Answer:
x=0.87 or 86.67%
13 prescription claims out of 15 are paid
Step-by-step explanation:
First we organize the data:
prescription claims submitted: 4500
Prescriptions: 3900
The rate of claims paid is calculated using the following formula:
![x =\frac{Paid}{Total}](https://tex.z-dn.net/?f=x%20%3D%5Cfrac%7BPaid%7D%7BTotal%7D)
We know that:
Paid = 3900
Total = 4500
So
![x =\frac{3900}{4500}](https://tex.z-dn.net/?f=x%20%3D%5Cfrac%7B3900%7D%7B4500%7D)
![x =\frac{39}{45}](https://tex.z-dn.net/?f=x%20%3D%5Cfrac%7B39%7D%7B45%7D)
![x =\frac{13}{15}](https://tex.z-dn.net/?f=x%20%3D%5Cfrac%7B13%7D%7B15%7D)
x=0.87
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 .
So first of all you would add the two together
so it would be m23.
138÷23 would equal 6,
therefore m would equal six
to check it substitute m for 6 and you should get 138!