Suppose that 15% of people dont show up for a flight, and suppose that their decisions are independent. how many tickets can you sell for a plane with 144 seats and be 99% sure that not too many people will show up.
the book says to do this by using the normal distribution function and that the answer is selling 157 tickets.

157 tickets

Explanation:

The people not showing up for the flight can be treated as from Binomial distribution.

The binomial distribution B(n, p) is approximately close to the normal i.e. N(np, np(1 − p))  for large 'n' and for 'p' and neither too close to 0 nor 1 .

Now, Let us assume 'n' = n

and we are given

p=0.15

So now B(n,0.15n) follows Normal distribution

u=n

= 0.15n

We have to calculate P(X<144) with 99% accuracy

P(X<144) = P(Z<z)

where;

z=

z score for 99% is 2.33

i.e.  

solving this we will get one root nearly equal to 157 and other root as 133

Hence the answer is 157.

domain is all the x values included in the

0 ≤ x ≤ 2

range is all the y values of the graph

-2 ≤ y ≤ 2

***keep in mind that there are different ways to show domain and range

hope this !

~just a girl in love with shawn mendes

yes the total will be same because in both of them the expression.is rearranged but their sign remains the same hence it hss just been shuffled