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Magazine Chapter 3: Problem 2
If \(X\) is \(N(75,100)\), find \(P(X<60)\) and \(P(70
Short Answer
Expert verified
The probability that X is less than 60 is 0.0668, and the probability that X is between 70 and 100 is 0.6853.
Step by step solution
01
Calculate the z-score for X
In order to calculate z-score, the formula \(Z = (X-\mu)/\sigma\) is utilized, where \mu is the mean and \sigma is the standard deviation. In this case, \(X=60\), \(\mu=75\), and \(\sigma=\sqrt{100}=10\). Hence, the z-score \(Z=(60-75)/10=-1.5\).
02
Use the z-score to find P(X
The z-score can be used to find the probability by using either standard normal distribution table or the pnorm command in R. In this case, since Z = -1.5, we look this up in the standard normal distribution table or use pnorm(-1.5) in R and we find that P(Z<-1.5) = 0.0668. Thus, P(X<60)= P(Z<-1.5)=0.0668.
03
Calculate the z-scores for 70
As we did in the previous calculation, we will have to find z-scores for X=70 and X=100. Calculating the z-scores we get \(Z1=(70-75)/10=-0.5\) and \(Z2=(100-75)/10=2.5\).
04
Use the z-scores to find P(70
To find P(70<X<100), we need to calculate P(-0.5<Z<2.5) from the cumulative standard normal table or using R command. This can be given by P(Z<2.5) - P(Z<-0.5): Using the table or the pnorm function in R gives us 0.9938 and 0.3085 respectively. Subtracting these values yields P(-0.5<Z<2.5)= 0.9938 - 0.3085 = 0.6853. Thus, P(70 < X < 100) = P(-0.5<Z<2.5)= 0.6853.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Z-Score Calculation
Understanding the z-score is crucial for analyzing data relative to the mean of a group of values. A z-score, also known as a standard score, indicates how many standard deviations an element is from the mean. To calculate a z-score, you can use the formula:
\[ Z = \frac{X - \mu}{\sigma} \]
where \(X\) is the value in question, \(\mu\) is the mean of the data, and \(\sigma\) is the standard deviation.
For instance, if a test score \(X\) is 60, the mean \(\mu\) is 75, and the standard deviation \(\sigma\) is the square root of the variance, say 10, then the z-score is calculated as \( Z = \frac{60 - 75}{10} = -1.5 \). This z-score indicates that the score is 1.5 standard deviations below the mean.
\[ Z = \frac{X - \mu}{\sigma} \]
where \(X\) is the value in question, \(\mu\) is the mean of the data, and \(\sigma\) is the standard deviation.
For instance, if a test score \(X\) is 60, the mean \(\mu\) is 75, and the standard deviation \(\sigma\) is the square root of the variance, say 10, then the z-score is calculated as \( Z = \frac{60 - 75}{10} = -1.5 \). This z-score indicates that the score is 1.5 standard deviations below the mean.
Standard Normal Distribution Table
A standard normal distribution table, often referred to as a z-table, is an essential tool used to determine the probability that a statistic is below, above, or between certain values. It represents the area to the left of a z-score in a standard normal distribution which is a special case of the normal distribution with a mean of 0 and a standard deviation of 1.
To use the table effectively, locate the z-score along the margins, and the corresponding value in the table will give you the cumulative probability. For a z-score of -1.5, you would look up this value in the z-table to find the probability of observing a value less than 60 given the mean and standard deviation from our example. This approach transforms our original question about a score less than 60 into a question about a z-score less than -1.5.
To use the table effectively, locate the z-score along the margins, and the corresponding value in the table will give you the cumulative probability. For a z-score of -1.5, you would look up this value in the z-table to find the probability of observing a value less than 60 given the mean and standard deviation from our example. This approach transforms our original question about a score less than 60 into a question about a z-score less than -1.5.
R pnorm Command
For those preferring a computational approach, the 'pnorm' function in R programming language offers an efficient way to calculate the cumulative probability for the normal distribution. This function provides the cumulative distribution function (CDF), which is the probability that a normally distributed random variable will be less than or equal to a certain value.
The syntax is straightforward: simply use \( pnorm(z, mean = 0, sd = 1) \) where \(z\) is the z-score, \(mean\) is the average, and \(sd\) is the standard deviation. If the mean and standard deviation are not provided, pnorm uses the standard normal distribution by default. Thus, for a z-score of -1.5, the command \( pnorm(-1.5) \) will give you the cumulative probability associated with that z-score.
The syntax is straightforward: simply use \( pnorm(z, mean = 0, sd = 1) \) where \(z\) is the z-score, \(mean\) is the average, and \(sd\) is the standard deviation. If the mean and standard deviation are not provided, pnorm uses the standard normal distribution by default. Thus, for a z-score of -1.5, the command \( pnorm(-1.5) \) will give you the cumulative probability associated with that z-score.
Cumulative Normal Distribution
The cumulative normal distribution is the probability that a normal random variable is less than or equal to a particular value. It adds up all probabilities from the left-end of the distribution up to the given point, capturing the idea of 'less than or equal to.'
In practical terms, if you want to know the probability that a student scores less than 60 on a test with mean 75 and standard deviation 10 (given our example), you are actually looking for the cumulative probability up to the z-score of -1.5. Whether using a z-table or the pnorm function in R, this cumulative value helps us understand the likelihood of such an event occurring in a normally distributed dataset.
In practical terms, if you want to know the probability that a student scores less than 60 on a test with mean 75 and standard deviation 10 (given our example), you are actually looking for the cumulative probability up to the z-score of -1.5. Whether using a z-table or the pnorm function in R, this cumulative value helps us understand the likelihood of such an event occurring in a normally distributed dataset.
