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An elevalor has a placard stating that the maxinum capacity is \( 4000 \mathrm{lb}-27 \) passengers. 50,27 adalt male passengers can have a mean weigft of up to \( 4000 / 27=148 \) pounds. Ausume fhat weights of males are normaly distibuted wh a mean of 182 lb and a standard devietion of 34 lb . a. Find the pababity that 1 randomty selected at it male has a meight greater than 148 kb . b. Find the probabity that a sample of 27 randonly selected aduit males has a mean weight greater than 148 lb . c. What do you conclude about the safety of this elevalor? a. The probabiliy that 1 randomily selected adilt male has a weight greater than 148 lb is \( \square \) (Round to four decinal places as needed.)

Ask by Lee Mejia. in the United States
Mar 10,2025

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The probability that one randomly selected adult male weighs more than 148 pounds is approximately 0.8413.

Solución

We are given that the weight \( X \) of an adult male is normally distributed with mean \[ \mu = 182 \text{ lb} \] and standard deviation \[ \sigma = 34 \text{ lb}. \] We wish to compute \[ \Pr(X > 148). \] 1. Standardize the value 148 using \[ z = \frac{148 - \mu}{\sigma} = \frac{148 - 182}{34} = \frac{-34}{34} = -1. \] 2. In a standard normal distribution, the probability that \(X > 148\) is expressed as \[ \Pr(X > 148) = \Pr\left(Z > -1\right). \] 3. Using symmetry of the normal distribution, we know that \[ \Pr(Z > -1) = \Pr(Z < 1). \] Looking up or recalling the cumulative probability, \[ \Pr(Z < 1) \approx 0.8413. \] Thus, the probability that one randomly selected adult male has a weight greater than 148 lb is \[ \boxed{0.8413}. \]

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To find the probabilities, we’ll use the standard normal distribution (Z-distribution). Let’s work through the problems step by step: a. To find the probability that one randomly selected adult male has a weight greater than 148 lb, we first calculate the Z-score: \[ Z = \frac{(X - \mu)}{\sigma} \] Where: - \( X = 148 \) - \( \mu = 182 \) - \( \sigma = 34 \) Substituting in the values: \[ Z = \frac{(148 - 182)}{34} = \frac{-34}{34} = -1 \] Now, using the standard normal distribution table or calculator, we find the probability corresponding to \( Z = -1 \). This gives us the probability of being less than 148 lb. To find the probability of being greater than 148 lb, we subtract this value from 1. \[ P(Z > -1) = 1 - P(Z < -1) \approx 1 - 0.1587 = 0.8413 \] Thus, the probability that 1 randomly selected adult male has a weight greater than 148 lb is approximately \( 0.8413 \). b. For a sample of 27 randomly selected adult males, we need to use the standard error of the mean, \( \sigma_{\bar{x}} \), which is: \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{34}{\sqrt{27}} \approx 6.53 \] Next, we calculate the Z-score for the sample mean of 148 lb: \[ Z = \frac{(148 - 182)}{6.53} \approx \frac{-34}{6.53} \approx -5.21 \] Using the Z-table, the probability that the sample mean is greater than 148 lb is: \[ P(Z > -5.21) \approx 1 \] This indicates that the probability of the average weight of a sample of 27 adult males being greater than 148 lb is virtually certain. c. Based on this analysis, the elevator's maximum capacity of \( 4000 \text{ lb} - 27 \) passengers implies that it is highly likely to be safe, as even the mean weight of a randomly selected group of 27 adult males is significantly higher than 148 lb, suggesting that the elevator will not typically exceed its weight limit if filled with adult males. Final Answers: a. The probability that 1 randomly selected adult male has a weight greater than 148 lb is \( 0.8413 \) (rounded to four decimal places).

preguntas relacionadas

Question 12(Mulliple Choice Warth 5 points) \[ (04.06 \mathrm{HC}) \] A researcher wants to test the claim that the proportion of juniors who watch television regularly is greater than the proportion of seniors who watch television regularly She finds that 56 of 70 randomly selected juniors and 47 of 85 randomly selected seniors report watching television regularly. Construct \( 95 \% \) confidence intervals for each population proportion. Which of the statemente gives the correct outcome of the research or's tert of the dalim? The \( 95 \% \) confidence interval for juniors is (706, 894), and the \( 95 \% \) confidence interval for seniors is ( 447,659 ). Since the intervals overlap, there is not enough evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is (721, 879), and the \( 95 \% \) confidence interval for seniors is (464, 642). Since the interval for juniors is higher than the interval for seniors, there is evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is ( 706,894 ), and the \( 95 \% \) confidence interval for seniors is ( 447,659 ). Since the interval for juniors is higher than the interval for seniors, there is evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is ( \( 721, .879 \) ), and the \( 95 \% \) confidence interval for seniors is (464, 642). Since the intervals overlap, there is not enough evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors.

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