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At 07:57 AM 12/1/2006, Omar Farook wrote:
> Kindly , may I know buy using SPSS , how can I find the following
> probabilities for a given binomial random variable X with n = 25 and
> p =0.6?
> P(18<X<21).
> P(X≥12).
> P(18≤X≤21).
Not sure what those last two are (what are characters ≥ and
≤?), but SPSS has functions for probabilities for the binomial
distribution: PDF.BINOM and CDF.BINOM. By chance, I just ran up some
code to create a binomial probability table, to check something from
another posting. The following is SPSS draft output, 14.0.1:
NEW FILE.
INPUT PROGRAM.
. NUMERIC N (F4)
/P (F6.3)
/K (F4)
/P_K P_0$K P_K$N (PCT7.2).
. VAR LABELS
N 'Total number of questions'
K 'Number of right answers'
P 'Probability of answering a questioh right'
P_K 'Probability of this # of right answers'
P_0$K 'Probability of this many, or fewer'
P_K$N 'Probability of at least this many'
. LEAVE N K P.
. COMPUTE N = 15.
. COMPUTE P = 0.2.
. LOOP K = 0 TO N.
. COMPUTE P_K = PDF.BINOM(K,N,P).
. COMPUTE P_0$K = CDF.BINOM(K,N,P).
. COMPUTE P_K$N = P_K + (1 - P_0$K).
* Kludge, because the SPSS 'PCT' formats are simply wrong: .
* the number that displays as '20%' is 20, not 0.2. .
. DO REPEAT PROB = P_K P_0$K P_K$N.
. COMPUTE PROB=PROB*100.
. END REPEAT.
. END CASE.
. END LOOP.
END FILE.
END INPUT PROGRAM.
LIST.
List
|-----------------------------|---------------------------|
|Output Created |30-NOV-2006 22:59:45 |
|-----------------------------|---------------------------|
N P K P_K P_0$K P_K$N
15 .200 0 3.52% 3.52% 100.00%
15 .200 1 13.19% 16.71% 96.48%
15 .200 2 23.09% 39.80% 83.29%
15 .200 3 25.01% 64.82% 60.20%
15 .200 4 18.76% 83.58% 35.18%
15 .200 5 10.32% 93.89% 16.42%
15 .200 6 4.30% 98.19% 6.11%
15 .200 7 1.38% 99.58% 1.81%
15 .200 8 .35% 99.92% .42%
15 .200 9 .07% 99.99% .08%
15 .200 10 .01% 100.00% .01%
15 .200 11 .00% 100.00% .00%
15 .200 12 .00% 100.00% .00%
15 .200 13 .00% 100.00% .00%
15 .200 14 .00% 100.00% .00%
15 .200 15 .00% 100.00% .00%
Number of cases read: 16 Number of cases listed: 16
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