Problem 14
Question
Test Scores The scores on an exam are normally distributed, with a mean of 85 and a standard deviation of \(5 .\) What percent of the scores are from 85 to 95\(?\)
Step-by-Step Solution
Verified Answer
Approximately 34% of the scores are between 85 and 95.
1Step 1: Understand the normal distribution
In a normal distribution, the mean, median and mode all coincide. The distribution is symmetric about the mean. The standard deviation determines the spread of the distribution. In a normal distribution, about 68% of values are within ±1 standard deviation from the mean, 95% are within ±2 standard deviations, and about 99.7% are within ±3 standard deviations.
2Step 2: Identify the given values
The given values are a mean of 85, a standard deviation of 5, and we're asked to determine the percentage of values between 85 and 95.
3Step 3: Calculate the percentage
Since 95 is one standard deviation above the mean in our current normal distribution, we will use the rule stating that approximately 34% of the data falls between the mean and one standard deviation above the mean in a normal distribution. Therefore, approximately 34% of scores are between 85 and 95.
Key Concepts
Mean and Standard Deviation68-95-99.7 RuleSymmetry in Distributions
Mean and Standard Deviation
When analyzing a normal distribution, two key metrics stand out: the mean and the standard deviation. The mean is essentially the average of all the data points in a set. Imagine adding up all the scores from an exam and then dividing that total by the number of scores. That's your mean. It shows you the central tendency or the common score in the dataset.
The standard deviation, on the other hand, measures how spread out those scores are. It tells you, on average, how far each score is from the mean. If most of the scores cluster tightly around the mean, the standard deviation will be small. If the scores are spread over a wide range, the standard deviation will be larger.
For instance, in our exercise, the mean is 85, and the standard deviation is 5. This shows that the average score is 85, and typically, scores deviate by 5 points from this average, either higher or lower. A normal distribution with a low standard deviation indicates scores are close to 85, while a higher deviation suggests varied scores.
The standard deviation, on the other hand, measures how spread out those scores are. It tells you, on average, how far each score is from the mean. If most of the scores cluster tightly around the mean, the standard deviation will be small. If the scores are spread over a wide range, the standard deviation will be larger.
For instance, in our exercise, the mean is 85, and the standard deviation is 5. This shows that the average score is 85, and typically, scores deviate by 5 points from this average, either higher or lower. A normal distribution with a low standard deviation indicates scores are close to 85, while a higher deviation suggests varied scores.
68-95-99.7 Rule
The 68-95-99.7 Rule, also known as the Empirical Rule, plays a crucial role in understanding normal distributions. This rule tells us how data tends to distribute itself around the mean.
• **68%** of the data lies within one standard deviation from the mean, which includes both above and below it.
• **95%** within two standard deviations.
• **99.7%** within three standard deviations.
This rule provides a quick estimate of data spread, without complex calculations.
In our scenario with a mean of 85 and a standard deviation of 5, this means:
• One standard deviation above the mean (85 + 5), so up to 90, includes 34% of the data.
• Since we're looking at scores from 85 to 95 (which is two standard deviations), approximately 34% lies between 85 (the mean) and 90 (one standard deviation), and another 34% from 90 to 95 (two standard deviations).
The rule simplifies the process. Instead of manual calculations, it swiftly indicates that about 68% lies within one deviation (85-95).
• **68%** of the data lies within one standard deviation from the mean, which includes both above and below it.
• **95%** within two standard deviations.
• **99.7%** within three standard deviations.
This rule provides a quick estimate of data spread, without complex calculations.
In our scenario with a mean of 85 and a standard deviation of 5, this means:
• One standard deviation above the mean (85 + 5), so up to 90, includes 34% of the data.
• Since we're looking at scores from 85 to 95 (which is two standard deviations), approximately 34% lies between 85 (the mean) and 90 (one standard deviation), and another 34% from 90 to 95 (two standard deviations).
The rule simplifies the process. Instead of manual calculations, it swiftly indicates that about 68% lies within one deviation (85-95).
Symmetry in Distributions
Symmetry in normal distributions signifies balance and uniformity in data. Imagine folding your distribution graph (bell curve) in half along the mean line. If one side perfectly mirrors the other, you've got a symmetric distribution; this indicates that the mean, median, and mode are all located at the center.
This symmetry isn’t just elegant; it aids in predictions and probability assessments. It implies there's an equal amount of data on either side of the mean.
When applying this concept to our test scores with a mean of 85 and standard deviation of 5, you know that the same proportion of scores deviate above 85 as do below it, assuming perfect data with no skew.
This means knowing just half the distribution allows you to accurately predict the other half, making calculations more intuitive. Symmetry simplifies data analysis, assuring us that as long as data is normally distributed, everything mirrors around the center point (mean).
This symmetry isn’t just elegant; it aids in predictions and probability assessments. It implies there's an equal amount of data on either side of the mean.
When applying this concept to our test scores with a mean of 85 and standard deviation of 5, you know that the same proportion of scores deviate above 85 as do below it, assuming perfect data with no skew.
This means knowing just half the distribution allows you to accurately predict the other half, making calculations more intuitive. Symmetry simplifies data analysis, assuring us that as long as data is normally distributed, everything mirrors around the center point (mean).
Other exercises in this chapter
Problem 13
Suppose \(A\) and \(B\) are independent events, with \(P(A)=0.60\) and \(P(B)=0.25\) Find each probability. a. \(P(A \text { and } B)\) b. \(P(A | B)\) c. What
View solution Problem 13
Graph the probability distribution described by each function. $$ P(x)=\frac{2 x+1}{15} \text { for } x=1,2, \text { and } 3 $$
View solution Problem 14
A data set has mean 25 and standard deviation \(5 .\) Find the \(z\) -score of each value. $$ 11 $$
View solution Problem 14
Identify the outlier of each set of values. Then describe how its value affects the mean of the data. $$ 947 \quad 757 \quad 103 \quad 619 \quad 661 \quad 582 \
View solution