Answer :
Answer:0.1514
Step-by-step explanation:
Let n be the no of flights on time =145
Let p be the probability that a flight be on time=0.64
Check the condition for normal approximation to the binomial distribution
[tex]np\geq 10[/tex]
[tex]92.8\geq 10[/tex]
Also
[tex]n(1-p)\geq 10[/tex]
[tex]52.2\geq 10[/tex]
Both the conditions satisfied ,so the normal distribution can be used to approximate probability
[tex]mean\mu =92.8[/tex]
Standard deviation[tex]\left ( \sigma \right )=\sqrt{np\left ( 1-p\right )}[/tex]
[tex]\sigma =5.779[/tex]
[tex]P\left ( X>99\right )=1-P\left ( x\leqslant 99\right )[/tex]
[tex]P\left ( X>99\right )=1-P\left ( \frac{x-\mu }{\sigma }\leq \frac{99-92.8}{5.779}\right )[/tex]
[tex]P\left ( X>99\right )=1-P\left [ 1.072\right ][/tex]
[tex]P=1-08485\approx 0.1514[/tex]