Question 1175558: The probability that a passenger’s bag will be mishandled on a U.S. airline is .0046. During spring break, suppose that 500 students fly from Minnesota to various southern destinations.
(a) What is the expected number of mishandled bags? [2]
(b) What is the approximate probability of no mishandled bags? What is the probability of more than two mishandled bags? [4]
(c) Would you expect the approximation to be accurate (cite a rule of thumb)?
Found 2 solutions by Boreal, ewatrrr:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! E(X)=np=500*0.0046=2.3
Normal approximation would not be too accurate given thet np is not > 10. Also, normality will not apply to this skewed distribution.
V(X)=np(1-p)=2.3*0.9954=2.289
sd=sqrt(V)=1.51
for no mishandled bags (approximation) it is z=(0-2.289)/1.51 or -1.51 or probability 0.0668. Note: this is z < 0 strictly speaking, but < 0 doesn't exist for lost bags, so the approximation for normality will break down here.
exact value is 0.9954^50=0.0997
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more than 2 approximation is z>(2.5-2.3)/1.51 using the continuity correction factor or z>0.13
that probability is 0.4483
exact value: we know 0, and 1 would be 500*0.0046*0.9954^499=0.2304
2 would be 500C2*0.0046^2*0.9954^498=0.2657
those three probabilities add to 0.5958
The complement, and the answer, is 1-0.5958 or 0.4042
Answer by ewatrrr(24785)  (Show Source):
You can put this solution on YOUR website!
Hi
Binomial Distribution: p = .0046 q = .9954 and n = 500
a) expected number of mishandled bags = .0046*500 = 2.3 0r 2
b) P(x = 0) = binompdf(0, 500, .0046) = .0997
P(x > 2) = 1 - P(x≤ 2) = 1 - binomcdf(500, .0046, 2) = 1-.5959 = .4041
c) The general rule of thumb to use normal approximation to binomial distribution
is that the sample size n is sufficiently large... if np ≥ 5.
Here: np < 5
Wish You the Best in your Studies.
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