approximations of distributions 2

• Jun 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

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

• Jun 21st 2010, 08:06 PM
TKHunny
Which side of 17,5 do you want? 1 - 0,587064 = 0,412936 That looks pretty close.
• Jun 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

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

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

CB
• Jun 22nd 2010, 03:50 AM
Apprentice123
Not is P(X >= x) = (P X >= x-0,5) ???
• Jun 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)

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

CB