Question 730800: Suppose 80% of all college students who own cell phones send and receive text messages. In a random sample of 10 cell-phone owners, what is the probability that A) 8 of these student send and receive text messages and B) at least 7 of these students send and receive text messages?
I know that p=.8, and that the n for A is 8. But I can't figure out the rest.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Both parts will use the binomial distribution probability formula, which is
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
for both parts A and B, n = 10 and p = 0.8
A)
For this part A) only, k = 8
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 8) = (10 C 8)*(0.8)^(8)*(1-0.8)^(10-8)
P(X = 8) = (10 C 8)*(0.8)^(8)*(0.2)^(10-8)
P(X = 8) = (45)*(0.8)^(8)*(0.2)^2
P(X = 8) = (45)*(0.167772)*(0.04)
P(X = 8) = 0.30199
So the probability that exactly 8 students sent and received text messages is roughly 0.30199
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B)
For this part, we need to find P(X = 7), P(X = 8), P(X = 9), and P(X = 10). We will add them up to calculate P(X >= 7)
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P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 7) = (10 C 7)*(0.8)^(7)*(1-0.8)^(10-7)
P(X = 7) = (10 C 7)*(0.8)^(7)*(0.2)^(10-7)
P(X = 7) = (120)*(0.8)^(7)*(0.2)^3
P(X = 7) = (120)*(0.209715)*(0.008)
P(X = 7) = 0.201327
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P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 8) = (10 C 8)*(0.8)^(8)*(1-0.8)^(10-8)
P(X = 8) = (10 C 8)*(0.8)^(8)*(0.2)^(10-8)
P(X = 8) = (45)*(0.8)^(8)*(0.2)^2
P(X = 8) = (45)*(0.167772)*(0.04)
P(X = 8) = 0.30199
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P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 9) = (10 C 9)*(0.8)^(9)*(1-0.8)^(10-9)
P(X = 9) = (10 C 9)*(0.8)^(9)*(0.2)^(10-9)
P(X = 9) = (10)*(0.8)^(9)*(0.2)^1
P(X = 9) = (10)*(0.134218)*(0.2)
P(X = 9) = 0.268435
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P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 10) = (10 C 10)*(0.8)^(10)*(1-0.8)^(10-10)
P(X = 10) = (10 C 10)*(0.8)^(10)*(0.2)^(10-10)
P(X = 10) = (1)*(0.8)^(10)*(0.2)^0
P(X = 10) = (1)*(0.107374)*(1)
P(X = 10) = 0.107374
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So in short, we found this
P(X = 7)= 0.201327
P(X = 8)= 0.30199
P(X = 9)= 0.268435
P(X = 10)= 0.107374
They add to...
P(X >= 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)
P(X >= 7) = 0.201327 + 0.30199 + 0.268435 + 0.107374
P(X >= 7) = 0.879126
So the probability that at least 7 of these students send and receive text messages is approximately 0.879126
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