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جستجو

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جستجوهای پرتکرار

میتونی لایو بذاری!

a) State three conditions that must be satisfied for a situation to be modelled by a binomial distribution.

On any day, there is a probability of 0.3 that Julie’s train is late.

b) Nine days are chosen at random. Find the probability that Julie’s train is late on more than 7 days or fewer than 2 days.

c) 90 days are chosen at random. Find the probability that Julie’s train is late on more than 35 days or fewer than 27 days.

پاسخ تشریحی :
نمایش پاسخ

a) constant/given prob, independent trials, fixed/given no. of trials, only two outcomes

b) $P\left( {8,{\text{ }}9,{\text{ }}0,{\text{ }}1} \right) = $

${}^9{C_8}{\left( {0.3} \right)^8}\left( {0.7} \right) + {\left( {0.3} \right)^9} + {\left( {0.7} \right)^9} + {}^9{C_1}\left( {0.3} \right){\left( {0.7} \right)^8}$

$ = 0.196$

c) mean $ = 90 \times 0.3 = 27$

var $ = 18.9$

$P\left( {X > 35} \right) = 1 - \Phi \left( {\frac{{35.5 - 27}}{{\sqrt {18.9} }}} \right)$

$ = 1 - \Phi \left( {1.955} \right) = 0.0253$

${\text{P}}\left( {X < 27} \right) = \Phi \left( {\frac{{26.5 - 27}}{{\sqrt {18.9} }}} \right) = 1 - \Phi \left( {0.115} \right)$

$ = 0.4542$

Total prob $ = 0.480$ accept $0.48$

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