Rodgrigo,
I think you might find that p=0.1, not 0.1856. There are only ten possible digits, unless I am missing a few ! I suspect the question does not mean random selection from the sequence 1,2,3,4,5,6,7,8,9,10,11....625 like you have done, but selection from a random ordering of length n=625 from the digits 0 to 9 (eg 5211973060425783...).
Graham
-----Original Message-----
From: A UK-based worldwide e-mail broadcast system mailing list [mailto:[log in to unmask]] On Behalf Of Rodrigo Briceņo
Sent: 08 October 2012 16:55
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Subject: normal binomial
Dear All stat users. I have had problems finding the correct way to answer this exercise.
Among 625 random digits, find the probability that the digit 7 appears:
(a) between 50 and 60 times inclusive, (b) between 60 and 70 times inclusive.
The answers provided are: a) 0.3518 b) 0.5131
I proposed to identify how many numbers from a set of 625 have the digit 7. If I'm ok the number is 116. So applyin the normal approximation to binomial I got that the average is 116 and the standard deviation is 94.47 (116*0.1856*0.8144)
The question a: P(50<X<60)= P(49.5<X<60.5)= P(Z<-0.58)-P(Z<-0.70)=
0.281-0.242=0.039
As you may see I'm possibly missing something to have the correct answer.
--
Rodrigo Briceņo
Economist
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