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Chapter 2: Problem 69

If \(X\) is normally distributed with mean 1 and variance 4 , use the tables to find \(P\\{2

Short Answer

Expert verified
The probability that a normally distributed random variable \(X\) with mean 1 and variance 4 falls between 2 and 3 is approximately 0.1498, or 14.98%.

Step by step solution

01

Identify the given parameters.

The problem states that \(X\) is normally distributed with a mean (\(\mu\)) of 1 and a variance (\(\sigma^2\)) of 4. From the given variance, we can find the standard deviation (\(\sigma\)) by taking the square root of the variance. Therefore, $$ \sigma = \sqrt{4} = 2. $$
02

Standardize the random variable using Z-scores.

To calculate the Z-scores, we need to subtract the mean from the given values and divide by the standard deviation. The Z-score formula is given by: $$ Z = \frac{X - \mu}{\sigma} $$ We will apply this formula to both the lower (\(X_1 = 2\)) and upper (\(X_2 = 3\)) boundaries of our desired range. Our goal is to find \(P\{Z_1 < Z < Z_2\}\), with \(Z_1\) and \(Z_2\) being the Z-scores corresponding to \(X_1\) and \(X_2\). Calculating these Z-scores, we have: $$ Z_1 = \frac{X_1 - \mu}{\sigma} = \frac{2 - 1}{2} = 0.5 $$ $$ Z_2 = \frac{X_2 - \mu}{\sigma} = \frac{3 - 1}{2} = 1 $$
03

Find the probabilities in the Z-table.

Now, using a standard normal table (Z-table), we can lookup the probabilities corresponding to the calculated Z-scores \(Z_1 = 0.5\) and \(Z_2 = 1\). These probabilities represent the area under the curve to the left of each Z-score. According to the Z-table: \(P\{Z < 0.5 \} = 0.6915\) and \(P\{ Z < 1\} = 0.8413\)
04

Calculate the probability for the desired range.

To find the probability \(P\{2<X<3\}\), we can subtract the probability of \(Z_1\) from the probability of \(Z_2\). Using the probabilities we found in the Z-table: $$ P\{2 < X < 3\} = P\{0.5 < Z < 1\} = P\{Z < 1\} - P\{Z < 0.5\} = 0.8413 - 0.6915 = 0.1498 $$ So, the probability of the random variable \(X\) falling between 2 and 3 is approximately 0.1498, or 14.98%.

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Key Concepts

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

Z-score
The Z-score helps in understanding how far a specific data point is from the mean of a dataset, measured in terms of standard deviations. In the context of a normal distribution, this score represents the standardization of any given point of data. Thanks to this score, different data points can be compared on the same scale:
  • The Z-score is calculated by subtracting the mean from the data point and dividing by the standard deviation: \( Z = \frac{X - \mu}{\sigma} \).
  • This transformation converts any normal distribution to the standard normal distribution, making it straightforward to work with probability tables.
The Z-score allows us to translate a random variable from its original distribution into a universal format. This makes it easier to find probabilities associated with specific data points using standard normal probability tables.
Standard deviation
The standard deviation is a key measure in statistics that indicates the extent of variation or dispersion of a set of values:
  • It reflects how much the values in a dataset deviate from the mean.
  • This measurement is essential in defining how spread out the values are in a normal distribution.
For example, in our problem, the standard deviation, denoted by \( \sigma \), was derived from the given variance \( \sigma^2 \) by taking the square root, resulting in a value of 2. Understanding standard deviation is crucial for analyzing data spread and plays an important role in calculating Z-scores, which require this value for normalization.
Probability tables
Probability tables, or Z-tables, are indispensable tools for finding the probabilities associated with specific Z-scores.
  • These tables show the cumulative probability of a standard normal distribution up to any given Z-score.
  • When you look up a Z-score in a probability table, you get the probability of a random variable being less than or equal to that Z-score.
In the step-by-step solution, probability tables were used to determine the probabilities associated with Z-scores 0.5 and 1. This was key to calculating the likelihood of a value falling within a specified range of the normal distribution.
Variance
Variance is fundamental in statistics for quantifying how much a set of numbers differs from the mean:
  • The variance \( \sigma^2 \) is the mean of the squared deviations from the average value of the dataset.
  • It directly relates to the standard deviation as the square root of variance results in the standard deviation.
In the example problem, the variance was 4, leading to a standard deviation of 2. This variance tells us about the dataset's dispersion. Knowing variance is useful for understanding the extent to which data points can fluctuate, which then feeds into further analysis calculations like the Z-score for probability assessments.

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Most popular questions from this chapter

Let \(a_{1}

Let \(X\) and \(Y\) be independent normal random variables, each having parameters \(\mu\) and \(\sigma^{2}\). Show that \(X+Y\) is independent of \(X-Y\). Hint: Find their joint moment generating function.

A television store owner figures that 50 percent of the customers entering his store will purchase an ordinary television set, 20 percent will purchase a color television set, and 30 percent will just be browsing. If five customers enter his store on a certain day, what is the probability that two customers purchase color sets, one customer purchases an ordinary set, and two customers purchase nothing?

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