Geometric mean of n numbers is defined as the nth root of the product of n numbers.
GM=x1×x2×x3...xn−−−−−−−−−−−−−−√nGM=x1×x2×x3...xnn
Where −
· nn = Total numbers.
· xixi = numbers.
Problem Statement:
Determine the geometric mean of following set of numbers.
|
1 |
3 |
9 |
27 |
81 |
Solution:
Step 1: Here n = 5
GM=x1×x2×x3...xn−−−−−−−−−−−−−−−√n=1×3×9×27×81−−−−−−−−−−−−−−−−√5=33×33×34−−−−−−−−−−√5=310−−−√5=325−−−√5=95−−√5=9GM=x1×x2×x3...xnn=1×3×9×27×815=33×33×345=3105=3255=955=9
Thus geometric mean of given numbers is 99.
The geometric distribution is a special case of the negative binomial distribution. It deals with the number of trials required for a single success. Thus, the geometric distribution is a negative binomial distribution where the number of successes (r) is equal to 1.
P(X=x)=p×qx−1P(X=x)=p×qx−1
Where −
· pp = probability of success for single trial.
· qq = probability of failure for a single trial (1-p)
· xx = the number of failures before a success.
· P(X−x)P(X−x) = Probability of x successes in n trials.
Problem Statement:
In an amusement fair, a competitor is entitled for a prize if he throws a ring on a peg from a certain distance. It is observed that only 30% of the competitors are able to do this. If someone is given 5 chances, what is the probability of his winning the prize when he has already missed 4 chances?
Solution:
If someone has already missed four chances and has to win in the fifth chance, then it is a probability experiment of getting the first success in 5 trials. The problem statement also suggests the probability distribution to be geometric. The probability of success is given by the geometric distribution formula:
P(X=x)=p×qx−1P(X=x)=p×qx−1
Where −
· p=30%=0.3p=30%=0.3
· x=5x=5 = the number of failures before a success.
Therefore, the required probability:
P(X=5)=0.3×(1−0.3)5−1,=0.3×(0.7)4,≈0.072≈7.2%