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A worker in the automobile industry works an average of 43.7 hours per week. If the distribution is approximately normal with a standard deviation of 1.6 hours, what is the probability that a randomly selected automobile worker works less than 40 hours per week?

Short Answer

Expert verified
The probability that a worker works less than 40 hours per week is about 0.0104 or 1.04%.

Step by step solution

01

Identify the Given Values

From the problem, we know the mean of the distribution is \( \mu = 43.7 \) hours, and the standard deviation is \( \sigma = 1.6 \) hours. We need to find the probability that a worker works less than 40 hours per week.
02

Convert to Z-score

We use the Z-score formula: \[ z = \frac{X - \mu}{\sigma} \] where \( X = 40 \) is the number of hours we are analyzing. Calculating the Z-score gives us: \[ z = \frac{40 - 43.7}{1.6} = \frac{-3.7}{1.6} = -2.3125 \]
03

Look Up Z-score in Standard Normal Distribution Table

Use the standard normal distribution table to find the probability associated with the Z-score of -2.3125. The table gives the probability of \( P(Z < -2.3125) \).
04

Interpret the Probability

By consulting a Z-table or using a calculator for \( Z = -2.3125 \), we find that \( P(Z < -2.3125) \) is approximately 0.0104.

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Key Concepts

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

Understanding the Normal Distribution
In statistics, the normal distribution is a continuous probability distribution that is symmetrical around its mean, creating a bell-shaped curve. This curve is a crucial concept because it models many natural phenomena and helps simplify the calculation of probabilities.
  • The mean, median, and mode of a normal distribution are equal, positioned at the center of the curve.
  • Its shape is determined by two parameters: the mean ( ext backslash ( ext mu ext backslash ) and the standard deviation ( ext backslash ( ext backslash ) ).
  • A smaller standard deviation results in a steeper curve, while a larger standard deviation leads to a flatter curve.
The characteristics of the normal distribution make it instrumental in calculating probabilities since many statistical tests rely on it. With the assumption of normality, we can easily use mathematical tools like the Z-score to determine the likelihood of observing certain values within a dataset.
What is a Z-Score?
A Z-score, also known as a standard score, is a statistic that indicates how many standard deviations an element is from the mean of the distribution. This conversion is essential for translating different data points into a standard normal distribution.
  • Z-scores help us understand where a particular value lies within a distribution.
  • A Z-score of 0 means the value is equal to the mean, while positive or negative values indicate deviation above or below the mean.
Z-scores make it easier to compare scores from different datasets because they measure relative position based on the mean and standard deviation.To convert a raw score to a Z-score, use the formula:\[ z = \frac{X - \mu}{\sigma} \]where \( X \) is the raw score, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. For example, if we calculated a Z-score of -2.3125 for our problem, it indicates that 40 hours per week is 2.3125 standard deviations below the mean of 43.7 hours.
The Role of Standard Deviation
Standard deviation is a measure in statistics that quantifies the amount of variation or dispersion present in a set of data. It's a key component in understanding the spread of values within a dataset.
  • A small standard deviation means that the values tend to be close to the mean.
  • Larger standard deviations indicate that the values are spread out over a wider range.
Mathematically, the standard deviation is usually denoted as \( \sigma \) and is calculated as the square root of the variance.Understanding standard deviation is essential when working with normal distributions, as it defines the width of the curve. In the context of our exercise, knowing that the average workweek is 43.7 hours with a standard deviation of 1.6 hours helps us determine the likelihood of deviations, such as working less than 40 hours. The lower a value's Z-score, relative to the standard deviation, the less likely it is for that value to occur under normal circumstances.

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Most popular questions from this chapter

Assume that the sample is taken from a large population and the correction factor can be ignored. Monthly Precipitation for Miami The mean precipitation for Miami in August is 8.9 inches. Assume that the standard deviation is 1.6 inches and the variable is normally distributed. a. Find the probability that a randomly selected August month will have precipitation of less than 8.2 inches. (This month is selected from August months over the last 10 years.) b. Find the probability that a sample of 10 August months will have a mean of less than 8.2 inches. c. Does it seem reasonable that a randomly selected August month will have less than 8.2 inches of rain? d. Does it seem reasonable that a sample of 10 months will have a mean of less than 8.2 months?

According to the World Almanac, \(72 \%\) of households own smartphones. If a random sample of 180 households is selected, what is the probability that more than 115 but fewer than 125 have a smartphone?

Assume that the sample is taken from a large population and the correction factor can be ignored. Systolic Blood Pressure Assume that the mean systolic blood pressure of normal adults is 120 millimeters of mercury \((\mathrm{mm} \mathrm{Hg})\) and the standard deviation is 5.6 . Assume the variable is normally distributed. a. If an individual is selected, find the probability that the individual's pressure will be between 120 and \(121.8 \mathrm{~mm} \mathrm{Hg} .\) b. If a sample of 30 adults is randomly selected, find the probability that the sample mean will be between 120 and \(121.8 \mathrm{~mm} \mathrm{Hg} .\) c. Why is the answer to part \(a\) so much smaller than the answer to part \(b\) ?

Check to see whether the correction factor should be used. If so, be sure to include it in the calculations. Home Values A study of 800 homeowners in a certain area showed that the average value of the homes was \(\$ 82,000,\) and the standard deviation was \(\$ 5000 .\) If 50 homes are for sale, find the probability that the mean of the values of these homes is greater than \(\$ 83,500\)

A mail order company has an \(8 \%\) success rate. If it mails advertisements to 600 people, find the probability of getting fewer than 40 sales.

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