Problem 23
Question
A normal distribution has a mean of 100 and a standard deviation of \(10 .\) Find the probability that a value selected at random is in the given interval. at least 100
Step-by-Step Solution
Verified Answer
The probability that a value selected at random is at least 100 is 0.50 or 50%.
1Step 1: Determine the Z-score for the value
Calculate the Z-score for the value of 100 using the formula: \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the value from the normal distribution, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. For \(X = 100\), \(\mu = 100\), and \(\sigma = 10\), the Z-score is \(Z = \frac{100 - 100}{10} = 0\).
2Step 2: Use the Z-score to find the probability
Since the Z-score is 0, which corresponds to the mean, the probability of a value being at least the mean is 50% plus the probability of a value being above the mean. For a standard normal distribution, the total area under the curve is 1, so half of it is to the right of the mean. Therefore, the probability is 0.50 or 50%.
Key Concepts
Z-score CalculationStandard DeviationNormal Distribution Mean
Z-score Calculation
Understanding the Z-score in statistics is crucial for interpreting how individual points relate to a normal distribution. The Z-score, also known as a standard score, measures the number of standard deviations a data point is from the mean of the distribution. It's a way of standardizing scores on different scales so they can be compared directly.
The formula for calculating a Z-score is: \(Z = \frac{X - \mu}{\sigma}\), where \(X\) represents the value we're looking at, \(\mu\) is the mean of the distribution, and \(\sigma\) is the standard deviation. When \(X\) is equal to the mean, the Z-score is 0, indicating that the value is exactly average. In the case of our exercise, the calculated Z-score for a value of 100 is zero since it is the mean of the distribution.
The formula for calculating a Z-score is: \(Z = \frac{X - \mu}{\sigma}\), where \(X\) represents the value we're looking at, \(\mu\) is the mean of the distribution, and \(\sigma\) is the standard deviation. When \(X\) is equal to the mean, the Z-score is 0, indicating that the value is exactly average. In the case of our exercise, the calculated Z-score for a value of 100 is zero since it is the mean of the distribution.
Standard Deviation
The standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means that most numbers are close to the mean (average) of the data set, whereas a high standard deviation means that the values are spread out over a wider range.
In a normal distribution, almost all values (about 68%) lie within one standard deviation of the mean, and nearly all values (about 95%) are within two standard deviations. In the context of our example, with a mean of 100 and a standard deviation of 10, a value that falls between 90 and 110 would be within one standard deviation of the mean, which is typical for the majority of observations in a normal distribution.
In a normal distribution, almost all values (about 68%) lie within one standard deviation of the mean, and nearly all values (about 95%) are within two standard deviations. In the context of our example, with a mean of 100 and a standard deviation of 10, a value that falls between 90 and 110 would be within one standard deviation of the mean, which is typical for the majority of observations in a normal distribution.
Normal Distribution Mean
The mean of a normal distribution is quite literally the 'average' around which the entire distribution is centered. It's the peak of the bell curve, where most values cluster around. In a perfectly symmetrical normal distribution, 50% of the values fall below the mean and 50% fall above it.
In practical terms, when given a mean value, like 100 in our normal distribution example, it represents the point at which the distribution splits into two equal halves. Since the exercise asks for the probability of selecting a value at least equal to the mean, it inherently means that we're looking for the sum of all probabilities from the mean to the furthest end of the distribution, which accounts for half of the distribution's area or 50%. The mean is essentially the balance point of the distribution where the total weight of the values on either side is equal.
In practical terms, when given a mean value, like 100 in our normal distribution example, it represents the point at which the distribution splits into two equal halves. Since the exercise asks for the probability of selecting a value at least equal to the mean, it inherently means that we're looking for the sum of all probabilities from the mean to the furthest end of the distribution, which accounts for half of the distribution's area or 50%. The mean is essentially the balance point of the distribution where the total weight of the values on either side is equal.
Other exercises in this chapter
Problem 23
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