Question

According to an​ airline, flights on a certain route are on time 80​% of the time. Suppose 24 flights are randomly selected and the number of​ on-time flights is recorded. ​(a) Explain why this is a binomial experiment. ​(b) Find and interpret the probability that exactly 16 flights are on time. ​(c) Find and interpret the probability that fewer than 16 flights are on time. ​(d) Find and interpret the probability that at least 16 flights are on time. ​(e) Find and interpret the probability that between 14 and 16 ​flights, inclusive, are on time.

Answer 1

Solution-:

Given:

n = 24, p = 0.80

Let, X- be the number of on-time flights is recored among the 24 flights

(a) This situation is siutable for binomial distribution with parameters n=24 and p=0.80

$\therefore X\rightarrow B(n=24,p=0.80)$  

The p.m.f. of X is given by,

24 P(X = z) = *p"*(1 - p)24-- C

I= 0, 1, 2, .24

:0 <p< 1;q = 1 — р

  $=0$

: Other wise

We find all probability in following table:

	n=24	p=0.80	q=0.20
x	nCx	p^x	q^(24-x)	P[X=x]
0	1	1	1.67772E-17	0.0000
1	24	0.8	8.38861E-17	0.0000
2	276	0.64	4.1943E-16	0.0000
3	2024	0.512	2.09715E-15	0.0000
4	10626	0.4096	1.04858E-14	0.0000
5	42504	0.32768	5.24288E-14	0.0000
6	134596	0.262144	2.62144E-13	0.0000
7	346104	0.209715	1.31072E-12	0.0000
8	735471	0.167772	6.5536E-12	0.0000
9	1307504	0.134218	3.2768E-11	0.0000
10	1961256	0.107374	1.6384E-10	0.0000
11	2496144	0.085899	8.192E-10	0.0002
12	2704156	0.068719	4.096E-09	0.0008
13	2496144	0.054976	2.048E-08	0.0028
14	1961256	0.04398	1.024E-07	0.0088
15	1307504	0.035184	0.000000512	0.0236
16	735471	0.028147	0.00000256	0.0530
17	346104	0.022518	0.0000128	0.0998
18	134596	0.018014	0.000064	0.1552
19	42504	0.014412	0.00032	0.1960
20	10626	0.011529	0.0016	0.1960
21	2024	0.009223	0.008	0.1493
22	276	0.007379	0.04	0.0815
23	24	0.005903	0.2	0.0283
24	1	0.004722	1	0.0047

(b) We find, P[exactly 16 flights are on time]

$=P[X=16]$

$=0.0530$ (Using above table)

Therefore, the required probability is 0.0530.

This is interpreted as about a chance is exactly 16 flights out of 24 being on time.

(c) P[that fewer than 16 flights are on time]

  $=P[X<16]$

$=P[X\leq 15]$

$=P[X=0]+P[X=1]+P[X=2]+P[X=3]+P[X=4]+P[X=5]+P[X=6]+P[X=7]+P[X=8]+P[X=9]+P[X=10]+P[X=11]+P[X=12]+P[X=13]+P[X=14]+P[X=15]$

  $=0+0+0+0+0+0+0+0+0+0+0+0.0002+0.0008+0.0028+0.0088+0.0236$

(Using above table)

$=0.0362$

Therefore, the required probability is 0.0362.

This is interpreted as about a chance of fewer than 16 flights out of 24 being on time.

(d) P[at least 16 flights are on time]

$=P[X\geq 16]$

  $=1-\{P[X=0]+P[X=1]+P[X=2]+P[X=3]+P[X=4]+P[X=5]+P[X=6]+P[X=7]+P[X=8]+P[X=9]+P[X=10]+P[X=11]+P[X=12]+P[X=13]+P[X=14]+P[X=15]\}$

$=1-0.0362$ (using (b) )

  $=0.9638$

Therefore, the required probability is 0.9638.

This is interpreted as about a chance at least 16 flights out of 24 being on time.

(e) P[14 and 16 ​flights are on time]

  $=P[14\leq X\leq 16]$

  $=P[X=14]+P[X=15]+P[X=16]$

$=0.0088+0.0236+0.0530$

  $=0.0854$

This can be interpreted as about a $8.54\%\approx 9\%$ % chance that between 14 and 16 (inclusive of both 14 and 16) flights are on time.