This question is not complete, the complete question is;
When confronted with an in-flight medical emergency, pilots and crew can consult staff physicians at a global response center located in Arizona. If the global response center is called, there is a 4.8 percent chance that the flight will be diverted for an immediate landing. Suppose the response center is called 8,465 times in a given year.
a) What is the expected number of diversions
b) What is the probability of at least 400 diversions
c) What is the probability of fewer than 450 diversions?
Answer:
a) expected number of diversions μ is 406.32
b) the probability of at least 400 diversions is 0.6368
c) the probability of fewer than 450 diversions is 0.9861
Explanation:
Given the data in the question;
n = 8465
π = 4.8% = 0.048
so expected number of diversions μ = nπ = 8465 × 0.048 = 406.32
and
σ = √(π(1 - π ) = √(8465 × 0.048(1 - 0.048 ) = √( 406.32 × 0.952 )
σ = √386.81664
σ = 19.6677
b) the probability of at least 400 diversions
P(at least 400 diversion ) = P( X ≥ 400 )
= P( X ≥ 399.5 )
so
z = (X - μ) / σ
we substitute
z = (399.5 - 406.32 ) / 19.6677
z = -6.82 / 19.6677
z = -0.35
so
P(Z > -0.35 ) = 0.500 + p( -0.35 ≤ z ≤ 0 )
= 0.5000 + p( 0 ≤ z ≤ 0.35 )
= 0.5000 + 0.1368
= 0.6368
Therefore, the probability of at least 400 diversions is 0.6368
c) the probability of fewer than 450 diversions
P(fewer than 450 diversion ) = P( X ≤ 450 )
= P( X ≤ 449.5 )
so
z = (X - μ) / σ
we substitute
z = (449.5 - 406.32 ) / 19.6677
z = 43.18 / 19.6677
z = 2.20
so
P(Z ≤ 2.20 ) = 0.500 + p( 0 ≤ z ≤ 2.20 )
= 0.5000 + 0.4861
= 0.9861
Therefore, the probability of fewer than 450 diversions is 0.9861