7. Solar-heat installations successfully reduce the utility bill 60% of the time. What is the probability that at least 9 out of 10 solar-heat installations are successful and will reduce the utility bill? a. 0.0464 b. 0.9432 c. 0.0403 d. 0.8429
@jim_thompson5910
I used a binomial distr. on this problem
yes you'll need to use a binomial cdf
i did and I got 0.9987 but the solution key says it's a)
how did you get 0.9987 ?
oops wrong problem.. I mean 0.09473
i use p = 0.60 and y = 9 and n = 10
which calculator did you use?
I used the 89 and yes I know about nCr(10,9)*p^y*q^(n-y)
That is what I used..
oh shoot I did it over again and got c.
that's the probability of getting exactly 9 out of 10 you aren't considering 8 out of 10, 7 out of 10, etc and then adding them up
sorry I meant to say just 9 out of 10 and 10 out of 10
can I still use the nCr command?
probability of getting exactly 9 out of 10 = 0.0403 probability of getting exactly 10 out of 10 = x the answer will be equal to `0.0403 + x`
oh okay i see
got it 0.04636
so we are looking at the ones that failed?
`so we are looking at the ones that failed? ` what do you mean?
you should get 0.0463574016 which rounds to 0.04636
I am sorry I got that but i rounded. nvm about my statement above it doesn't make sense. why do we look at 9 out of 10 and 10 out of 10?
`at least 9 out of 10` = (exactly 9 out of 10) + (exactly 10 out of 10)
saying "at least 9" means we have 9 or more so we have 9 or we have 10. We can't have any more than 10 because n = 10 is the sample size
I see so P(Y>=9)
correct
\(\Large P(Y \ge 9)\) is the same as \(\Large P(Y = 9) + P(Y = 10)\)
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