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In Exercises 5.15 to 5.17 , find the specified areas for a normal distribution. For a \(\mathrm{N}(60,10)\) distribution (a) The area to the right of 65 (b) The area to the left of 48

Short Answer

Expert verified
The area to the right of 65 is 0.3085 and the area to the left of 48 is 0.1151.

Step by step solution

01

Compute z-scores

The first step is to calculate the z-scores for 65 and 48. The formula for calculating a z-score is \( z = \frac{x - \mu}{\sigma} \) where x is the number, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. \n For 65, z-score will be \( z = \frac{65 - 60}{10} = 0.5 \) \n For 48, z-score will be \( z = \frac{48 - 60}{10} = -1.2 \)
02

Find Area to the right of 65

To find the area to the right of 65 under the curve for this normal distribution, look up the z-score of 0.5 in the Z-table (standard Normal distribution table). This gives a value of 0.6915. Since this is the area to the left of 65, subtract this value from 1 to get the area to the right of 65. So, the area to the right of 65 is \(1 - 0.6915 = 0.3085\).
03

Find Area to the left of 48

To find the area to the left of 48 under the curve for this normal distribution, look up the z-score of -1.2 in the Z-table. This gives a value of 0.1151. Since this is the area to the left of 48, we do not need further computation. So, the area to the left of 48 is 0.1151.

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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 working with normal distributions. The z-score represents the number of standard deviations a data point is from the mean of the distribution. To calculate the z-score, you use the formula \[ z = \frac{x - \mu}{\sigma} \]where
  • \( x \) represents the value for which you are finding the z-score,
  • \( \mu \) is the mean of the distribution, and
  • \( \sigma \) is the standard deviation.
Let's say you have test scores with a mean (average) of 60 and a standard deviation of 10. To find out how well you did compared to the average if you scored 65, you would plug in your score for x, the mean for \( \mu \), and the standard deviation for \( \sigma \) into the formula, which in this case would yield a z-score of 0.5. This means your score is half a standard deviation above the mean.Conversely, if you scored 48, the same process finds that your z-score is -1.2. This signals that your score is 1.2 standard deviations below the mean. Z-scores can be positive or negative; a positive z-score indicates a value above the mean, while a negative z-score signifies a value below the mean.
Standard Deviation
The idea of standard deviation is pivotal in statistics, especially with normal distributions. It measures the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation suggests that the values are spread out over a wider range.

To provide a concrete example, if a class's test scores have a standard deviation of 10, this indicates a level of variance around the average score. When interpreting the z-score, this value of 10 is crucial because it tells you how significantly a score of 65 or 48 deviates from the mean. When referring back to the z-score section, you can see how the standard deviation is used to normalize the distribution's values, making it possible to assess where a score stands relative to others.
Z-Table Lookup
After calculating the z-score, the next step is to understand its significance by using the Z-table, a chart that displays the percentage of values expected to lie below a given z-score in a standard normal distribution, which has a mean of 0 and a standard deviation of 1.

To perform a Z-table lookup, you'll first need the calculated z-score. With z-scores of 0.5 and -1.2, you would look for these values in the Z-table. For instance, a z-score of 0.5 corresponds to an area of 0.6915. This value represents the probability of a data point falling to the left of the score associated with the z-score of 0.5. To find the probability of falling to the right, as needed in the example problem, you subtract this value from 1, resulting in 0.3085.

This process allows you to calculate the likelihood or percentage of scores falling above or below a specified value within a normal distribution, making the Z-table an essential tool for statistical analysis. Its usage extends from academic evaluations to professional settings such as quality control and finance.

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