Chapter 7: Problem 28
Assume that the random variable \(X\) is normally distributed, with mean
\(\mu=50\) and standard deviation \(\sigma=7 .\) Compute the following
probabilities. Be sure to draw a normal curve with the area corresponding to
the probability shaded.
\(P(56
Short Answer
Expert verified
0.1898
Step by step solution
01
- Convert to Standard Normal Distribution
Use the Z-score formula to convert the given X values to standard normal form. The Z-score is given by: \[ Z = \frac{X - \mu}{\sigma} \] For \( P(56 < X < 68) \), find the Z-scores for X = 56 and X = 68.
02
- Calculate Z-scores
Calculate the Z-scores for X = 56 and X = 68. For X = 56: \[ Z_1 = \frac{56 - 50}{7} = \frac{6}{7} \approx 0.86 \] For X = 68: \[ Z_2 = \frac{68 - 50}{7} = \frac{18}{7} \approx 2.57 \]
03
- Find Corresponding Probabilities
Use the Z-table to find the probabilities corresponding to \(Z_1\) and \(Z_2\). From the Z-table, \(P(Z < 0.86) \approx 0.8051\) and \(P(Z < 2.57) \approx 0.9949\).
04
- Calculate the Desired Probability
The probability \( P(56 < X < 68) \) can be found by subtracting the probability corresponding to \(Z_1\) from the probability corresponding to \(Z_2\). \[ P(56 < X < 68) = P(Z < 2.57) - P(Z < 0.86) = 0.9949 - 0.8051 = 0.1898 \]
05
- Draw and Shade the Normal Curve
Draw a normal distribution curve with mean \( \mu = 50 \) and shade the area between \( X = 56 \) and \( X = 68 \). Annotate the corresponding Z-scores on the graph.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Z-score
The Z-score is a fundamental concept in statistics, especially within the context of the normal distribution. The Z-score measures how many standard deviations a data point is from the mean of the data set.
The formula for calculating a Z-score is: \[ Z = \frac{X - \mu}{\sigma} \] Here, \( X \) represents the value of the interest, \( \mu \) is the mean of the distribution, and \( \sigma \) is the standard deviation.
There are several reasons why understanding Z-scores is important: \begin{itemize} \item Z-scores standardize different data points for easy comparison. \item They help in understanding where a data point lies within the distribution. \item They facilitate conversion of a normal distribution to the standard normal distribution. \end{itemize} For example, in the problem P(56 < X < 68), we first converted X values (56 and 68) to Z-scores: \[ Z_1 = \frac{56 - 50}{7} = 0.86 \] and \[ Z_2 = \frac{68 - 50}{7} = 2.57 \] These scores help in later steps for calculating probabilities.
The formula for calculating a Z-score is: \[ Z = \frac{X - \mu}{\sigma} \] Here, \( X \) represents the value of the interest, \( \mu \) is the mean of the distribution, and \( \sigma \) is the standard deviation.
There are several reasons why understanding Z-scores is important: \begin{itemize} \item Z-scores standardize different data points for easy comparison. \item They help in understanding where a data point lies within the distribution. \item They facilitate conversion of a normal distribution to the standard normal distribution. \end{itemize} For example, in the problem P(56 < X < 68), we first converted X values (56 and 68) to Z-scores: \[ Z_1 = \frac{56 - 50}{7} = 0.86 \] and \[ Z_2 = \frac{68 - 50}{7} = 2.57 \] These scores help in later steps for calculating probabilities.
Standard Normal Distribution
A standard normal distribution is a type of normal distribution that has been transformed to have a mean of 0 and a standard deviation of 1. This transformation allows us to use a special table called the Z-table to find probabilities associated with this distribution.
This is achieved through converting a normal distribution to a standard normal form using Z-scores. The transformation formula is: \[ Z = \frac{X - \mu}{\sigma} \] The standard normal distribution is important for a few reasons: \begin{itemize} \item It provides a universal scale for comparison. \item Simplifies the complex calculations involved in finding probabilities. \item Facilitates the use of the Z-table to obtain probability values. \end{itemize} For instance, transforming our problem (initially with a mean \( \mu = 50\) and a standard deviation \( \sigma = 7\)), we were able to use the Z-scores (0.86 and 2.57) to look up associated probabilities.
This is achieved through converting a normal distribution to a standard normal form using Z-scores. The transformation formula is: \[ Z = \frac{X - \mu}{\sigma} \] The standard normal distribution is important for a few reasons: \begin{itemize} \item It provides a universal scale for comparison. \item Simplifies the complex calculations involved in finding probabilities. \item Facilitates the use of the Z-table to obtain probability values. \end{itemize} For instance, transforming our problem (initially with a mean \( \mu = 50\) and a standard deviation \( \sigma = 7\)), we were able to use the Z-scores (0.86 and 2.57) to look up associated probabilities.
Probability Calculation
Probability calculation is used to find the likelihood of an event occurring within a specific range of values. In the context of a normal distribution, we often use the Z-table to simplify this process.
The Z-table contains probabilities up to a given Z-score. To find the probability between two points in the normal distribution, we: \begin{itemize} \item Convert the points to Z-scores. \item Find their corresponding probabilities using the Z-table. \item Subtract the smaller probability from the larger one. \end{itemize} From our problem, the Z-scores were 0.86 and 2.57. Their respective probabilities found from the Z-table were approximately 0.8051 and 0.9949.
So, the probability that X values lie between 56 and 68 was calculated as follows: \[ P(56 < X < 68) = P(Z < 2.57) - P(Z < 0.86) \] \[ = 0.9949 - 0.8051 \] \[ = 0.1898 \] Thus, the probability is about 18.98%.
The Z-table contains probabilities up to a given Z-score. To find the probability between two points in the normal distribution, we: \begin{itemize} \item Convert the points to Z-scores. \item Find their corresponding probabilities using the Z-table. \item Subtract the smaller probability from the larger one. \end{itemize} From our problem, the Z-scores were 0.86 and 2.57. Their respective probabilities found from the Z-table were approximately 0.8051 and 0.9949.
So, the probability that X values lie between 56 and 68 was calculated as follows: \[ P(56 < X < 68) = P(Z < 2.57) - P(Z < 0.86) \] \[ = 0.9949 - 0.8051 \] \[ = 0.1898 \] Thus, the probability is about 18.98%.
Shaded Area
Visualizing probability calculations often involves shading areas under the normal distribution curve. This shaded area represents the probability for the given range of values.
To create this visualization: \begin{itemize} \item Draw a normal distribution curve. \item Mark the mean value \( \mu = 50\). \item Identify the values X = 56 and X = 68 along the horizontal axis. \item Shade the area between these two points under the curve. \end{itemize} Annotate the Z-scores (0.86 and 2.57) below their corresponding points on the X-axis. This shaded region visually shows the probability we calculated, making abstract concepts more concrete and easier to understand.
The visual representation helps spot-check the steps undertaken to ensure that calculations align with the area under the curve.
To create this visualization: \begin{itemize} \item Draw a normal distribution curve. \item Mark the mean value \( \mu = 50\). \item Identify the values X = 56 and X = 68 along the horizontal axis. \item Shade the area between these two points under the curve. \end{itemize} Annotate the Z-scores (0.86 and 2.57) below their corresponding points on the X-axis. This shaded region visually shows the probability we calculated, making abstract concepts more concrete and easier to understand.
The visual representation helps spot-check the steps undertaken to ensure that calculations align with the area under the curve.