Answer:
The probability that the aircraft is overload = 0.9999
Yes , The pilot has to be take strict action .
Step-by-step explanation:
P.S - The exact question is -
Given - Before every flight, the pilot must verify that the total weight of the load is less than the maximum allowable load for the aircraft. The aircraft can carry 37 passengers, and a flight has fuel and baggage that allows for a total passenger load of 6,216 lb. The pilot sees that the plane is full and all passengers are men. The aircraft will be overloaded if the mean weight of the passengers is greater than 6216/37 = 168 lb. Assume that weight of men are normally distributed with a mean of 182.7 lb and a standard deviation of 39.6.
To find - What is the probability that the aircraft is overloaded ?
Should the pilot take any action to correct for an overloaded aircraft ?
Proof -
Given that,
Mean, μ = 182.7
Standard Deviation, σ = 39.6
Now,
Let X be the Weight of the men
Now,
Probability that the aircraft is loaded be
P(X > 168 ) = P(
)
= P( z >
)
= P( z > -0.371)
= 1 - P ( z ≤ -0.371 )
= 1 - P( z > 0.371)
= 1 - 0.00010363
= 0.9999
⇒P(X > 168) = 0.9999
As the probability of weight overload = 0.9999
So, The pilot has to be take strict action .
Answer:
0.68269
Step-by-step explanation:
When we are to find the z score for population where a random sample is picked, the z.score formula we use is
z = (x-μ)/Standard error, where
x is the raw score,
μ is the population mean
Standard error = σ/√n
σ is the population standard deviation
n = random number of samples
For : x = 38 minutes, μ = 40, σ = 10, n = 5
z = 38 - 40/10 /√25
= -2/10/5
= -2/2
= -1
Determining the probability value using z table
P(x = 38) = P(z = -1)
= 0.15866
For : x = 42 minutes, μ = 40, σ = 10, n = 25
z = 42 - 40/10 /√25
= 2/10/5
= 2/2
= 1
Determining the probability value using z table
P(x = 42) = P(z = 1)
= 0.84134
The probability that their average waiting time will be between 38 and 42 minutes is calculated as
P(-Z<x<Z)
= P(-1 < x < 1)
= P(z = 1) - P(z = -1)
= 0.84134 - 0.15866
= 0.68269
Therefore, the probability that their average waiting time will be between 38 and 42 minutes is 0.68269
Answer:
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Step-by-step explanation:
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