approximations of distributions 2

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June 21st 2010, 06:04 PM
Apprentice123

approximations of distributions 2

A given honest is launched 100 times in a row. What is the probability that in 18 or more of these releases occur side 2?

I find:
p = 1/6

$Z = \frac{17,5 - (100* \frac{1}{6})}{\sqrt{100 * \frac{1}{6} * \frac{5}{6}}} = 0,22 -> 0,587064$

But the answer is: 0,4129
June 21st 2010, 08:06 PM
TKHunny

Which side of 17,5 do you want? 1 - 0,587064 = 0,412936 That looks pretty close.
June 21st 2010, 10:19 PM
CaptainBlack

Quote:

Originally Posted by Apprentice123

A given honest is launched 100 times in a row. What is the probability that in 18 or more of these releases occur side 2?

I find:
p = 1/6

$Z = \frac{17,5 - (100* \frac{1}{6})}{\sqrt{100 * \frac{1}{6} * \frac{5}{6}}} = 0,22 -> 0,587064$

But the answer is: 0,4129

You have calculated the probability of fewer than 18 not 18 or more.

CB
June 22nd 2010, 03:50 AM
Apprentice123

Not is P(X >= x) = (P X >= x-0,5) ???
June 22nd 2010, 04:08 AM
CaptainBlack

Quote:

Originally Posted by Apprentice123

Not is P(X >= x) = (P X >= x-0,5) ???

What does that mean? (I'm betting whatever you intend it to mean the answer is no)

$P(X\ge x)=1-P(X<x)$

CB

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