he transport layer on the destination host O reassembles the application message if it has been fragmented O removes the transpo
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Answer:
The correct Answer is 0.0571
Explanation:
53% of U.S. households have a PCs.
So, P(Having personal computer) = p = 0.53
Sample size(n) = 250
np(1-p) = 250 * 0.53 * (1 - 0.53) = 62.275 > 10
So, we can just estimate binomial distribution to normal distribution
Mean of proportion(p) = 0.53
Standard error of proportion(SE) = =
= 0.0316
For x = 120, sample proportion(p) = =
= 0.48
So, likelihood that fewer than 120 have a PC
= P(x < 120)
= P( p^ < 0.48 )
= P(z < ) (z=
)
= P(z < -1.58)
= 0.0571 ( From normal table )