Problem 3
Question
The speeds of cars on a road are approximately normally distributed with a mean \(\mu=58 \mathrm{~km} / \mathrm{hr}\) and standard deviation \(\sigma=4 \mathrm{~km} / \mathrm{hr}\) (a) What is the probability that a randomly selected car is going between 60 and \(65 \mathrm{~km} / \mathrm{hr}\) ? (b) What fraction of all cars are going slower than 52 \(\mathrm{km} / \mathrm{hr} ?\)
Step-by-Step Solution
Verified Answer
(a) 26.84% (b) 6.68%
1Step 1: Identify the Distribution
The speeds of the cars follow a normal distribution with mean \( \mu = 58 \text{ km/hr} \) and standard deviation \( \sigma = 4 \text{ km/hr} \). We will use this information to find the probabilities for the given ranges.
2Step 2: Calculate the Z-scores for Part (a)
For the speed range between 60 and 65 km/hr, we calculate the Z-scores using the formula: \( Z = \frac{X - \mu}{\sigma} \). For 60 km/hr: \[ Z_{60} = \frac{60 - 58}{4} = 0.5 \] For 65 km/hr: \[ Z_{65} = \frac{65 - 58}{4} = 1.75 \]
3Step 3: Find the Probability for Part (a)
Using the standard normal distribution table: - The probability for \( Z = 0.5 \) is approximately 0.6915. - The probability for \( Z = 1.75 \) is approximately 0.9599. Thus, the probability that a car is going between 60 and 65 km/hr is: \[ P(60 < X < 65) = 0.9599 - 0.6915 = 0.2684 \]
4Step 4: Calculate the Z-score for Part (b)
For cars going slower than 52 km/hr, we find the Z-score for 52 km/hr: \[ Z_{52} = \frac{52 - 58}{4} = -1.5 \]
5Step 5: Find the Fraction for Part (b)
Using the standard normal distribution table, the probability for \( Z = -1.5 \) is approximately 0.0668. Thus, the fraction of cars going slower than 52 km/hr is: \[ P(X < 52) = 0.0668 \]
Key Concepts
ProbabilityZ-scoresStandard DeviationMean
Probability
In the context of normal distribution, probability helps us measure how likely an event is to occur within a specific range. When dealing with normally distributed data, the probability we seek is essentially the area under the bell curve. For example, if we want to know the probability of a randomly selected car traveling between 60 and 65 km/hr, we calculate the area under the curve between these two points. This is achieved using Z-scores and standard normal distribution tables.
The probability concept allows us to understand variations and expected outcomes in real-world situations like traffic speeds. It's important to remember that these probabilities range from 0 to 1, providing us with an understanding of how common or rare an event might be based on the data given.
The probability concept allows us to understand variations and expected outcomes in real-world situations like traffic speeds. It's important to remember that these probabilities range from 0 to 1, providing us with an understanding of how common or rare an event might be based on the data given.
Z-scores
Z-scores are a critical part of working with normal distributions. They help quantify how far away a particular value is from the mean, measured in standard deviations. In simple terms, a Z-score tells you how many standard deviations a data point (like a car's speed) is from the average speed.
For example, to find the Z-score for a speed of 65 km/hr, we use the formula: \( Z = \frac{X - \mu}{\sigma} \). Substituting the values, for 65 km/hr, \( Z = \frac{65 - 58}{4} = 1.75 \). This means the speed is 1.75 standard deviations above the mean.
Z-scores are crucial because they allow us to use standard normal distribution tables to determine the probability of each occurrence within our data set. They provide a standardized way to compare different data points.
For example, to find the Z-score for a speed of 65 km/hr, we use the formula: \( Z = \frac{X - \mu}{\sigma} \). Substituting the values, for 65 km/hr, \( Z = \frac{65 - 58}{4} = 1.75 \). This means the speed is 1.75 standard deviations above the mean.
Z-scores are crucial because they allow us to use standard normal distribution tables to determine the probability of each occurrence within our data set. They provide a standardized way to compare different data points.
Standard Deviation
Standard deviation is a measure of the spread of data points around the mean in a normal distribution. It tells us how much individual data points typically deviate from the average. In our case, the speeds of cars have a standard deviation of 4 km/hr.
This means that most car speeds will fall within 4 km/hr of the mean (58 km/hr), with fewer cars dramatically faster or slower. Standard deviation helps us understand the variability within a data set.
This means that most car speeds will fall within 4 km/hr of the mean (58 km/hr), with fewer cars dramatically faster or slower. Standard deviation helps us understand the variability within a data set.
- A small standard deviation indicates data points are close to the mean.
- A large standard deviation indicates wide variability and data points spread over a larger range.
Mean
In the study of normal distribution, the mean is a central concept. It represents the average of all data points. For the speeds of cars on a road, the mean is 58 km/hr, indicating it as the central point of the data set.
The mean divides the bell-shaped curve into two equal halves in a normal distribution. It is essential because it's the point around which we measure standard deviation and calculate Z-scores. A normal distribution assumes data clusters around the mean, with fewer instances as you move further away.
Understanding the mean allows us to gauge the central tendency of data, facilitating predictions and assessments based on the given data. It's a foundational element that ties together the whole process of working with normally distributed data.
The mean divides the bell-shaped curve into two equal halves in a normal distribution. It is essential because it's the point around which we measure standard deviation and calculate Z-scores. A normal distribution assumes data clusters around the mean, with fewer instances as you move further away.
Understanding the mean allows us to gauge the central tendency of data, facilitating predictions and assessments based on the given data. It's a foundational element that ties together the whole process of working with normally distributed data.
Other exercises in this chapter
Problem 3
A person who travels regularly on the \(9: 00 \mathrm{am}\) bus from Oakland to San Francisco reports that the bus is almost always a few minutes late but rarel
View solution Problem 4
The distribution of IQ scores can be modeled by a normal distribution with mean 100 and standard deviation \(15 .\) (a) Write the formula for the density functi
View solution Problem 5
Let \(p(t)=-0.0375 t^{2}+0.225 t\) be the density function for the shelf life of a brand of banana which lasts up to 4 weeks. Time, \(t\), is measured in weeks
View solution