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Chapter 7: Problem 42

General Electric manufactures a decorative Crystal Clear 60 -watt light bulb that it advertises will last 1500 hours. Suppose that the lifetimes of the light bulbs are approximately normally distributed, with a mean of 1550 hours and a standard deviation of 57 hours. (a) What proportion of the light bulbs will last less than the advertised time? (b) What proportion of the light bulbs will last more than 1650 hours? (c) What is the probability that a randomly selected GE Crystal Clear 60 -watt light bulb will last between 1625 and 1725 hours? (d) What is the probability that a randomly selected GE Crystal Clear 60 -watt light bulb will last longer than 1400 hours?

Short Answer

Expert verified
(a) 0.1908 (b) 0.0392 (c) 0.0923 (d) 0.9957

Step by step solution

01

- Convert Hours to Z-Scores

To find the proportion of light bulbs lasting less than a certain time, more than a certain time, or between certain times, convert the times to Z-scores using the formula: \[ Z = \frac{X - \mu}{\sigma} \]where \(X\) is the time, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
02

- Proportion of Bulbs Lasting Less than 1500 Hours

Calculate the Z-score for 1500 hours: \[ Z = \frac{1500 - 1550}{57} = \frac{-50}{57} \approx -0.877 \]Using standard normal distribution tables or a calculator, find the proportion corresponding to \(Z = -0.877\). The probability is approximately 0.1908.
03

- Proportion of Bulbs Lasting More than 1650 Hours

Calculate the Z-score for 1650 hours: \[ Z = \frac{1650 - 1550}{57} = \frac{100}{57} \approx 1.754 \]Using standard normal distribution tables or a calculator, find the probability for \(Z = 1.754\). This is approximately 0.9608. To find the proportion lasting more than 1650 hours, calculate \(1 - 0.9608 = 0.0392\).
04

- Probability of Lasting Between 1625 and 1725 Hours

Find the Z-scores for 1625 and 1725 hours:\[ Z_{1625} = \frac{1625 - 1550}{57} = \frac{75}{57} \approx 1.316 \]\[ Z_{1725} = \frac{1725 - 1550}{57} = \frac{175}{57} \approx 3.070 \]Find the probabilities for these Z-scores from standard normal distribution tables or a calculator: \( P(Z_{1625}) \approx 0.9066 \) and \( P(Z_{1725}) \approx 0.9989 \). The probability of lasting between 1625 and 1725 hours is: \[ P(1625 < X < 1725) = P(Z_{1725}) - P(Z_{1625}) = 0.9989 - 0.9066 = 0.0923 \]
05

- Probability of Lasting More than 1400 Hours

Calculate the Z-score for 1400 hours: \[ Z = \frac{1400 - 1550}{57} = \frac{-150}{57} \approx -2.632 \]Using standard normal distribution tables or a calculator, find the probability for \(Z = -2.632\). This is approximately 0.0043. To find the proportion lasting more than 1400 hours, calculate \(1 - 0.0043 = 0.9957\).

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Key Concepts

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

The Role of Mean in Normal Distribution
Array
Importance of Standard Deviation
Array

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Most popular questions from this chapter

Use a normal probability plot to assess whether the sample data could have come from a population that is normally distributed. Customer Service A random sample of weekly work logs at an automobile repair station was obtained, and the average number of customers per day was recorded. $$\begin{array}{lllll} \hline 26 & 24 & 22 & 25 & 23 \\ \hline 24 & 25 & 23 & 25 & 22 \\ \hline 21 & 26 & 24 & 23 & 24 \\ \hline 25 & 24 & 25 & 24 & 25 \\ \hline 26 & 21 & 22 & 24 & 24 \\ \hline \end{array}$$

Draw a normal curve and label the mean and inflection points. $$ \mu=50 \text { and } \sigma=5 $$

Steel rods are manufactured with a mean length of 25 centimeters \((\mathrm{cm}) .\) Because of variability in the manufacturing process, the lengths of the rods are approximately normally distributed, with a standard deviation of \(0.07 \mathrm{~cm} .\) (a) What proportion of rods has a length less than \(24.9 \mathrm{~cm} ?\) (b) Any rods that are shorter than \(24.85 \mathrm{~cm}\) or longer than \(25.15 \mathrm{~cm}\) are discarded. What proportion of rods will be discarded? (c) Using the results of part (b), if 5000 rods are manufactured in a day, how many should the plant manager expect to discard? (d) If an order comes in for 10,000 steel rods, how many rods should the plant manager manufacture if the order states that all rods must be between \(24.9 \mathrm{~cm}\) and \(25.1 \mathrm{~cm} ?\)

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