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The lightbulbs used to provide exterior lighting for a large office building have an average lifetime of \(700 \mathrm{hr}\). If length of life is approximately normally distributed with a standard deviation of \(50 \mathrm{hr}\), how often should all the bulbs be replaced so that no more than \(20 \%\) of the bulbs will have already burned out?7.122 The lightbulbs used to provide exterior lighting for a large office building have an average lifetime of \(700 \mathrm{hr}\). If length of life is approximately normally distributed with a standard deviation of \(50 \mathrm{hr}\), 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
The light bulbs should be replaced approximately every 658 hours in order to ensure that no more than 20% have burned out.

Step by step solution

01

Identify the given values

In this case, the given values are the average or mean (\(\mu\)) which is 700 hours, the standard deviation (\(\sigma\)) which is 50 hours, and the percentile required (P) which is \(20% = 0.20\). Thus, we know that \(\mu = 700\), \(\sigma = 50\), and \(P = 0.20\).
02

Use the normal distribution table or Z-Score formula to find the corresponding Z-Score

The value that corresponds to \(P = 0.20\) can be found in a standard normal distribution or Z-Score table, which gives a Z-Score of approximately -0.84.
03

Use the Z-Score formula to find the corresponding value

The Z-Score formula is represented as \(Z = \frac{(X - \mu)}{\sigma}\), where X represents the values of the distribution. To find the X value, we rearrange the formula and substitute the known values: \(X = Z \cdot \sigma + \mu = -0.84 \cdot 50 + 700 = 658\).

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Key Concepts

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

Z-Score
The Z-Score is a statistical measure that tells you how many standard deviations a given data point is from the mean of the distribution. The Z-Score formula is represented as \(Z = \frac{(X - \mu)}{\sigma}\), where \(X\) is a value from the distribution, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

For our lightbulb example, the Z-Score helps us determine the lifespan at which no more than 20% of the lightbulbs are expected to fail. By using the Z-Score, we can calculate a specific hour value (\(X\)) around which to plan our replacement schedule. Essentially, the Z-Score provides a way to standardize different data points for comparison against a normal distribution.
Standard Deviation
Standard deviation, denoted by \(\sigma\), measures the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean (average), while a high standard deviation indicates that the values are spread out over a wider range.

In the case of our lightbulbs, a standard deviation of \(50\) hours tells us that the lifetimes of the bulbs are typically within \(50\) hours above or below the average lifetime of \(700\) hours. Understanding standard deviation is crucial for making predictions about a data set, as it allows us to assess the reliability and variability of the measurements.
Percentiles
Percentiles are used in statistics to give you a value below which a given percentage of data points in a dataset falls. The 20th percentile, for example, is the value below which 20% of the observations may be found.

In context with the lightbulb problem, we're interested in the lifespan that corresponds to the 20th percentile. This implies replacing the lightbulbs before 20% of them fail, ensuring continued lighting performance. Percentiles are a key concept in managing quality control and setting benchmarks in various industries.
Normal Distribution Table
A normal distribution table, commonly known as the 'Z-table', lists the cumulative probabilities associated with Z-Scores and is used to find the percentage of data that falls within certain Z-Scores in a standard normal distribution. Since the normal distribution is symmetrical, the table provides probabilities for the middle to the extreme end (from 0 to positive/negative infinity).

By using the Z-Score in step 2, the table helped us identify that approximately 20% of the bulbs have lifetimes shorter than the calculated value. This table is an essential tool for statisticians and researchers working with normally distributed data. In our exercise, the normal distribution table directly informed the replacement schedule for the lightbulbs.

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Most popular questions from this chapter

In a study of warp breakage during the weaving of fabric (Technometrics [1982]: 63), 100 pieces of yarn were tested. The number of cycles of strain to breakage was recorded for each yarn sample. The resulting data are given in the following table: $$ \begin{array}{rrrrrrrrrr} 86 & 146 & 251 & 653 & 98 & 249 & 400 & 292 & 131 & 176 \\ 76 & 264 & 15 & 364 & 195 & 262 & 88 & 264 & 42 & 321 \\ 180 & 198 & 38 & 20 & 61 & 121 & 282 & 180 & 325 & 250 \\ 196 & 90 & 229 & 166 & 38 & 337 & 341 & 40 & 40 & 135 \\ 597 & 246 & 211 & 180 & 93 & 571 & 124 & 279 & 81 & 186 \\ 497 & 182 & 423 & 185 & 338 & 290 & 398 & 71 & 246 & 185 \\ 188 & 568 & 55 & 244 & 20 & 284 & 93 & 396 & 203 & 829 \\ 239 & 236 & 277 & 143 & 198 & 264 & 105 & 203 & 124 & 137 \\ 135 & 169 & 157 & 224 & 65 & 315 & 229 & 55 & 286 & 350 \\ 193 & 175 & 220 & 149 & 151 & 353 & 400 & 61 & 194 & 188 \end{array} $$ a. Construct a frequency distribution using the class intervals 0 to \(<100,100\) to \(<200\), and so on. b. Draw the histogram corresponding to the frequency distribution in Part (a). How would you describe the shape of this histogram? c. Find a transformation for these data that results in a more symmetric histogram than what you obtained in Part (b).

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 receive an A? Explain.

Twenty-five percent of the customers entering a grocery store between 5 P.M. and 7 P.M. use an express checkout. Consider five randomly selected customers, and let \(x\) denote the number among the five who use the express checkout. a. What is \(p(2)\), that is, \(P(x=2)\) ? b. What is \(P(x \leq 1)\) ? c. What is \(P(2 \leq x)\) ? (Hint: Make use of your computation in Part (b).) d. What is \(P(x \neq 2) ?\)

The amount of time spent by a statistical consultant with a client at their first meeting is a random variable having a normal distribution with a mean value of \(60 \mathrm{~min}\) and a standard deviation of \(10 \mathrm{~min}\). a. What is the probability that more than \(45 \mathrm{~min}\) is spent at the first meeting? b. What amount of time is exceeded by only \(10 \%\) of all clients at a first meeting? c. If the consultant assesses a fixed charge of \(\$ 10\) (for overhead) and then charges \(\$ 50\) per hour, what is the mean revenue from a client's first meeting?

Let \(x\) denote the amount of gravel sold (in tons) during a randomly selected week at a particular sales facility. Suppose that the density curve has height \(f(x)\) above the value \(x\), where $$ f(x)=\left\\{\begin{array}{ll} 2(1-x) & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array}\right. $$ The density curve (the graph of \(f(x)\) ) is shown in the following figure: Use the fact that the area of a triangle \(=\frac{1}{2}\) (base)(height) to calculate each of the following probabilities: a. \(P\left(x<\frac{1}{2}\right)\) b. \(P\left(x \leq \frac{1}{2}\right)\) c. \(P\left(x<\frac{1}{4}\right)\) d. \(P\left(\frac{1}{4}

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