Consider the probability that greater than 99 out of 158 flights will be on-time. Assume the probability that a given flight will be on-time is 64%. Approximate the probability using the normal distribution. Round your answer to four decimal places

Answer:0.1514Step-by-step explanation: Let n be the no of flights on time =145Let  p be the probability that a flight be on time=0.64Check 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]