the data.
A normal probability plot is a scatterplot of (normal score, observed values) pairs.

73. Consider a random sample with n = 5.
To find the appropriate normal scores for a sample of size 5, divide the standard normal curve into 5 equal-area regions.
Each region has an area equal to 0.2.
Why are these regions not the same width?
Each region should have 20% of the area in them – the higher the curve the less width needed for 20%.

74. Consider a random sample with n = 5.
Next – find the median z-score for each region.
These are the normal scores that we would plot our data against.
Why is the median not in the “middle” of each region?
We use technology (calculators or statistical software) to compute these normal scores.
The median should divide the area into half. In this case, the first median would be at the 10th percentile, the second at the 30th percentile, etc.
1.28
-1.28
-.524
.524

75. Ways to Assess Normality
Such as curvature which would indicate skewness in the data
Some of the most frequently used statistical methods are valid only when x1, x2, …, xn has come from a population distribution that at least is approximately normal. One way to see whether an assumption of population normality is plausible is to construct a normal probability plot of the data.
A normal probability plot is a scatterplot of (normal score, observed values) pairs.
A strong linear pattern in a normal probability plot suggest that population normality is plausible.
On the other hand, systematic departure from a straight-line pattern indicates that it is not reasonable to assume that the population distribution is normal.
Or outliers

76. Sketch a scatterplot by pairing the smallest normal score with the smallest observation from the data set & so on
The following data represent egg weights (in grams) for a sample of 10 eggs.
Let’s construct a normal probability plot. Since the values of the normal scores depend on the sample size n, the normal scores when n = 10 are below:
Since the normal probability plot is approximately linear, it is plausible that the distribution of egg weights is approximately normal.

77. Using the Correlation Coefficient to Assess Normality
The correlation coefficient, r, can be calculated for the n (normal score, observed value) pairs.
If r is too much smaller than 1, then normality of the underlying distribution is questionable.
Consider these points from the weight of eggs data:
(-1.539, 52.53) (-1.001, 52.66) (-.656,52.86) (-.376,53.00) (-.123, 53.04) (.123,53.07) (.376,53.16) (.656,53.23) (1.001,53.26) (1.539,53.50)
Calculate the correlation coefficient for these points.
r = .986
Since
r > critical r,
then it is plausible that the sample of egg weights came from a distribution that was approximately normal.
Values to Which r Can be Compared to Check for Normality n 5 10 15 20 25 30 40 50 60 75
Critical r
.832
.880
911
.929
.941
.949
.960
.966
.971
.976
How smaller is “too much smaller than 1”?

78. Transforming Data to Achieve Normality
When the data is not normal, it is common to use a transformation of the data.
For data that shows strong positive skewness (long upper tail), a logarithmic transformation usually applied.
Square root, cube root, and other transformations can also be applied to the data to determine which transformation best normalizes the data.

79. Consider the data set in Table 7
Consider the data set in Table 7.4 (page 463) about plasma and urinary AGT levels.
A histogram of the urinary AGT levels is strongly positively skewed.
A logarithmic transformation is applied to the data. The histogram of the log urinary AGT levels is more symmetrical.

80. Using the Normal Distribution to Approximate a Discrete Distribution
Suppose this bar is centered at x = 6. The bar actually begins at 5.5 and ends at Theses endpoints will be used in calculations.
This is called a continuity correction.
Suppose the probability distribution of a discrete random variable x is displayed in the histogram below.
Often, a probability histogram can be well approximated by a normal curve. If so, it is customary to say that x has an approximately normal distribution.
The probability of a particular value is the area of the rectangle centered at that value. 6

81. Normal Approximation to a Binomial Distribution
Let x be a random variable based on n trials and success probability p, so that:
If n and p are such that:
np > 10 and n (1 – p) > 10
then x has an approximately normal distribution.

82. Can the distribution of x be approximated by a normal distribution?
Premature babies are born before 37 weeks, and those born before 34 weeks are most at risk. A study reported that 2% of births in the United States occur before 34 weeks. Suppose that 1000 births are randomly selected and that the number of these births that occurred prior to 34 weeks, x, is to be determined.
np = 1000(.02) = 20 > 10
n(1 – p) = 1000(.98) = 980 > 10
Find the mean and standard deviation for the approximated normal distribution.
Since both are greater than 10, the distribution of x can be approximated by a normal distribution
Can the distribution of x be approximated by a normal distribution?

83. Premature Babies Continued . . .
m = 20 and s = 4.427
What is the probability that the number of babies in the sample of 1000 born prior to 34 weeks will be between 10 and 25 (inclusive)?
P(10 < x < 25) =
Look up these values in the table and subtract the probabilities.
To find the shaded area, standardize the endpoints.
= .8836