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The light bulbs used to provide exterior lighting for a large office building have an average lifetime of 700 hours. If lifetime is approximately normally distributed with a standard deviation of 50 hours, how often should all the bulbs be replaced so that no more than \(20 \%\) of the bulbs will have already burned out?

Short Answer

Expert verified
To ensure that no more than \(20\%\) of the light bulbs have burned out, they should be replaced every 658 hours. This is determined using the given normal distribution of lifetime hours with a mean of 700 hours and a standard deviation of 50 hours, and finding the number of hours corresponding to a cumulative probability of \(20\%\).

Step by step solution

01

Identifying the normal distribution parameters

The problem states that the lifetime of the light bulbs is normally distributed with a mean (µ) of 700 hours and a standard deviation (σ) of 50 hours. Thus, we have the following parameters: Mean (µ) = 700 hours Standard deviation (σ) = 50 hours
02

Finding the z-score corresponding to 20% probability

We need to find the z-score corresponding to the \(20\%\) cumulative probability. You can either look this up in a standard normal distribution table, use a calculator with a built-in function, or use a software tool. The z-score corresponding to \(P(z) = 0.20\) is approximately \(-0.84\).
03

Applying z-score to the normal distribution formula

Now, we apply the z-score we found to the normal distribution formula: \(z = \frac{x - \mu}{\sigma}\) Where \(x\) is the number of hours after which no more than \(20\%\) of the bulbs will have burned out, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. Replace the formula with values: \(-0.84 = \frac{x - 700}{50}\)
04

Solving for x

Now solve for \(x\) using the equation from Step 3: \begin{align*} -0.84 \times 50 &= x - 700 \\ -42 &= x - 700 \\ x &= (-42) + 700 \\ x &= 658 \end{align*} So, \(x \approx 658\) hours.
05

Conclusion

To ensure that no more than \(20\%\) of the light bulbs have burned out, they should be replaced every 658 hours.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Distribution Parameters
Understanding the normal distribution parameters is crucial when dealing with problems related to the likelihood of certain events, such as the lifetime of light bulbs in our example. The normal distribution is a symmetrical, bell-shaped curve that is defined by two parameters: the mean \( \( \mu \) \) and the standard deviation \( \( \sigma \) \).

The mean or average \( \( \mu \) \) is the center of the distribution, where the highest point of the bell curve lies. It tells us the average value around which the data points are clustered. In our bulb lifetime example, this is 700 hours. The standard deviation \( \( \sigma \) \) indicates how much the individual data points deviate from the mean on average. A smaller standard deviation means that the data points are close to the mean, while a larger one signifies more spread. For the light bulbs, this is 50 hours, showing moderate spread around the mean lifetime.

With these parameters, we can describe the entire distribution of bulb lifetimes and predict probabilities for different lifespans, which enables us to make decisions on maintenance schedules.
Z-Score
The z-score is a statistical measure that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. If a z-score is 0, it indicates that the data point's score is identical to the mean score.

In the case of the light bulb example, finding the z-score corresponding to the 20% cumulative probability required us to look it up in a standard normal distribution table or use a calculator with a statistical function. The z-score of approximately \( -0.84 \) suggests that the time by which 20% of the light bulbs fail is 0.84 standard deviations below the mean lifetime of 700 hours.

This value is critical for finding the exact time to replace the bulbs before reaching the 20% failure threshold. By converting a z-score back into raw score using the mean and standard deviation, we get the practical information to make decisions—like when to replace the bulbs.
Cumulative Probability
Cumulative probability refers to the likelihood that a random variable is less than or equal to a specified value. It's a fundamental concept in statistics and probability, often visualized on a graph as the area under the curve to the left of a z-score on the normal distribution curve.

Interpreting the cumulative probability helps in determining how often certain events will occur. For instance, in our light bulb problem, we aimed to determine how long the bulbs should last so that no more than 20% of them would fail. Using cumulative probability, we translated this percentage into a z-score and ultimately into the number of hours after which the bulbs should be replaced.

Cumulative probabilities are not only theoretical but have real-world applications in business, engineering, and other fields, guiding decision-making processes based on statistical evidence. In the light bulb example, using this concept ensures efficient maintenance and cost savings by minimizing premature replacements while avoiding a high frequency of bulb outages.

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Most popular questions from this chapter

A pizza shop sells pizzas in four different sizes. The 1000 most recent orders for a single pizza resulted in the following proportions for the various sizes: $$ \begin{array}{lcccc} \text { Size } & 12 \text { in. } & 14 \text { in. } & 16 \text { in. } & 18 \text { in. } \\ \text { Proportion } & 0.20 & 0.25 & 0.50 & 0.05 \end{array} $$ With \(x=\) the size of a pizza in a single-pizza order, the given table is an approximation to the population distribution of \(x\). a. Write a few sentences describing what you would expect to see for pizza sizes over a long sequence of single-pizza orders. b. What is the approximate value of \(P(x<16)\) ? c. What is the approximate value of \(P(x \leq 16)\) ?

Suppose that your statistics professor tells you that the scores on a midterm exam were approximately normally distributed with a mean of 78 and a standard deviation of 7 . The top \(15 \%\) of all scores have been designated A's. Your score is \(89 .\) Did you earn an \(\mathrm{A}\) ? Explain.

The distribution of the number of items produced by an assembly line during an 8 -hour shift can be approximated by a normal distribution with mean value 150 and standard deviation 10 . a. What is the approximate probability that the number of items produced is at most \(120 ?\) b. What is the approximate probability that at least 125 items are produced? c. What is the approximate probability that between 135 and 160 (inclusive) items are produced?

A machine that cuts corks for wine bottles operates in such a way that the distribution of the diameter of the corks produced is well approximated by a normal distribution with mean \(3 \mathrm{~cm}\) and standard deviation \(0.1 \mathrm{~cm} .\) The specifications call for corks with diameters between 2.9 and \(3.1 \mathrm{~cm}\). A cork not meeting the specifications is considered defective. (A cork that is too small leaks and causes the wine to deteriorate; a cork that is too large doesn't fit in the bottle.) What proportion of corks produced by this machine are defective?

An appliance dealer sells three different models of freezers having 13.5,15.9 , and 19.1 cubic feet of storage space. Consider the random variable \(x=\) the amount of storage space purchased by the next customer to buy a freezer. Suppose that \(x\) has the following probability distribution: \(x\) \(\begin{array}{lll}13.5 & 15.9 & 19.1\end{array}\) \(p(x)\) \(\begin{array}{lll}0.2 & 0.5 & 0.3\end{array}\) a. Calculate the mean and standard deviation of \(x\). (Hint: See Example \(6.15 .\) ) b. Give an interpretation of the mean and standard deviation of \(x\) in the context of observing the outcomes of many purchases.

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