Chapter 5
5.1
Random Variable = the numerical value of an outcome (outcomes happen by chance)
Continuous Random Variable = random variable that can have fraction amounts like on a number
line
Example: Height of person, weight of backpack, temperature, distance traveled, time, volume
Continuous Distribution = distribution of a continuous random variable. Like a histogram but all the
possible values are graphed, not just bars on intervals. These are graphed above the x axis which
shows the values of the random variable. x-axis is the variable X. y-axis is frequency or count.
Example:
Continuous Probability Distribution = probability distribution of a continuous random variable.
These are graphed above the x axis which shows the values of the random variable. y-axis is percent.
Mean of Continuous Random Variable μ
Variance of Continuous Random Variable σ2
Standard Deviation of Continuous Random Variable
σ
σ = √𝝈𝟐
Normal Distribution = Normal Curve = a probability distribution that
 Is a bell shaped curve
 Is symmetric about the mean (no skewness)
 has total area under the curve = 1. All the probabilities of all the possible points add up to 1.0
units of area.
 Has mean = median = mode = μ (at the center of the curve)
 Has tails at the right and left end that approach but never touch the x axis.
 Has inflection points at +1σ and -1σ
Symmetric = equal shape on both sides of the middle.
Inflection points = where the curve changes from cupped up to cupped down. This happens at +1 σ
and -1 σ.
Concave up = cupped up = curved up
Concave down = cupped down = curved down
Measure of Central Tendency = Mean, Median, Mode
Measure of Spread = Variance, Standard Deviation, Range
When σ and σ2 are large, the curve is spread wider and not as tall.
When σ and σ2 are small, the curve is narrower and taller.
Remember: the total area under the curve = 1.0, so if it is not as tall, it must be wider.
Two curves with the same standard deviation will be the same shape, same spread, same height.
Two curves with the same mean will have same center.
Standard score = z-score
z can be positive or negative.
When z is positive, the data value is above the mean.
When z is negative, the data value is below the mean.
z tells the number of standard deviations you are above or below the mean.
Z = 0 is at the mean.
Ex: The test scores for a civil service exam have a mean of 152 and standard deviation of 7. Find the standard
z-score for a person with a score of 161.
𝑧=
161 − 152
= 1.29
7
z = 1.29
The data value 161 is a little more than 1 standard deviation above the mean
The Standard Normal Curve = Standard Normal Distribution
 x-axis is z-scores.
 y-axis is percent.
 has mean of 0 and a standard deviation of 1.
 Total area under the curve is 1.
 Data usually goes a little past -3 σ to the left and a little past +3 σ to the right.
 Inflection points are at +1σ and -1σ
Empirical Rule
About 68% of the data lies within 1 standard deviation of the mean
About 95% of the data lies within 2 standard deviations of the mean
About 99.7% of the data lies within 3 standard deviations of the mean
Cumulative area
The cumulative area at a certain z-score is the total area under the curve from 0 to the z-score.
This is area to the LEFT of the z-score.
The cumulative area for z = -3.49 is close to 0.
The cumulative area for z = 0 is 0.500.
The cumulative area for z = +3.49 is close to 1.0.
Find cumulative area from a table of values
Example:
Find the cumulative area for a z-score of –1.25
On the Standard Normal Distribution Table (Table 4),
read down the z column on the left to z = –1.2
and across to the column under .05.
The value in the cell is 0.1056, the cumulative area.
Interpret: There is 10.56% of the data to the left of the z-score of -1.25.
–3 –2 –1 0 1 2 3
The probability that z is at most –1.25 is 0.1056.
𝑃(𝑧 ≤ −1.25) = 0.1056
To find the area to the right of a z-score, use 1 - probability
The probability that z is greater than –1.25 is 1.00 - 0.1056 = 0.8944.
𝑃(𝑧 > −1.25) = 0.8944
To find the probability z is between two given values, find the cumulative areas for each and
subtract the smaller area from the larger.
Find P(–1.25 < z < 1.17).
1. P(z < 1.17) = 0.8790
2. P(z < –1.25) = 0.1056
3. P(–1.25 < z < 1.17) = 0.8790 – 0.1056 = 0.7734
-3 -2 -1 0 1 2 3
5.2 Finding probabilities with data and normal distributions
Finding probability from data value and μ and σ
1. Find z-score
2. Look up z-score to get cumulative area up to that z-score
3. Decide which are you need based on the problem.
Example:
Find the probability the computer owners will upgrade in fewer than 2 years, when μ = 2.4
and σ = 0.5
1. Find z-score
𝑧=
2−2.4
0.5
= −0.80
2. Look up z-score to get cumulative area up to that z-score
P(z ≤ -0.80) = 0.2119
3. Decide which are you need based on the problem.
We want the area left of the z score. Fewer than 2 years. So 21.19% of computer owners
will upgrade in fewer than 2 years.