The Normal Distribution

NORMAL DISTRIBUTION

An airport is studying the noise levels of jets during takeoff as they pass over a neighborhood. They find that at the present time the mean noise level is 103 decibels and the standard deviation is 5.4 decibels. The distribution of noise levels for all jets during takeoff over this neighborhood has a normal distribution.

A. What proportion of these jets have a noise level of 95 decibels or less when taking off over this neighborhood?

B. What is the probability one jet would have a noise level that is between 100 and 110 decibels?

C. The airport would like to have 99% of all jets taking off over this neighborhood to have a noise level of 100 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.

SOLUTION

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

A. The value of x = 95 must first be transformed to a z-score using the formula.

The normal curves shown below have x = 95, z = -1.48, and the area from the normal table corresponding to this z-score marked. The desired area corresponding to decibels levels less than 95 is shaded.

Since the area under a normal curve on each side of the mean is .5, the proportion of jets with a decibel level less than 95 is .5000 - .4306 = .0694.

B. Transforming the x-values of 100 and 110 to z-scores yields

The normal distributions below 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.

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 .2123 + .4032 = .6155.

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

Substituting x = 100, z = 2.33, and s = 5.4 into the following formula and solving for m produces the desired value.

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.

COMMENTS ABOUT THE SOLUTION

When working any problem dealing with the normal distribution, first define in words what "x is". The variable x is always the data being described. Be sure to include the units of measurement for x, and the value of m and s if known.

Always sketch a normal curve to help find the solution to the problem. Shade desired areas under this curve, mark appropriate values of x, mark corresponding z-scores, and label areas from the normal table when they are found. Often more than one curve should be sketched when there is more than one part to the problem.

There are three different types of values in every "normal problem". First, there are x-values. For each value of x there is a corresponding z-score. For each z-score there is a corresponding area under the normal curve. The normal table contains areas between the mean, and the z-score. All of these three types of values are in one-to-one correspondence. It is extremely important to keep straight whether any value is an area, an x-value, or a z-score. A properly labeled normal curve is of invaluable help for doing this.

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