Question

According to the Bureau of Transportation​ Statistics, 81.9​% of American Airlines flights were on time in 2017. Assume this percentage still holds true for American Airlines. For the next

44 flights from American​ Airlines, use the normal approximation to the binomial distribution to complete parts a through d.

a.) Determine the probability that fewer than 34 flights will arrive on time.

b.) Determine the probability that exactly 32 flights will arrive on time.

c.) Determine the probability that 25​, 26​, 27​, or 28 flights will arrive on time.

d.) Determine the probability that 28​, 29​, 30​, 31​, or 32 flights will arrive on time.

Answer 1

(a)

n = 44

p = 0.819

q = 1 - p = 0.181

To find P(X<34):

Applying Continuity Correction, we get:

To find P(X<33.5):

Z = (33.5 - 36.036)/2.5539

= - 0.9930

Table of Area Under Standard Normal Curve gives area = 0.3389

So,

P(X<34) = 0.5 - 0.3389 = 0.1611

So,

Answer is:

0.1611

(b)

To find P(X=32):

Applying Continuity Correction, we get:

To find P(31.5<X<32.5):

Case 1: For X from 31.5 to mid value

Z = (31.5 - 36.036)/2.5539

= - 1.7761

Table of Area Under Standard Normal Curve gives area = 0.4625

Case 2: For X from 32.5 to mid value

Z = (32.5 - 36.036)/2.5539

= - 1.3845

Table of Area Under Standard Normal Curve gives area = 0.4162

So,

P(X=32) = 0.4625 - 0.4162 = 0.0463

So,

Answer is:

0.0463

(c)

To find P(24<X=29):

Applying Continuity Correction, we get:

To find P(23.5<X<29.5):

Case 1: For X from 23.5 to mid value

Z = (23.5 - 36.036)/2.5539

= - 4.9086

Table of Area Under Standard Normal Curve gives area = 0.5 nearly

Case 2: For X from 29.5 to mid value

Z = (29.5 - 36.036)/2.5539

= - 2.5592

Table of Area Under Standard Normal Curve gives area = 0.4948

So,

P(24<X=29) = 0.5 - 0.4948 = 0.0052

So,

Answer is:

0.0052

(d)

To find P(27<X=33):

Applying Continuity Correction, we get:

To find P(26.5<X<33.5):

Case 1: For X from 26.5 to mid value

Z = (26.5 - 36.036)/2.5539

= - 3.7339

Table of Area Under Standard Normal Curve gives area = 0.5 nearly

Case 2: For X from 33.5 to mid value

Z = (33.5 - 36.036)/2.5539

= - 0.9930

Table of Area Under Standard Normal Curve gives area = 0.3389

So,

P(27<X=33) = 0.5 - 0.3389 = 0.1611

So,

Answer is:

0.1611