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A governments department of transportation reported that in 2009, airline Aled al domestic airlines in on-time arrivals for
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Answer #1

Probability that the all domestic airlines will be in time = 0.84

this is a binomial distribution.

a) Probability that in the next 6 flights, exactly 4 will be on time,

\small P(X = 4)

\small = \binom{6}{4}*0.84^{4}*(1-0.84)^{6-4} = 0.1912

b) Probability that in the next 6 flights, two or fewer will be on time,

\small P(X \leq 2)

  \small =P(X=0)+P(X=1)+P(X=2)

\small = \binom{6}{0}*0.84^{0}*(1-0.84)^{6-0} +\binom{6}{0}*0.84^{1}*(1-0.84)^{6-1}+\binom{6}{2}*0.84^{2}*(1-0.84)^{6-2}

= 0.0075

c) Probability that in the next 6 flights, at least 4 flights will be on time,

  \small P(X \geq 4)

\small =P(X=4)+P(X=5)+P(X=6)

\small = \binom{6}{4}*0.84^{4}*(1-0.84)^{6-4} +\binom{6}{5}*0.84^{5}*(1-0.84)^{6-5}+\binom{6}{6}*0.84^{6}*(1-0.84)^{6-6}

= 0.9440

d) Mean and standard deviation of the dstribution,

Mean = n*p = 6*0.84 = 5.04

Variance = n*p*(1-p) = 6*0.84*(1-0.84) = 0.8064

Standard deviation = \tiny \sqrt{0.8064} = 0.8980

e) Histogram of the distribution,

  1596217066364_image.png

****If you have any queries or doubts please comment below. if you're satisfied please give a like. Thank you!

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