Statistics – Lab Week 4
Name:_______________________
MATH221
Statistical Concepts:
 Probability
 Binomial Probability Distribution
Calculating Binomial Probabilities
 Open a new MINITAB worksheet.
 We are interested in a binomial experiment with 10 trials. First, we will make the
probability of a success ¼. Use MINITAB to calculate the probabilities for this
distribution. In column C1 enter the word ‘success’ as the variable name (in the
shaded cell above row 1. Now in that same column, enter the numbers zero through
ten to represent all possibilities for the number of successes. These numbers will end
up in rows 1 through 11 in that first column. In column C2 enter the words ‘one
fourth’ as the variable name. Pull up Calc > Probability Distributions > Binomial
and select the radio button that corresponds to Probability. Enter 10 for the Number
of trials: and enter 0.25 for the Event probability:. For the Input column: select
‘success’ and for the Optional storage: select ‘one fourth’. Click the button OK and
the probabilities will be displayed in the Worksheet.
 Now we will change the probability of a success to ½. In column C3 enter the words
‘one half’ as the variable name. Use similar steps to that given above in order to
calculate the probabilities for this column. The only difference is in Event
probability: use 0.5.
 Finally, we will change the probability of a success to ¾. In column C4 enter the
words ‘three fourths’ as the variable name. Again, use similar steps to that given
above in order to calculate the probabilities for this column. The only difference is in
Event probability: use 0.75.
Plotting the Binomial Probabilities
1. Create plots for the three binomial distributions above. Select Graph > Scatter Plot
and Simple then for graph 1 set Y equal to ‘one fourth’ and X to ‘success’ by clicking
on the variable name and using the “select” button below the list of variables. Do this
two more times and for graph 2 set Y equal to ‘one half’ and X to ‘success’, and for
graph 3 set Y equal to ‘three fourths’ and X to ‘success’. Paste those three scatter
plots below.
Scatterplot of One Fourth vs Sucess
0.30
0.25
One Fourth
0.20
0.15
0.10
0.05
0.00
0
2
6
4
8
10
8
10
Sucess
Scatterplot of One half vs Sucess
0.25
One half
0.20
0.15
0.10
0.05
0.00
0
2
4
6
Sucess
Scatterplot of Three Fourths vs Sucess
0.30
Three Fourths
0.25
0.20
0.15
0.10
0.05
0.00
0
2
4
6
8
10
Sucess
Calculating Descriptive Statistics
 Open the class survey results that were entered into the MINITAB worksheet.
2. Calculate descriptive statistics for the variable where students flipped a coin 10 times.
Pull up Stat > Basic Statistics > Display Descriptive Statistics and set Variables: to
the coin. The output will show up in your Session Window. Type the mean and the
standard deviation here.
Mean: 4.600
Standard deviation: 1.429
Short Answer Writing Assignment – Both the calculated binomial probabilities and the
descriptive statistics from the class database will be used to answer the following
questions.
3. List the probability value for each possibility in the binomial experiment that was
calculated in MINITAB with the probability of a success being ½. (Complete
sentence not necessary)
P(x=0)
P(x=1)
P(x=2)
P(x=3)
P(x=4)
P(x=5)
0.000977
0.009766
0.043945
0.117188
0.205078
0.246094
P(x=6)
P(x=7)
P(x=8)
P(x=9)
P(x=10)
0.205078
0.117188
0.043945
0.009766
0.000977
4. Give the probability for the following based on the MINITAB calculations with the
probability of a success being ½. (Complete sentence not necessary)
P(x≥1)
P(x>1)
P(4<x ≤7)
0.999023
0.989257
0.56836
P(x<0)
0
P(x≤4)
0.376954
P(x<4 or x≥7) 0.343752
5. Calculate the mean and standard deviation (by hand) for the MINITAB created
binomial distribution with the probability of a success being ½. Either show work or
explain how your answer was calculated. Mean = np, Standard Deviation = npq
Mean: np = 10(1/2)= 5
Standard deviation:
npq  10(1 / 2)(1  1 / 2)  2.5  1.581
6. Calculate the mean and standard deviation (by hand) for the MINITAB created
binomial distribution with the probability of a success being ¼ and compare to the
results from question 5. Mean = np, Standard Deviation = npq
Mean: np = 10(1/4)= 2.5
Standard deviation:
npq  10(1 / 4)(1  1 / 4)  1.875  1.369
Comparison: The mean is lower than in the question 5 (2.5<5) and the Standard deviation
is lower too (1.369 < 1.581)
7. Calculate the mean and standard deviation (by hand) for the MINITAB created
binomial distribution with the probability of a success being ¾ and compare to the
results from question 6. Mean = np, Standard Deviation = npq
Mean: np = 10(3/4)= 7.5
Standard deviation:
npq  10(3 / 4)(1  3 / 4)  1.875  1.369
Comparison: The mean is higher than in the question 5 (7.5 > 5) and the Standard
deviation is lower (1.369 < 1.581)
8. Explain why the coin variable from the class survey represents a binomial
distribution.
We have two possible outcomes if we flip a coin (Head or Tail), we flipped 10 coins
(n=10) and the events are independent.
So the coin variable (X=number of success) has a binomial distribution with n = 10 and
P(success) = p
9. Give the mean and standard deviation for the coin variable and compare these to the
mean and standard deviation for the binomial distribution that was calculated in
question 5. Explain how they are related. Mean = np, Standard Deviation = npq
For the coin variable:
Mean = 4.600 and Standard deviation = 1.429
For question 5) (variable has a binomial distribution with n=10 and p =0.5)
Mean =5 and Standard deviation = 1.581
Means and standard deviations are similar, is reasonable to think that the value of p is 0.5
(a fair coin)