Mean = 10 and S.D = 0.22

P ( X >= 5 ) = P ( Z >= 5-10/0.22)

therefore = P ( Z >= -0.23 )

now you need to change that (-0.23)

you do 1 - P ( Z <= 0.23) >>>0.23 in the table look for P which is 0.5910

now do 1 - 0.5910 = 0.409 is the answer !

Hope it is clear .

Reply 5

Robbie242

Original post by charly2012

Mean = 10 and S.D = 0.22

P ( X >= 5 ) = P ( Z >= 5-10/0.22)

therefore = P ( Z >= -0.23 )

now you need to change that (-0.23)

you do 1 - P ( Z <= 0.23) >>>0.23 in the table look for P which is 0.5910

now do 1 - 0.5910 = 0.409 is the answer !

Hope it is clear .

How did you come to the conclusion that mean = 10? n=10, n is not the mean. It's Binomial distribution, not normal distribution, those are binomial parameters, not normal parameters.
If a random variable X follows a binomial distribution then X~B(n,p)

(edited 11 years ago)

Reply 6

charly2012

1

oh my bad . i thought it was normal distribution

Reply 7

.raiden.

OP

10

Original post by Robbie242

1-P(x\leq 5)=1-P(x=0,1,2,3,5,)

although this is long it does work I'm sure there's a faster method I'm missing.

I was thinking of this too, really was hoping there was an easier way or maybe its just a typical stupid textbook question lol thanks

Reply 8

.raiden.

OP

10

Original post by charly2012

oh my bad . i thought it was normal distribution

i could actually use normal distribution by normal approximation, but since this is hypothesis testing it would take very long to test other values :frown: