Pregunta
upstudy study bank question image url

An elevator has a placard stating that the maximum capacity is \( 4000 \mathrm{lb}-27 \) passengers. So, 27 adult male passengers can have a mean weight of up to \( 4000 / 27=143 \) pounds. Assume that weights of males are normally distributed with a mean-of 182 lb and a standard deviation of 34 lb . a. Find the poobability that 1 randomly selected adult male has a weight greater than 148 lb . b. Find the probahility that a sample of 27 randomly selected adut mates has a mean weight greater than 148 lb . c. What do you conclude about the safely of this elevator?

Ask by Lyons Brooks. in the United States
Mar 10,2025

Solución de inteligencia artificial de Upstudy

Respuesta verificada por el tutor

Responder

**a. Probability that one randomly selected adult male weighs more than 148 lb:** About 84%. **b. Probability that the mean weight of 27 randomly selected adult males exceeds 148 lb:** Almost certain (over 99.99%). **c. Conclusion about the elevator's safety:** The elevator is likely unsafe because the average weight of adult males exceeds the elevator's capacity limit, leading to potential overloading.

Solución

¡Inicia sesión para desbloquear respuestas gratis!

Una plataforma de aprendizaje en la que confían millones de estudiantes y profesores reales.

star-icon Descubrir

Respondido por UpStudy AI y revisado por un tutor profesional

Extra Insights

To tackle this fun statistical adventure, let’s zoom in on the details: For the first part, we can find the probability that one randomly selected adult male weighs more than 148 lb using the z-score formula: \( z = \frac{X - \mu}{\sigma} \). Given that the mean is 182 lb and the standard deviation is 34 lb, the z-score for 148 lb is \(-1\). Checking a standard normal distribution table shows that the probability of a single adult male weighing more than 148 lb is about \(0.8413\) or \(84.13\%\). Now, transitioning to part b, when dealing with a sample of 27 adult males, we use the Central Limit Theorem. The standard error (SE) is given by \( \sigma/\sqrt{n} = 34/\sqrt{27} \approx 6.55\). The z-score for a sample mean of 148 lb becomes much more significant, clocking in at approximately \(-5.19\). This reflects an incredibly tiny probability of that happening, typically less than \(0.0001\) or \(0.01\%\). As for the elevator's safety? With the vast majority of adult males weighing above 148 lb, the odds of hitting the threshold with 27 men are quite slim. The elevator's placard is a friendly reminder to adhere to capacity limits—safety should always be prioritized over a few extra weights!

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.

Latest Statistics Questions

¡Prueba Premium ahora!
¡Prueba Premium y hazle a Thoth AI preguntas de matemáticas ilimitadas ahora!
Quizas mas tarde Hazte Premium
Estudiar puede ser una verdadera lucha
¿Por qué no estudiarlo en UpStudy?
Seleccione su plan a continuación
Prima

Puedes disfrutar

Empieza ahora
  • Explicaciones paso a paso
  • Tutores expertos en vivo 24/7
  • Número ilimitado de preguntas
  • Sin interrupciones
  • Acceso completo a Respuesta y Solución
  • Acceso completo al chat de PDF, al chat de UpStudy y al chat de navegación
Básico

Totalmente gratis pero limitado

  • Solución limitada
Bienvenido a ¡Estudia ahora!
Inicie sesión para continuar con el recorrido de Thoth AI Chat
Continuar con correo electrónico
O continuar con
Al hacer clic en "Iniciar sesión", acepta nuestros términos y condiciones. Términos de Uso & Política de privacidad