Title | Worksheet-11-Binomial distribution solutions |
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Author | Linderson Johns |
Course | Finite Mathematics |
Institution | Central Washington University |
Pages | 4 |
File Size | 144.6 KB |
File Type | |
Total Downloads | 68 |
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Answer key for worksheet 11 above, shows questions, solution steps and answers. Correction for any wrong questions for worksheet 11...
Math 130
Worksheet 11 BINOMIAL PROBABILITY DISTRIBUTION
Name: Key You must show clear formula work or explanations for full credit. Circle the final answer. Do not show all the calculation steps. BINOMIAL EXPERIMENT criteria: ♦ INDEPENDENCE Repeat independent trial for A FIXED NUMBER OF TIMES ♦ TWO OUTCOMESEach trial results in either success (S) or failure (F) ♦ CONSTANT PROBABILITY OF SUCCESSThe probability of success is fixed and remains unchanged from trial to trial. RANDOM VARIABLE: X = number of successes observed in n trials X = 0, 1, 2, 3, …., n (Discrete Random Variable) PRELIMINARY KNOWLEDGE •
RULE 1: Independent events If events A and B are independent, then P(AB)=P(A)⋅P(B) In general, if events A1,A2,A3,…,An are independent, then P(A1A2A3….An) = P(A1) ⋅P(A2) ⋅⋅⋅ P(An)
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RULE 2 Mutually exclusive events If events A1,A2,A3,…,An are mutually exclusive, then P(A1 or A2 or A3 or ….or An) = P(A1) + P(A2) + …. + P(An)
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COMBINATION FORMULA C(n,x) = the number of ways to choose x objects from n distinct objects when the order does not matter.
EXAMPLE1: A coin is DAMAGED so that the probability of seeing HEADS is 0.25. Flip this coin 3 times and observe what happens. Let’s use S (success) to denote heads and F (failure) to denote tails.
Q1. Should we consider these 3 tosses INDEPENDENT? Yes/ No Q2. Should we consider these 3 tosses IDENTICAL trials ? Yes/ No Q3. HOW MANY possible outcomes are there for EACH FLIP (TRIAL)? __2______ Q4. Does this experiment (tossing a damaged coin 3 times) satisfy the BINOMIAL EXPERIMENT CRITERIA? Yes Q5. Use a Tree Diagram to construct the SAMPLE SPACE. List all the complete outcomes in the set notation. 0.25 H3 0.25 H2 0.75 T3 0.25 H1 0.25 H3 0.75 T2 0.75 T3 0.25 H3 0.25 H2 0.75 T3 0.75 T1 0.25 H3 0.75 T2 0.75 T3
Outcomes: (H1H2H3), (H1H2T3), (H1T2H3), (H1T2T3), (T1H2H3), (T1H2T3), (T1T2H3), (T1T2T3) Q6. Let the HEADS stand for SUCCESS. If the random variable X = NUMBER OF SUCCESSES in these 3 trials, what are the possible values for X? X = 0,1,2, or 3
Q7. Construct the probability distribution for X. Hint: P(0) = P(FFF), P(1) = P(SFF or FSF or FFS), P(2) = P(SSF or SFS or FSS) ) P(0) = (0.75)(0.75)(0.75) = 27/64 P(1) = (0.25)(0.75)(0.75) + (0.75)(0.25)(0.75) + (0.75)(0.75)(0.25) = 27/64
P(2) = (0.25)(0.25)(0.75) + (0.25)(0.75)(0.25) + (0.75)(0.25)(0.25) = 9/64 P(3) = (0.25)(0.25)(0.25) = 1/64
Fill out this Probability Distribution table for the experiment: X
0
1
2
3
27/64
27/64
9/64
1/64
P(x)
Q8. Calculate the POPULATION MEAN of X by using the general formula. The general formula is the one that works for any probability distributions including Binomial.
Mean = 0(27/64) + 1(27/64) + 2(9/64) + 3(1/64) = 0.75
Q9. Calculate the POPULATION VARIANCE of X by using the general formula? Var(X) = (0-0.75)2(27/64) + (1-0.75)2(27/64) + (2-0.75)2(9/64) + (3-0.75)2(1/64) = (243/1024) + (27/1024) + (225/1024) + (81/1024) = 0.5625
Q10. Calculate the POPULATION STANDARD DEVIATION of X? SD= (Var(X))1/2
SD= (0.5625)1/2
SD= 0.75
In addition to the tree diagram for the probability distribution and general formulas for the mean and standard deviation, there are special ways to efficiently get the answers as below: BINOMIAL PROBABILITY DISTRIBUTION FORMULA
Let p = probability of success for each trial
n = total number of trials X = total number of successes observed Then State the special formula and calculate the answer. P(X=x) = _______C(n,x)pxqn-x_____________________________ P(0) =
C(3,0) 0.250 x 0.753 = 27/64
P(1) =
C(3,1) 0.251 x 0.752 = 27/64
P(2) =
C(3,2) 0.252 x 0.751 = 9/64
P(3) =
C(3,3) 0.253 x 0.750 = 1/64
µ = ________E(x) = np_____= (3)(0.25) = 0.75______________________ σ2 = _________npq = (3)(0.25)(0.75) = 0.5625_____________________ σ = _________(npq)1/2 = ((3)(0.25)(0.75))1/2 = 0.75__________________
Note: The above formulas are solely for the binomial probability distribution, a special type, but NOT USEFUL for other kinds of probability distribution derived from other NON-BINOMIAL random experiments....