Chapter 5, Problem 96CE

High levels of cockpit noise in an aircraft can damage the hearing of pilots who are exposed to this hazard for many hours. Cockpit noise in a jet aircraft is mostly due to airflow at hundreds of miles per hour. This 3 × 3 contingency table shows 61 observations of data collected by an airline pilot using a handheld sound meter in a certain aircraft cockpit. Noise level is defined as “low” (under 88 decibels), “medium” (88 to 91 decibels), or “high” (92 decibels or more). There are three flight phases (climb, cruise, descent). (a) Calculate each probability (i–vi) and explain in words what it means. (b) Do you see evidence that noise level depends on flight phase? Explain. (c) Where else might ambient noise be an ergonomic issue? (Hint: Search the web.)

1. i. P(B)
2. ii. P(L)
3. iii. P(H | C)
4. iv. P(H | D)
5. v. P(L and B)
6. vi. P(L and C)

To determine

Calculate each probability (i-vi) and explain it meaning in words.

Answer to Problem 96CE

i. The probability $P\left(B\right)$ is 0.4098 and the meaning is the probability of a climb is 0.4098.

ii. The probability $P\left(L\right)$ is 0.2295 and the meaning is the probability of a low noise level is 0.2295.

iii. The probability $P\left(H|C\right)$ is 0.3750 and the meaning is the probability of high noise given that the person is in the cruise phase is 0.3750.

iv. The probability $P\left(H|D\right)$ is 0.5 and the meaning is the probability of high noise given that the person is in the descent phase is 0.5.

v. The probability $P\left(L\text{ and }B\right)$ is 0.0984 and the meaning is the probability of low noise and climbing is 0.0984.

vi. The probability $P\left(L\text{ and }C\right)$ is 0.0328 and the meaning is the probability of low noise and cruising is 0.0328.

Explanation of Solution

Calculation:

The given table shows that cockpit noise levels in the jet aircraft.

The given contingency table is,

	Flight Phase
Noise Level	Climb (B)	Cruise (C)	Descent (D)	Row Total
Low (L)	6	2	6	14
Medium (M)	18	3	8	29
High (H)	1	3	14	18
Column Total	25	8	28	61

For (i)$P\left(B\right)$:

The formula for finding the probability $P\left(B\right)$ is,

$P\left(B\right)=\frac{\text{Frequency for the class }B}{\text{Total frequencies in the distribution}}$

Substitute 25 for ‘Frequency for the class B’ and 61 for ‘Total frequencies in the distribution’

$\begin{array}{c}P\left(B\right)=\frac{25}{61}\\ =0.4098\end{array}$

Therefore, the probability $P\left(B\right)$ is 0.4098 and the meaning is the probability of a climb is 0.4098.

For (ii) $P\left(L\right)$:

The formula for finding the probability $P\left(L\right)$ is,

$P\left(L\right)=\frac{\text{Frequency for the class }L}{\text{Total frequencies in the distribution}}$

Substitute 14 for ‘Frequency for the class L’ and 61 for ‘Total frequencies in the distribution’

$\begin{array}{c}P\left(L\right)=\frac{14}{61}\\ =0.2295\end{array}$

Therefore, the probability $P\left(L\right)$ is 0.2295 and the meaning is the probability of a low noise level is 0.2295.

For (iii) $P\left(H|C\right)$:

The formula for finding the probability $P\left(H|C\right)$ is,

$P\left(H|C\right)=\frac{\text{Frequency for the class }H\text{ and }C}{\text{Frequency for the class }C}$

Substitute 3 for ‘Frequency for the class H and C’ and 8 for ‘Frequency for the class C’,

$\begin{array}{c}P\left(H|C\right)=\frac{3}{8}\\ =0.3750\end{array}$

Therefore, the probability $P\left(H|C\right)$ is 0.3750 and the meaning is the probability of high noise given that the person is in the cruise phase is 0.3750.

For (iv) $P\left(H|D\right)$:

The formula for finding the probability $P\left(H|D\right)$ is,

$P\left(H|D\right)=\frac{\text{Frequency for the class }H\text{ and }D}{\text{Frequency for the class }D}$

Substitute 14 for ‘Frequency for the class H and D’ and 28 for ‘Frequency for the class D’,

$\begin{array}{c}P\left(H|D\right)=\frac{14}{28}\\ =0.5\end{array}$

Therefore, the probability $P\left(H|D\right)$ is 0.5 and the meaning is the probability of high noise given that the person is in the descent phase is 0.5.

For (v) $P\left(L\text{ and }B\right)$:

The formula for finding the probability $P\left(L\text{ and }B\right)$ is,

$P\left(L\text{ and }B\right)=\frac{\text{Frequency for the class }L\text{ and }B}{\text{Total frequencies in the distribution}}$

Substitute 6 for ‘Frequency for the class L and B’ and 61 for ‘Total frequencies in the distribution’,

$\begin{array}{c}P\left(L\text{ and }B\right)=\frac{6}{61}\\ =0.0984\end{array}$

Therefore, the probability $P\left(L\text{ and }B\right)$ is 0.0984 and the meaning is the probability of low noise and climbing is 0.0984.

For (vi) $P\left(L\text{ and }C\right)$:

The formula for finding the probability $P\left(L\text{ and }C\right)$ is,

$P\left(L\text{ and }C\right)=\frac{\text{Frequency for the class }L\text{ and }C}{\text{Total frequencies in the distribution}}$

Substitute 2 for ‘Frequency for the class L and C’ and 61 for ‘Total frequencies in the distribution’,

$\begin{array}{c}P\left(L\text{ and }C\right)=\frac{2}{61}\\ =0.0328\end{array}$

Therefore, the probability $P\left(L\text{ and }C\right)$ is 0.0328 and the meaning is the probability of low noise and cruising is 0.0328.

To determine

Check whether there is evidence that noise level depends on flight phase. Also, explain the reason.

Answer to Problem 96CE

Yes, there is evidence that noise level depends on flight phase because $P\left(H\text{ and }C\right)\ne P\left(H\right)P\left(C\right)$.

Explanation of Solution

Calculation:

Special law of multiplication:

If two events A and B are independent, then

$P\left(A\cap B\right)=P\left(A\right)P\left(B\right)$

Consider noise level as High and flight phase type as cruise.

The formula for checking independence is,

$\begin{array}{c}P\left(H\text{ and }C\right)=P\left(H\right)P\left(C\right)\\ \left(\frac{\text{Frequency for the class }H\text{ and }C}{\text{Total frequencies in the distribution}}\right)=\left(\begin{array}{l}\left(\frac{\text{Frequency for the class }H}{\text{Total frequencies in the distribution}}\right)\times \\ \left(\frac{\text{Frequency for the class }C}{\text{0Total frequencies in the distribution}}\right)\end{array}\right)\\ \frac{3}{61}=\frac{18}{61}\times \frac{8}{61}\\ 0.0492=\frac{144}{3,721}\end{array}$

                                                        $0.0492\ne 0.0387$

Here, it is observed that $P\left(H\text{ and }C\right)\ne P\left(H\right)P\left(C\right)$. Therefore, there is evidence that noise level depends on flight phase.

To determine

Identify the ambient noise is an ergonomic issue.

Explanation of Solution

Answers may vary: One of the answers is given below:

Some of the ambient noises are an ergonomic issues are,

• • The noise from the heavy machines running in a factory.
• • The excess sound created by vehicles in the traffic signals.