Baseball Games Back Calculation
Jun/100
Binomial probability Calculations. Need help understanding.?
Alright, what’s going to happen is i’ll provide the questions, I can get all the correct variable, I just can’t input the correct calculation in my calculator to get the correct answer.
Q:
A statistics practionor working for major league baseball determined the probability that a hitter will be out on ground balls .75. In a game where there are 20 ground balls, find the probability that all of them are outs.
SO:
P= .75 x or k= 20 n=20
The answer should come out to be = .00317
I inputted the calculation into this site as well http://faculty.vassar.edu/lowry/binomialX.html , and got the correct answer, provided by the back of the text book.
So my question is how does one calculate this?
Any help would be greatly appreciated.
What type of calculator do you have?
Most of the TI calculators, the 83, 84, 89 and 92 have a nCr function. use this function for the binomialpmf function in the 83/84 to find the probabilities.
Also note that in this case, as I will type out in a minute will simplify quite nicely.
Let X be the number of outs made. X has the binomial distribution with n = 20 trials and success probability p = 0.75
In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, …, n
P[X = x] = 0 for any other value of x.
The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n – x failures.
Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.
X ~ Binomial( n , p )
the mean of the binomial distribution is n * p = 15
the variance of the binomial distribution is n * p * (1 – p) = 3.75
the standard deviation is the square root of the variance = √ ( n * p * (1 – p)) = 1.936492
The Probability Mass Function, PMF,
f(X) = P(X = x) is:
P( X = 0 ) = 9.094947e-13
P( X = 1 ) = 5.456968e-11
P( X = 2 ) = 1.555236e-09
P( X = 3 ) = 2.799425e-08
P( X = 4 ) = 3.569266e-07
P( X = 5 ) = 3.426496e-06
P( X = 6 ) = 2.569872e-05
P( X = 7 ) = 0.0001541923
P( X = 8 ) = 0.0007516875
P( X = 9 ) = 0.00300675
P( X = 10 ) = 0.009922275
P( X = 11 ) = 0.02706075
P( X = 12 ) = 0.06088669
P( X = 13 ) = 0.1124062
P( X = 14 ) = 0.1686093
P( X = 15 ) = 0.2023312
P( X = 16 ) = 0.1896855
P( X = 17 ) = 0.1338956
P( X = 18 ) = 0.06694781
P( X = 19 ) = 0.02114141
P( X = 20 ) = 0.003171212
P(X = 20 ) = 20! / (20! 0!) * 0.75 ^ 20 * (0.25)^(0)
= 1 * 0.75 ^ 20 * 1
= 0.75 ^ 20
= 0.003171212
Adam Bold – Back to Basics: Rebuilding Your Personal Wealth
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