Question 730163: 40% of house holds have adult children living at home. A random sample of 10 households were selected at random and asked if they had adult children living at home and the number of YES responses were recorded.
PLEASE HELP ME WITH THESE QUESTIONS.
(a) find the probability that exactly 5 households had adult children living at home.
(b) Find the probability that between 4 and 8 houses inclusive have adult children living at home.
(c) Find the probability that more than 7 households have adult children living at home.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! All 3 parts are dealing with the binomial distribution
In each case
n = 10
p = 0.4
a)
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 5) = (10 C 5)*(0.4)^(5)*(1-0.4)^(10-5)
P(X = 5) = (10 C 5)*(0.4)^(5)*(0.6)^(10-5)
P(X = 5) = (252)*(0.4)^(5)*(0.6)^5
P(X = 5) = (252)*(0.01024)*(0.07776)
P(X = 5) = 0.200658
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b)
Use the same formula to find P(X = 4), P(X = 5), ..., all the way up to P(X = 8)
The steps will be the same as in part a (for each individual probability). However, the value of k will change. Doing that will give you this
P(X = 4)= 0.250823
P(X = 5)= 0.200658
P(X = 6)= 0.111477
P(X = 7)= 0.042467
P(X = 8)= 0.010617
Add them up to get...
P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
0.250823 + 0.200658 + 0.111477 + 0.042467 + 0.010617
0.616042
So the answer to part b) is 0.616042
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c)
Same as part b), but now we want P(X = 8),..., up to P(X = 10).
P(X = 8)= 0.010617
P(X = 9)= 0.001573
P(X = 10)= 0.000105
P(X = 8) + P(X = 9) + P(X = 10)
0.010617 + 0.001573 + 0.000105
0.012295
So the answer to part c) is 0.012295
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