Business Mathematics_case_2

One fine Monday, Mr. Rajeev Khanna, Head of Operations Department at I.G. airport caught Mr. Shailesh Saxena, a manager of the Quality control department, at the reception counter. “Could you come in & see me for a few minutes? We have got a problem with our ISO:14000 certification and Shailesh, perhaps, you could give me some advice.”

“Certainly,” Mr. Saxsena smiled. “Anytime this afternoon would be fine.” Mr. Saxsena knew that the decibel level of planes during takeoff and landing was a source of concern in the Operations department since last 2 weeks.

Mr. Saxsena knocked the door and came in. Mr. Khanna expressed his views of checking the current decibel level of the noise created by the planes during takeoff. Mr. Saxsena informed Mr. Khanna that their department has already carried out a routine check and put forward the inspection report.

On studying the inspection report, it was found that the noise levels of jets during takeoff as they pass over a neighbourhood, at the mean noise level was 103 decibels and the standard deviation was 5.4 decibels.

Study Questions:
What proportion of these jets did Mr. Saxsena found to have a noise level of 95 decibels or less when takeoff and landing over that neighbourhood?
What was the chance that Mr. Saxsena gave to Mr. Khanna, of one jet having a noise level between 100 and 110 decibels?

Mr. Khanna would like to have 99% of all jets taking off and landing over the neighbourhood to have a noise level of 100 decibels or less. What would the mean noise level have to be in order to achieve this goal? Assume the standard deviation would remain the same.

When working with any problem that deals with the normal distribution, we first define what "x is?".
The variable x is always the data that is being described. We should also include the units of measurement for x, and the value of m and s, if they are known.

• The variable of interest is x.
x = the noise level of jets,
It is measured in decibels, as the planes take off and land over the neighbourhood.
The distribution of noise levels is normal with a mean of m = 103 and a standard deviation of s = 5.4

The value of x = 95 must first be transformed to a Z-score using the formula:
Z = (x - m)
s
= (95 - 103)
5.4
= - 1.48

The value of x = 95 must first be transformed to a Z-score using the formula:
Z = (x - m)
s
= (95 - 103)
5.4
= - 1.48

Diagram

The normal curve shown has z = -1.48 and the area from the normal table corresponding to this Z-score marked. The desired area corresponding to decibel levels less than 95 is shaded.

Diagram

Since the area under a normal curve on each side of the mean is 0.5, the proportion of jets with a decibel level less than 95 is
0.5000 - 0.4306
= 0.0694.

• Transforming the x-values of 100 to Z-scores yields
Z = (x - m)
s
= (100 - 103)
5.4
= - 0.56

Transforming the x-values of 110 to Z-scores yields
Z = (x - m)
s
= (110 - 103)
5.4
= 1.30

The normal distributions have the desired probability (area under the curve) shaded, and the important values appropriately marked. The areas corresponding to the calculated z-scores as found in the normal table are also shown.

Diagram

The shaded area is the total of the two values found in the normal table.
The probability one jet would have a noise level that is between 100 and 110 decibels is equal to
0.2123 + 0.4032 = 0.6155.

The figures below are used to find the value of Z for which approximately 99% of the normal curve is below.

Diagram


Substituting x = 100, z = 2.33, and s = 5.4 into the following formula and solving for m produces the desired value.
x = m + z s
100 = m + 2.33 (5.4)
m = 87.418
The mean decibel level would need to be lowered to about 87.4 decibels in order to have 99% of all jets taking off over this neighborhood to have a noise level of 100 or less.