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Chapter 6: Problem 66

Most users of automatic garage door openers activate their openers at distances that are normally distributed with a mean of 30 feet and a standard deviation of 11 feet. To minimize interference with other remote-controlled devices, the manufacturer is required to limit the operating distance to 50 feet. What percentage of the time will users attempt to operate the opener outside its operating limit?

Short Answer

Expert verified
Answer: 3.44%

Step by step solution

01

Identify given information

We are given that the distances for automatic garage door openers are normally distributed with a mean (渭) of 30 feet and a standard deviation (蟽) of 11 feet. We want to find the percentage of the time users attempt to operate the opener beyond 50 feet.
02

Calculate the Z-score

To determine the probability of a user attempting to operate the garage door opener beyond 50 feet, we need to find the Z-score for 50 feet. The formula for the Z-score is: Z = (X - 渭) / 蟽 Where X is the value we're interested in (50 feet in this case), 渭 is the mean, and 蟽 is the standard deviation. Plugging the values into the formula, we get: Z = (50 - 30) / 11 鈮� 1.82
03

Use the Z-score to find the percentage

Now, we will use the Z-score to find the percentage of times a user will attempt to operate the garage door opener outside its operating limit (beyond 50 feet). Since we want the percentage of users operating the opener beyond 50 feet, we need to look for the area to the right of our Z-score. You can either use a standard normal distribution table or an online calculator to find the area to the right of our Z-score (1.82). We find that the area to the right of our Z-score is 鈮�0.0344 or 3.44%. Finally, we can conclude that users will attempt to operate the garage door opener outside its operating limit of 50 feet about 3.44% of the time.

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Key Concepts

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

Z-score calculation
When working with normal distributions, the Z-score is an essential tool to determine how far a specific data point is from the mean. It essentially helps us standardize different data points to see their relation to the mean, expressed in terms of standard deviations. Here's how you can calculate it:

  • The formula for Z-score is given by \( Z = \frac{X - \mu}{\sigma} \)
  • \( X \) is the data point you're examining (e.g., the distance from which a garage door opener is activated).
  • \( \mu \) is the mean of the distribution (in this case, 30 feet).
  • \( \sigma \) is the standard deviation (11 feet here).
By calculating the Z-score, we can identify how many standard deviations away a particular value is from the average. In our exercise, calculating the Z-score for a distance of 50 feet results in \( Z = \frac{50 - 30}{11} \approx 1.82 \). This tells us that 50 feet is about 1.82 standard deviations above the mean.
Probability calculation
Once you have the Z-score, you can determine probabilities, which is a key part of statistical analysis. The Z-score indicates where a particular value falls on the normal distribution, allowing us to calculate the probability of that value occurring.

  • To find this probability, we look up the Z-score in a standard normal distribution table or use a calculator capable of this function.
  • These tools tell us the probability that a value is less than the Z-score we calculated.
  • Since the exercise asks for the probability of attempting beyond 50 feet, we are interested in the right-side tail of the distribution.
For our Z-score of 1.82, the area to the right in a standard normal distribution table gives us the probability that a user could operate the opener beyond 50 feet. This probability turns out to be approximately 3.44%, indicating that such an event occurs relatively infrequently.
Statistical analysis
Statistical analysis bridges raw data and meaningful insights. In this scenario, the analysis involves understanding the probability of an event happening under a normal distribution.

  • We start by defining the problem with known statistical parameters: mean and standard deviation.
  • Next, we identify the limits or constraints in the problem (a maximum operating distance of 50 feet).
  • Using the Z-score and probability calculation, we assess the likelihood of events outside these constraints.
  • We conclude whether the event (in this case, trying to open the garage door from beyond 50 feet) is common or rare.
The calculated probability of 3.44% is a product of statistical analysis. This information could guide decisions about manufacturing devices, setting acceptable operational parameters, or informing users of the likelihood of unsuccessful operation outside the designed limits.

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Most popular questions from this chapter

The discharge of suspended solids from a phosphate mine is normally distributed, with a mean daily discharge of 27 milligrams per liter \((\mathrm{mg} / \mathrm{l})\) and a standard deviation of \(14 \mathrm{mg} / \mathrm{l}\). What proportion of days will the daily discharge exceed \(50 \mathrm{mg} / \mathrm{l} ?\)

Students very often ask their professors whether they will be "curving the grades." The traditional interpretation of "curving grades" required that the grades have a normal distribution, and that the grades will be assigned in these proportions: $$ \begin{array}{l|lllll} \text { Letter Grade } & \mathrm{A} & \mathrm{B} & \mathrm{C} & \mathrm{D} & \mathrm{F} \\ \hline \text { Proportion of Students } & 10 \% & 20 \% & 40 \% & 20 \% & 10 \% \end{array} $$ a. If the average "C" grade is centered at the average grade for all students, and if we assume that the grades are normally distributed, how many standard deviations on either side of the mean will constitute the "C" grades? b. How many deviations on either side of the mean will be the cutoff points for the "B" and "D" grades?

The life span of oil-drilling bits depends on the types of rock and soil that the drill encounters, but it is estimated that the mean length of life is 75 hours. Suppose an oil exploration company purchases drill bits that have a life span that is approximately normally distributed with a mean equal to 75 hours and a standard deviation equal to 12 hours. a. What proportion of the company's drill bits will fail before 60 hours of use? b. What proportion will last at least 60 hours? c. What proportion will have to be replaced after more than 90 hours of use?

In Exercise \(5.75,\) a cross between two peony plants-one with red petals and one with streaky petals-produced offspring plants with red petals \(75 \%\) of the time. Suppose that 100 seeds from this cross were collected and germinated, and \(x\), the number of plants with red petals, was recorded. a. What is the exact probability distribution for \(x ?\) b. Is it appropriate to approximate the distribution in part a using the normal distribution? Explain. c. Use an appropriate method to find the approximate probability that between 70 and 80 (inclusive) offspring plants have red flowers. d. What is the probability that 53 or fewer offspring plants had red flowers? Is this an unusual occurrence? e. If you actually observed 53 of 100 offspring plants with red flowers, and if you were certain that the genetic ratio 3: 1 was correct, what other explanation could you give for this unusual occurrence?

Calculate the area under the standard normal curve to the left of these values: a. \(z=1.6\) b. \(z=1.83\) c. \(z=.90\) d. \(z=4.18\)
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