Chapter 6.4, Problem 1AC

Applying the Concepts 6–4

Mountain Climbing Safety

Assume one of your favorite activities is mountain climbing. When you go mountain climbing, you have several safety devices to keep you from falling. You notice that attached to one of your safety hooks is a reliability rating of 97%. You estimate that throughout the next year you will be using this device about 100 times. Answer the following questions.

1. Does a reliability rating of 97% mean that there is a 97% chance that the device will not fail any of the 100 times?

2. What is the probability of at least one failure?

3. What is the complement of this event?

4. Can this be considered a binomial experiment?

5. Can you use the binomial probability formula? Why or why not?

6. Find the probability of at least two failures.

7. Can you use a normal distribution to accurately approximate the binomial distribution? Explain why or why not.

8. Is correction for continuity needed?

9. How much safer would it be to use a second safety hook independent of the first?

See page 368 for the answers.

To determine

To check: If reliability of 97% means that there is a 97% chance that the device will not fail any of the 100 items.

Answer to Problem 1AC

No.

Explanation of Solution

Given info:

Reliability rate is 97%.

Number of times device used is 100.

Calculations:

The random variable X denotes the number of times device fails out of 100 times.

Since reliability rate 97% means that, on average the device will not fail for 97% of time. Therefore, reliability of 97% doesn’t means that there is a 97% chance that the device will not fail any of the 100 items.

To determine

The probability of at least one failure.

Answer to Problem 1AC

Probability of at least one failure is 0.95.

Explanation of Solution

Given info:

Reliability rate is 97%.

Number of times device used is 100.

Calculation:

The random variable X denotes the number of times device fails out of 100 times.

Therefore, probability of at least one failure is equal to 1 – Probability of zero failures.

$\begin{array}{c}\text{At least one failure }=1-{\left(\text{No failure}\right)}^{100}\\ =1-{\left(0.97\right)}^{100}\\ =1-\text{0}\text{.04755}\\ =\text{0}\text{.95}\end{array}$

Therefore, the probability of at least one failure is 0.95.

To determine

The complement of the event at least one failure.

Answer to Problem 1AC

Event of zero times failure of safety hook out of 100 trials.

Explanation of Solution

Given info:

Reliability rate is 97%.

Number of times device used is 100.

Justification:

The random variable X denotes the number of times device fails out of 100 times.

If A denotes the event that the safety hook fails at least once. Then the complement of the event is the event that the safety hook does not fails out of 100 times used.

To determine

The given experiment is binomial experiment or not.

Answer to Problem 1AC

Given experiment is binomial experiment.

Explanation of Solution

Given info:

Reliability rate is 97%.

Number of times device used is 100.

Justification:

The random variable X denotes the number of times device fails out of 100 times and there are two possible outcomes device working and device not working. Also, the reliability rate of safety hook in each time its use is constant i.e 0.97, therefore given experiment is binomial experiment of number of times the safety hooks fails out of 100 times with probability that safety hook fails in one trial is 0.03. So, given experiment is binomial experiment.

To determine

Whether the binomial formula can be used or not.

Answer to Problem 1AC

Yes, the binomial formula can be used.

Explanation of Solution

Justification:

The probability that the safety hook fails x number of times out 100 with probability that it fails in one trial being 0.03 is ${{\displaystyle C}}_x^{100}{(0.03)}^x{(1-0.03)}^{100-x}$. Thus, the binomial formula can be used but as the number of trial is 100, the calculation may be cumbersome.

To determine

The probability of at least two failures.

Answer to Problem 1AC

The probability of at least two failures is 0.81.

Explanation of Solution

Justification:

Calculation:

Probability of at least 2 failures is,

$\begin{array}{c}P\left(X\ge 2\right)=1-P\left(X=0\right)-P\left(X=1\right)\\ =1-C_0^{100}{(}^{0.03}{(}^1-C_1^{100}{(}^{0.03}{(}^1\\ =1-0.048-0.147\\ =0.81\end{array}$

Therefore, the probability of at least 2 failures of safety hook is 0.81.

To determine

Whether normal distribution can be used to accurate approximate the binomial distribution.

Answer to Problem 1AC

No, normal distribution cannot be used to accurate approximate the binomial distribution.

Explanation of Solution

Justification:

Conditions:

$\begin{array}{c}np=\left(100\right)\left(0.03\right)\\ =3\\ nq=\left(100\right)\left(0.97\right)\\ =97\end{array}$

Here, np is less than 5.

Hence, normal distribution cannot be used to accurately approximate the binomial distribution.

To determine

Correction for continuity is needed.

Answer to Problem 1AC

No, continuity correction is not needed.

Explanation of Solution

Justification:

From part (7), the normal approximation is not applied. Hence, the continuity correction is not required.

To determine

How much safer it would be to use a second safety hook independent of the first.

Answer to Problem 1AC

Probability of failure after second hook is 0.0009.

Explanation of Solution

Given Info:

Reliability rate is 97%.

Number of times device used is 100.

Second safety hook is used independent of first safety hook.

Calculation:

Reliability rate of first safety hook is 97%.

Thus,

$\begin{array}{c}\text{Probability of failure if first safety hook}=1-\frac{\text{reliability}\text{rate}}{100}\\ =1-0.97\\ =0.03\end{array}$

Since, second safety hook is used independent of first safety hook, Therefore probability of failure if both safety hooks are used is

$\begin{array}{c}\text{(Probability}\text{that}\text{first}\text{hook}\text{fails)(Probability}\text{that}\text{second}\text{hook}\text{fails)}=\text{(0}\text{.03)(0}\text{.03)}\\ =0.0009\end{array}$

Therefore, probability of failure if both the safety hooks are used independently is 0.0009.