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Suppose that the lifetimes of light bulbs are approximately normally distributed, with a mean of 57 hours and a stande deviation of 3.5 hours. With this information, answer the following questions. (a) What proportion of light bulbs will last more than 60 hours? (b) What proportion of light bulbs will last 52 hours or less? (c) What proportion of light bulbs will last between 57 and 61 hours? (d) What is the probability that a randomly selected light bulb lasts less than 45 hours? (a) The proportion of light bulbs that last more than 60 hours is (Round to four decimal places as needed.)

Ask by Herrera Wagner. in the United States
Mar 17,2025

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Answer

The proportion of light bulbs lasting more than 60 hours is approximately 0.1949.

Solution

\[ \textbf{Step 1: Define the problem.} \] We are given that the lifetime \( X \) of a light bulb is normally distributed with mean \( \mu = 57 \) hours and standard deviation \( \sigma = 3.5 \) hours. We need to find the proportion of bulbs that last more than \( 60 \) hours. \[ \textbf{Step 2: Standardize the value.} \] We use the transformation to the standard normal variable \( Z \): \[ Z = \frac{X - \mu}{\sigma} \] For \( X = 60 \): \[ Z = \frac{60 - 57}{3.5} = \frac{3}{3.5} \approx 0.8571 \] \[ \textbf{Step 3: Compute the desired probability.} \] We need to find: \[ P(X > 60) = P\left(Z > 0.8571\right) \] This is equivalent to: \[ P\left(Z > 0.8571\right) = 1 - P\left(Z \le 0.8571\right) \] Using a standard normal table or calculator, we find: \[ P\left(Z \le 0.8571\right) \approx 0.8051 \] Thus: \[ P(X > 60) \approx 1 - 0.8051 = 0.1949 \] \[ \textbf{Final Answer:} \] The proportion of light bulbs that last more than \( 60 \) hours is approximately \( \boxed{0.1949} \).

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To find the proportion of light bulbs that last more than 60 hours, you can use the Z-score formula: \[ Z = \frac{(X - \mu)}{\sigma} \] where \( X \) is the value you're interested in (60 hours), \( \mu \) is the mean (57 hours), and \( \sigma \) is the standard deviation (3.5 hours). 1. Calculate the Z-score: \[ Z = \frac{(60 - 57)}{3.5} = \frac{3}{3.5} \approx 0.8571 \] 2. Using a standard normal distribution table, find the area to the left of \( Z = 0.8571 \) (which is approximately 0.8051). Therefore, the area to the right (the proportion lasting more than 60 hours) is: \[ 1 - 0.8051 = 0.1949 \] Rounded to four decimal places, the proportion of light bulbs that last more than 60 hours is **0.1949**. For the questions that follow: (b) To find the proportion of light bulbs that last 52 hours or less, calculate the Z-score for 52 hours: \[ Z = \frac{(52 - 57)}{3.5} = \frac{-5}{3.5} \approx -1.4286 \] Using the Z-table, the area to the left of \( Z = -1.4286 \) is approximately 0.0764. Therefore, the proportion is **0.0764**. (c) To find the proportion of light bulbs that last between 57 and 61 hours, calculate the Z-scores: - For 57 hours: \( Z = 0 \) (mean) - For 61 hours: \[ Z = \frac{(61 - 57)}{3.5} = \frac{4}{3.5} \approx 1.1429 \] Using the Z-table, the area for \( Z = 1.1429 \) is approximately 0.8729. Thus, for the range from 57 to 61: \[ 0.8729 - 0.5 = 0.3729 \] So the proportion is **0.3729**. (d) To find the probability that a randomly selected light bulb lasts less than 45 hours: \[ Z = \frac{(45 - 57)}{3.5} = \frac{-12}{3.5} \approx -3.4286 \] Using the Z-table, the area to the left of \( Z = -3.4286 \) is close to 0 (approximately 0.0003). Therefore, the probability is **0.0003**. Hope this helps with your light bulb lifetime inquiries!

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