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Statistics final review- please help me learn how to do this problem?
A certain flight arrives on time 76 percent of the timre. Suppose 117 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that
A exactly 98 flights are on time
B at least 98 flights are on time
C fewer than 75 flights are on time
D between 75 and 101 flights are on time. The answers are .0122, .0314, .0009, .9958. I just need to know how they got them. Thanks!!!
@Bee- could you slow it down in the beginning? i dont see how you got 4.62 thanks1
@Bee- could you slow it down in the beginning? i dont see how you got 4.62 thanks1
@Bee- could you slow it down in the beginning? i dont see how you got 4.62 thanks1
@Bee- could you slow it down in the beginning? i dont see how you got 4.62 thanks1
2 Answers
- BeeFreeLv 78 years agoFavorite Answer
negative
Here is a link to the table of "continuity corrections" . You need to "memorize" these for your test ...
http://people.richland.edu/james/lecture/m170/ch07...
p = 0.76
mean = np = 117(0.76) = 88.92
s.d. = sqrt(npq) = sqrt[(117)(0.76)(1-0.76)] = 4.62
Now use the Normal Approximation with continuity corrections ...
A) P(X = 98) = P(97.5 < X < 98.5) ... now convert to z-values:
P[(97.5 - 88.92) / (4.62) < z < (98.5 - 88.92) / (4.62)]
P( 1.86 < z < 2.07) = .0122
Use the same method and the table of "continuity corrections" to solve B, C and D.
Hope that helps
- ?Lv 45 years ago
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