1. Mean, Variance, and Standard Deviation for the Binomial Distribution
Section 5-4
Mean, Variance, and Standard Deviation for the Binomial Distribution

2. For Any Discrete Probability Distribution: Formulas
Mean
Variance
Standard deviation
These formulas were introduced in section 5-2, pages
These formulas will produce the mean, standard deviation, and variance for any probability distribution. The difficult part of these formulas is that one must have all the P(x) values for the random variables in the distribution.

3. Binomial Distribution: Formulas
Mean µ = n • p
Variance σ2 = n • p • q
Standard deviation
Where
n = number of fixed trials
p = probability of success in one of the n trials
q = probability of failure in one of the n trials
The obvious advantage to the formulas for the mean, standard deviation, and variance of a binomial distribution is that you do not need the values in the distribution table. You need only the n, p, and q values.
A common error students tend to make is to forget to take the square root of (n)(p)(q) to find the standard deviation, especially if they used L1 and L2 lists in a graphical calculator to find the standard deviation with non-binomial distributions. The square root is built into the programming of the calculator and students do not have to remember it. These formulas do require the student to remember to take the square root of the three factors.
Page 225 of Elementary Statistics, 10th Edition

4. Interpretation of Results
It is especially important to interpret results. The range rule of thumb suggests that values are unusual if they lie outside of these limits:
Maximum usual values = µ + 2σ
Minimum usual values = µ – 2σ
Page 225 of Elementary Statistics, 10th Edition.

5. Example 1: Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean µ and standard deviation σ. Also, use the range rule of thumb to find the minimum usual value
µ – 2σ the maximum usual value µ + 2σ. Random guesses are made for 50 SAT multiple choice questions, so n = 50 and p = 0.2.
µ = np = (50)(0.2) = 10
σ2 = npq = (50)(0.2)(0.8) = 8; σ = 2.828
minimum usual value = µ – 2σ = 10 – 2(2.828) = 4.3
maximum usual value µ + 2σ = (2.828) = 15.7

6. Example 2: Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean µ and standard deviation σ. Also, use the range rule of thumb to find the minimum usual value
µ – 2σ the maximum usual value µ + 2σ. In an analysis of test results from YSORT gender selection method, 152 babies are born and it is assumed that boys and girls are equally likely, so n = 152 and p = 0.5.
µ = np = (152)(0.5) = 76
σ2 = npq = (152)(0.5)(0.5) = 38; σ = 6.164
minimum usual value = µ – 2σ = 76 – 2(6.164) = 63.7
maximum usual value µ + 2σ = (6.164) = 88.3

7. Example 3: The midterm exam in a nursing course consists of 75 true/false questions. Assume that an unprepared student makes random guesses for each of the answers.
a) Find the mean and standard deviation for the number of correct answers for such students.
Let x = the number of correct answers.
Binomial problem: n = 75 and p = 0.5
µ = np = (75)(0.5) = 37.5
σ2 = npq = (75)(0.5)(0.5) = 18.75; σ = 4.330

8. b) Would it be unusual for a student to pass this exam by guessing and getting at least 45 correct answers? Why or why not?
minimum usual value = µ – 2σ = 37.5 – 2(4.33) = 28.8
maximum usual value µ + 2σ = (4.33) = 46.2
No. Since 45 is within the above limits, it would not be unusual for a student to pass by getting at least 45 correct answers.

9. Example 4: Mars, Inc. claims that 16% of its M&M plain candies are green. A sample of 100 M&Ms is randomly selected.
a) Find the mean and standard deviation for the numbers of green M&Ms in such groups of 100.
Let x = the number of green M&Ms.
Binomial problem: n = 100 and p = 0.16
µ = np = (100)(0.16) = 16
σ2 = npq = (100)(0.16)(0.84) = 13.44; σ = 3.666

10. b) Data set 18 in Appendix B consists of a random sample of 100 M&Ms in which 19 are green. Is this result unusual? Does it seem that the claimed rate of 16% is wrong?
minimum usual value = µ – 2σ = 16 – 2(3.666) = 8.7
maximum usual value µ + 2σ = (3.666) = 23.3
No. Since 19 is within the above limits, it would not be unusual for 100 M&Ms to include 19 green ones. This is not evidence that the claimed rate of 16% is wrong.

11. Example 5: A headline in USA Today states that “most stay at first jobs less than 2 years.” That headline is based on an Expeience.com poll of 320 college graduates. Among those polled, 78% stayed at their first full-time job less than 2 years.
a) Assuming that 50% is the true percentage of graduates who stay at their first job less than two years, find the mean and standard deviation of the numbers of such graduates in randomly selected groups of 320 graduates.
Let x = the number who stay at their first job less than two years.
Binomial problem: n = 320 and p = 0.5
µ = np = (320)(0.5) = 160
σ2 = npq = (320)(0.5)(0.5) = 80; σ = 8.944

12. b) Assuming that the 50% rate in part (a) is correct, find the range of usual values for the number of graduates among 320 who stay at their first job less than two years.
minimum usual value = µ – 2σ = 160 – 2(8.944) = 142.1
maximum usual value µ + 2σ = (8.944) = 177.9
c) Find the actual number of surveyed graduates who stayed at their first job less than two years. Use the range values from part (b) to determine whether that number is unusual. Does the result suggest that the headline is not justified?
x = 0.78(320) = Since 250 is not within the above limits, it would be unusual for 320 graduates to include 250 persons who stayed at their first job less than two years is the true proportion were 50%. Since 280 is greater than the above limits, the true proportion is most likely greater than 50%. The result suggests that the headline is justified.