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Chapter 5: Problem 5

Assume the given distributions are approximately normal. An electronic product takes an average of 3.4 hours to move through an assembly line. If the standard deviation is 0.5 hours, what is the probability that an item will take between 3 and 4 hours to move through the assembly line? (A) \(P(3

Short Answer

Expert verified
The probability that an item will take between 3 and 4 hours to move through the assembly line is 0.6730.

Step by step solution

01

- Understand the Problem

We need to find the probability that an item will take between 3 and 4 hours to move through the assembly line. This is for a normal distribution with a mean of 3.4 hours and a standard deviation of 0.5 hours.
02

- Standardize the Values

To convert the values to the standard normal distribution, we use the z-score formula: \[ z = \frac{x - \mu}{\sigma} \] Where \( x \) is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
03

- Calculate the z-scores

First, calculate the z-score for 3 hours: \[ z_1 = \frac{3 - 3.4}{0.5} = \frac{-0.4}{0.5} = -0.8 \]Next, calculate the z-score for 4 hours: \[ z_2 = \frac{4 - 3.4}{0.5} = \frac{0.6}{0.5} = 1.2 \]
04

- Interpret the Probability

We need to find the probability \( P(3 < X < 4) \), which is equivalent to finding the probability between z-scores -0.8 and 1.2. This is denoted as: \[ P(-0.8 < z < 1.2) \]
05

- Use Standard Normal Distribution Table

Using the standard normal distribution table (z-table), find the probabilities for the z-scores. \[ P(z < -0.8) = 0.2119 \] and \[ P(z < 1.2) = 0.8849 \]. Thus, the required probability is: \[ P(-0.8 < z < 1.2) = P(z < 1.2) - P(z < -0.8) = 0.8849 - 0.2119 = 0.6730 \]

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Key Concepts

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

Standard Normal Distribution
The standard normal distribution is a special type of normal distribution where the mean \( \mu \) is 0 and the standard deviation \( \sigma \) is 1. This helps us compare different normal distributions by standardizing their values. When any value from an original normal distribution is converted to this standard form, it鈥檚 called a z-score. Understanding the standard normal distribution is essential when working with probabilities and interpreting z-scores.
Z-score Calculation
To standardize a value from a normal distribution, we use the z-score formula: \[ z = \frac{x - \mu}{\sigma} \]. Here, \( x \) stands for the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. For example, if you want to find the z-score for 3 hours (with a mean of 3.4 and a standard deviation of 0.5), you would calculate: \[ z = \frac{3 - 3.4}{0.5} = -0.8 \]. This means 3 hours is 0.8 standard deviations below the mean. Similarly, the z-score for 4 hours is calculated as \[ z = \frac{4 - 3.4}{0.5} = 1.2 \]. This is 1.2 standard deviations above the mean.
Probability Interpretation
After converting to z-scores, we interpret probabilities using the standard normal distribution. For example, finding the probability that an item will take between 3 and 4 hours is equivalent to finding \[ P(-0.8 < z < 1.2) \]. With a z-table, you can find the areas under the curve. \ P(z < -0.8) \ is the area to the left of -0.8, and \ P(z < 1.2) \ is the area to the left of 1.2. The probability that the value falls between these z-scores is \[ P(-0.8 < z < 1.2) = P(z < 1.2) - P(z < -0.8) \], which equals 0.673 based on z-table values.
Standard Deviation
Standard deviation measures the spread of data points around the mean. In our example, the standard deviation is 0.5 hours, indicating that most assembly times are within 0.5 hours of the mean (3.4 hours). It's a crucial component in calculating z-scores and understanding the normal distribution. A smaller standard deviation means values are tightly clustered around the mean, while a larger standard deviation indicates a wider spread.

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Most popular questions from this chapter

Which of the following is a true statement? (A) The sampling distribution of \(\hat{p}\) has a mean equal to the population proportion \(p\). (B) The sampling distribution of \(\hat{p}\) has a standard deviation equal to \(\sqrt{n p(1-p)}\) (C) The sampling distribution of \(\hat{p}\) has a standard deviation that becomes larger as the sample size becomes larger. (D) The sampling distribution of \(\hat{p}\) is considered close toour sample is clearly less than \(10 \%\) of all butterfly larvae normal provided that \(n \geq 30\). (E) The sampling distribution of \(\hat{p}\) is always close to normal.

Which of the following is a true statement? (A) The larger the sample, the larger the spread in the sampling distribution is. (B) Bias relates to the spread of a sampling distribution. (C) Provided that the population size is significantly greater than the sample size, the spread of the sampling distribution does not depend on the population size. (D) Sample parameters are used to make inferences about population statistics. (E) Statistics from smaller samples have less variability.

Suppose a population is skewed right. For which of the following sample sizes would the sampling distribution of \(\bar{x}\) be closest to normal? (A) 10 (B) 30 (C) 50 (D) 100 (E) According to the central limit theorem, all give normal sampling distributions.

Assume the given distributions are approximately normal. A trucking firm determines that its fleet of trucks averages a mean of 12.4 miles per gallon with a standard deviation of 1.2 miles per gallon on cross-country hauls. What is the probability that one of the trucks averages fewer than 10 miles per gallon? (A) \(P(z<-2.4)\) (B) \(P(z<-2)\) (C) \(P(z<10)\) (D) \(P\left(z<\frac{10}{1.2}\right)\) (E) \(P\left(z<\frac{12.4}{1.2}\right)\)

A study is to be performed to estimate the proportion of voters who believe the economy is "heading in the right direction." Which of the following pairs of sample size and population proportion \(p\) will result in the smallest variance for the sampling distribution of \(\hat{p} ?\) (A) \(n=100\) and \(p=0.1\) (B) \(n=100\) and \(p=0.5\) (C) \(n=100\) and \(p=0.99\) (D) \(n=1000\) and \(p=0.1\) (E) \(n=1000\) and \(p=0.5\)
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