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Probability Normal Approximation to Binomial Distribution(pls help)?
It is claimed that 80% of the people in the United States watched the first moon-landing on TV. If this claim is true, what is the approximate probability that In a sample of 100 polled at random, the number of people who saw the landing will be
a) less than 70?
b) greater than 86?
c) between 76 and 86, inclusive?
1 Answer
- ?Lv 61 decade agoFavorite Answer
Let X be the number of people out of 100 who saw the landing. Using the Normal Distribution as an approximation,
X ~ Binomial distribution with:
n = 100
p = 0.8
mean = np = 80
variance = npq = 80 * (1-p) = 80 * .2 = 16
std dev = √variance = 4
Using the Z-table at Ref 1, the probability that P(X < 70)
= P(Z < (70 - 80) / 4)
= P(Z < -2.5)
= 1 - P(Z >= 2.5)
= 1 - .9938
= 0.0062 <<=====ANSWER for a)
The probability that P(X > 86)
= P(Z > (86 - 80) / 4)
= P(Z > 1.5)
= 1 - P(Z <= 1.5)
= 0.0668 <<=====ANSWER for b)
The probability that P(76 ≤ X ≤ 86)
= P( Z ≤ (86 - 80) / 4) - P(Z < (76 - 80)/4)
= P( Z ≤ 1.5) - P(Z < -1)
= 0.9332 - .3413
= 0.5919 <<=====ANSWER for c)
.
Source(s): 1. http://en.wikipedia.org/wiki/Standard_normal_table...
