The normal approximation to the binomial

[TyErd]

Homework Statement

I've attached the question

Homework Equations

Pr(X<=x)= (x + 0.5 - n*p) / sqrt(n*p*(1-p))

The Attempt at a Solution

okay so n=1150, p=0.02 , Pr(X<23)

=23 + 0.5 - 1150(0.02) / sqrt(1150*0.02*0.98)
=0.105316

is that bit right so far. Because it is less than i thought x might've been 22 instead of 23 so when i did that it equaled -0.105316

im not sure what do after that. Am i suppose to find that value in the normal distribution tables or something?

 

[Ray Vickson]

Homework Statement

I've attached the question

Homework Equations

Pr(X<=x)= (x + 0.5 - n*p) / sqrt(n*p*(1-p))

The Attempt at a Solution

okay so n=1150, p=0.02 , Pr(X<23)

=23 + 0.5 - 1150(0.02) / sqrt(1150*0.02*0.98)
=0.105316

is that bit right so far. Because it is less than i thought x might've been 22 instead of 23 so when i did that it equaled -0.105316

im not sure what do after that. Am i suppose to find that value in the normal distribution tables or something?

Your first equation makes no sense. Perhaps you meant to write
[tex]P(X \leq x) \approx P\left( Z \leq \frac{x + 1/2 - n p)}{\sqrt{n p (1-p)}} \right)? [/tex]
For n = 1150, p = 0.02 and x = 23, we have [itex] P(Z \leq 0.105316),[/itex] which is similar to what you said---although I am not sure, from what you wrote, that you really understand this. Of course, the question said "less than 23", which means <= 22, so you ought to use x = 22 in the formula. At this point you need to find the probability value from normal tables, or a spreadsheet, or by pressing a button on some brands of hand-held calculators.

RGV