Question

According to government data, 15% of employed women have never been married. Take a random sample of employed women and observe their marital status.
a) is X normal, binomial, or geometric? I think it's binomial
b) if 8 employed women are selected at random, what is the probability that 2 or fewer have never been married?
c) if 8 employed women are selected at random, what are the mean and standard deviation? mean = 1.2, SD = 1.0099

Answer 1

@jim_thompson5910 how does this look so far?
and how do I get b? Like in the last problem?

Answer 2

yep it's binomial because

a) each trial is independent
b) each trial has a yes or no question (in this case, married or not married)
c) the probability of each trial remains the same

Answer 3

Alright awesome!

Answer 4

b) if 8 employed women are selected at random, what is the probability that 2 or fewer have never been married?

in this case, Numtrial = 8 and x = 2. At the top, we're given p = 0.15

Plugging all that in will give you P(X <= 2) which is what they're asking for

You won't have to subtract from 1 in this case

Answer 5

If they wanted "more than 2", ie you to calculate P( X > 2 ), then you would subtract from 1

Answer 6

Oh okay.
Is 0.89478722 correct?

Answer 7

very good

Answer 8

Yeah!

Answer 9

as for part c)

the mean = 1.2, that's correct

but SD = 1.0099 is a bit off since it's really SD = 1.0099504938362 and if you round that to 4 decimal places, then you get SD = 1.0100

of course, you could round to 5 decimal places to get SD = 1.00995

Answer 10

Oh okay. Thanks :)

Answer 11

you're welcome

Answer 12

I have one final question.

Answer 13

what's that