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The length of similar components produced by a company are approximated by a normal distribution model with a mean of 8 cm and a standard deviation of 0.04 cm . Let the random variable X represent the length of one component chosen at random. Accurate to 2 decimal places, the cutoff of the top \( 45 \% \) has z-score: -0.13 0.14 0.13 -0.14 None of the above.

Ask by Coleman Welch. in the United States
Mar 09,2025

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Answer

The z-score cutoff for the top 45% is 0.13.

Solution

1. Let the random variable \( X \) represent the component length, and assume \( X \sim N(8, 0.04^2) \). 2. The cutoff for the top \(45\%\) corresponds to the value \( c \) such that \[ P(X \geq c) = 0.45. \] Equivalently, the cumulative probability up to that cutoff is \[ P(X \leq c) = 1 - 0.45 = 0.55. \] 3. Standardizing, we have the relationship \[ Z = \frac{X - \mu}{\sigma}, \] where \(\mu = 8\) and \(\sigma = 0.04\). Thus, the equation becomes \[ P\left(Z \leq \frac{c-8}{0.04}\right) = 0.55. \] 4. We must find a \( z \)-score \( z^* \) such that \[ P(Z \leq z^*) = 0.55. \] Consulting standard normal tables or using a calculator, we find that \( z^* \) is approximately \[ z^* \approx 0.13. \] 5. Therefore, the z-score cutoff accurate to two decimal places is \[ 0.13. \] The correct answer is \(0.13\).

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The Deep Dive

To find the z-score that corresponds to the top 45% of a normal distribution, we first need to consider that the top 45% corresponds to the bottom 55%. Using z-tables or standard normal distribution calculators, the z-score for the 55th percentile is approximately 0.13. Therefore, the correct answer is 0.13. Understanding how to interpret z-scores is key in statistics! A z-score helps you know how many standard deviations a data point is from the mean, making it a crucial tool in inferential statistics. So next time you're calculating something, don’t forget to visualize where your data points stand with respect to the mean! If you're looking to dive deeper into this topic, there are plenty of resources available that can give you a practical sense of z-scores, like online interactive calculators that let you experiment by adjusting mean and standard deviation values. Also, consider checking out textbooks or educational platforms that explain normal distribution with real-world examples!

Related Questions

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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