Question

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 1

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

$np\geq 10$

$92.8\geq 10$

Also

$n(1-p)\geq 10$

$52.2\geq 10$

Both the conditions satisfied ,so the normal distribution can be used to approximate probability

$mean\mu =92.8$

Standard deviation$\left ( \sigma \right )=\sqrt{np\left ( 1-p\right )}$

$\sigma =5.779$

$P\left ( X>99\right )=1-P\left ( x\leqslant 99\right )$

$P\left ( X>99\right )=1-P\left ( \frac{x-\mu }{\sigma }\leq \frac{99-92.8}{5.779}\right )$

$P\left ( X>99\right )=1-P\left [ 1.072\right ]$

$P=1-08485\approx 0.1514$