Question

In Mrs. Espinol’s class of 25 students, the average score on the final exam is 85, with a standard deviation of seven. Approximately what percent of the class passed the final exam if a passing score is above a 70 ? A. \(70 \%\) B. \(84 \%\) C. \(96 \%\) D. \(98 \%\)

Short answer

The approximate percentage of the class that passed is 98% (option D).

Step-by-step solution

- Understanding the Problem

Determine the average score, standard deviation, and passing score. The average (mean) score is 85, the standard deviation is 7, and the passing score is 70.

- Calculate the Z-Score

Use the formula for the Z-score: \[ Z = \frac{X - \mu}{\sigma} \]where:X is the passing score (70), \(\mu\) is the mean (85), and \(\sigma\) is the standard deviation (7).Substitute the values: \[ Z = \frac{70 - 85}{7} = \frac{-15}{7} \approx -2.14 \]

- Find the Corresponding Probability

Use a Z-table to find the probability corresponding to a Z-score of -2.14. This value indicates the percentage of students who scored below 70. In the Z-table, a Z-score of -2.14 corresponds to a probability of approximately 0.0162 or 1.62%.

- Calculate the Percentage Passed

Subtract the percentage of students who scored below 70 from 100% to find the percentage who passed. \[ 100\text{%} - 1.62\text{%} = 98.38\text{%} \]

- Determine the Approximate Percentage

Round the result to match the closest option provided in the exercise. 98.38% approximates to 98%.

Key concepts

average score calculation

To understand the concept of average score calculation, let's start with the basics. The average score, also known as the mean score, is calculated by adding up all the scores and then dividing by the number of scores. In Mrs. Espinol’s class, the mean score is given as 85. This means that if you sum up all the scores of the 25 students and then divide by 25, you get 85.
For example, if the individual scores of students are somewhat scattered, the average gives a central value that represents the overall performance.
Using the formula for average:\[\text{Average} = \frac{\text{Sum of all scores}}{\text{Number of scores}}\]
This formula simplifies the process of finding the central tendency in a set of data, which in this case is the exam scores.
Knowing the average is crucial because it provides a benchmark to which you can compare individual scores and understand the group's overall performance.

standard deviation

Standard deviation is a measure of how spread out the numbers in a data set are from the mean. In simple terms, it shows how much variation exists from the average score.
For Mrs. Espinol’s class, the standard deviation is 7, indicating that most students' scores are within 7 points of the mean (85).
To calculate standard deviation, follow these steps:

• Find the mean (average) of the data set.
• Subtract the mean from each score to find the deviation of each score.
• Square each deviation.
• Find the average of these squared deviations.
• Take the square root of this average.

Using these steps, we get the standard deviation. It helps to understand the dispersion of scores and is crucial in calculating the Z-score.

Z-score

The Z-score is a statistical measure that indicates how many standard deviations a particular score is from the mean.
In Mrs. Espinol’s class, to find out how many students passed the exam (scored above 70), we calculate the Z-score. The formula for Z-score is:\[ Z = \frac{X - \text{mean}}{\text{standard deviation}} \] where X is the passing score (70), the mean is 85, and the standard deviation is 7.
Substituting the values, we get:\[ Z = \frac{70 - 85}{7} = \frac{-15}{7} \text{ which is approximately } -2.14 \]
A Z-score of -2.14 means the score of 70 is 2.14 standard deviations below the mean. To find the percentage of students scoring below 70, we use the Z-table, which gives us a probability for this Z-score.
The Z-table shows that a Z-score of -2.14 corresponds to approximately 1.62%, meaning 1.62% of students scored below 70. Therefore, the percentage of students who passed the exam is:\[ 100\text{%} - 1.62\text{%} = 98.38\text{%} \].
Rounded to the nearest whole number, 98% of students passed, matching choice D in the original exercise.