1. Figure Determining probabilities or proportions for a normal distribution is shown as a two-step process with z-scores as an intermediate stop along the way. Note that you cannot move directly along the dashed line between X values and probabilities or proportions. Instead, you must follow the solid lines around the corner.

2. Step One: Convert to z-scores,
Example 6.6
What is the probability of selecting someone with an IQ of less then 120?
p(X<120) = ?
Step One: Convert to z-scores,
z = (X – μ) ∕ σ z = ( )/15 = +1.33
Step Two: look up z of in unit normal table
We want the area below 120 so find Proportion in column B
p(X<120) = p(z<1.33) = or %
Figure The distribution of IQ scores. The problem is to find the probability of proportion of the distribution corresponding to scores less than 120.

3. Figure 6-11 The distribution for Example 6.7.
Z-score of .30 is
Z-score of 0.70 is
What is the proportion of cars traveling between 55 and 65 miles per hour?
p(55< X >65) = ?
Step One: Covert scores to z-scores
Step Two: find z of 0.30 and 0.70 in Column A
find proportion in column D for each z score
because D gives proportion above and below mean add proportions.
= %
Figure The distribution for Example 6.7.

4. Figure 6-12 The distribution for Example 6.8.
Example 6.7 What proportion of cars are traveling between 65 and 75?
p(65< X <75) = ? OR p(0.70< z < 1.70)
p(z > 0.70) = from column D
p(z < 1.70) = from column D
Subtract proportion for 65 from proportion for 75
= %
Figure The distribution for Example 6.8.

5. Converting proportions to z-score
Example 6.10, how much commuting time in the top 10% ?
Look up 0.10 proportion in column C (or closest value)
find corresponding z score in column A
z = +1.28
Then convert z-sore to X value
X = μ + zσ
X = (10)
so X = 37.1 which is the cutoff for top 10%

6. Example 6.11
Find the range of commuting times for the middle 90%
Middle 90% is 45% above and below the mean.
Look up in column D (or closest score)
find z score of 1.65 in column A
convert z score to a X value using X = μ + zσ
X = (10) so X = Note positive Z score
But Wait!! You Are Not Done, this is only the upper score
X = (10) so X = Note negative Z score
Because we are looking for the range of scores for middle 90%
90 % is between and 40.8

7. Probability and the Binomial Distribution
Binomial distributions are formed by a series of observations (for example, 100 coin tosses) for which there are exactly two possible outcomes (heads and tails).
The two outcomes are identified as A and B, with probabilities of p(A) = p and p(B) = q.
For example the probability of getting heads or tails when flipping a coin.
Probability of getting heads: p = p(heads) = ½
Probability of getting tails: q = p(tails) = ½
p + q is always equal to 1.00

8. The probability of getting zero heads with n = 2 is 1/4 = 0.25
The binomial distribution shows the probability for each value of X, where X is the number of occurrences of A in a series of n observations.
In this example X is number of heads and n is the number of coin tosses so n = 2.
The probability of getting zero heads with n = 2 is 1/4 = 0.25
The probability of getting only one head with n = 2 is 2/4 = 0.50
Figure The binomial distribution showing the probability for the number of heads in 2 tosses of a balanced coin.

9. In this example in the graph (a) n =4 and in graph (b) n= 6.
The binomial distribution shows the probability for each value of X, where X is the number of occurrences of A in a series of n observations.
In this example in the graph (a) n =4 and in graph (b) n= 6.
The probability of getting only one head with n = 4 is 4/16 =
The probability of getting only one head with n = 6 is 6/64 =
Figure Binomial distributions showing probabilities for the number of heads (a) in 4 tosses of a balanced coin and (b) in 6 tosses of a balanced coin.

10. Figure 6-18 The relationship between the binomial distribution and the normal distribution. The binomial distribution is always a discrete histogram, and the normal distribution is a continuous smooth curve. Each X value is represented by a bar in the histogram or a section of the normal distribution.

11. Probability and the Binomial Distribution
Binomial distributions with a large number of outcomes closely approximated a normal distribution.
When pn and qn are both greater than 10
Mean: μ = pn
standard deviation: σ = npq.
In this situation, a z-score can be computed for each value of X and the unit normal table can be used to determine probabilities for specific outcomes.

12. n = 48, p = ¼ q = ¾ pn = ¼(48) =12 qn = ¾(48) =36
Figure 6-19 The binomial distribution (normal approximation) for example 6.13, predicting the suit of a card Because there are four suits, the probability of guessing the correct suit is p = ¼. So the probability of guessing the wrong suit is q = ¾. The number of correct predictions from a series of 48 trials, so n = 48. The shaded area corresponds to the probability of guessing correctly more than 15 times. Note that the score X = 15 is defined by its real limits, so the upper real limit is 15.5.
n = 48, p = ¼ q = ¾
pn = ¼(48) = qn = ¾(48) =36
pn = m = 12 and pnq = s = 3
z = (X – μ) ∕ σ
Z = (15.5 – 12)/3 = 1.17
Look up Z = p =
So probability of predicting the correct suit more than 15 times in 48 trials is 12.10%.

13. Probability and Inferential Statistics
Probability is important because it establishes a link between samples and populations.
For any known population it is possible to determine the probability of obtaining any specific sample.
In the Statistics course you will use this link as the foundation for inferential statistics.

14. Figure 6-20 A diagram of a research study
Figure A diagram of a research study. One individual is selected from the population and receives a treatment. The goal is to determine whether or not the treatment has an effect.

15. Figure 6-21 Using probability to evaluate a treatment effect
Figure Using probability to evaluate a treatment effect. Values that are extremely unlikely to be obtained from the original population are viewed as evidence of a treatment effect.