Problem 35 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

ACADEMIC TESTING The results of a calculus exam are normally distributed with a mean of \(72.3\) and a standard deviation of \(16.4\). Find the probability that a randomly chosen student's score is between 50 and 75 . If there are 82 students in the class, about how many have scores between 50 and \(75 ?\)

Step-by-Step Solution

Answer

The probability that a student's score is between 50 and 75 is 0.4767. Approximately 39 students scored between 50 and 75.

Step 1 - Understand the Problem

The problem gives a normal distribution with a mean \(\mu = 72.3\) and a standard deviation \(\sigma = 16.4\). We need to find the probability that a student's score is between 50 and 75.

Step 2 - Calculate the Z-Scores

Convert the raw scores into Z-scores using the formula: \[ Z = \frac{X - \mu}{\sigma} \] For X = 50: \[ Z_1 = \frac{50 - 72.3}{16.4} = -1.36 \] For X = 75: \[ Z_2 = \frac{75 - 72.3}{16.4} = 0.16 \]

Step 3 - Find the Probabilities

Use the standard normal distribution table to find the probabilities corresponding to the Z-scores. \[ P(Z \leq -1.36) = 0.0869 \] \[ P(Z \leq 0.16) = 0.5636 \]

Step 4 - Calculate the Probability Between Two Z-Scores

Subtract the smaller probability from the larger probability to find the probability that a student's score is between 50 and 75. \[ P(50 < X < 75) = P(Z \leq 0.16) - P(Z \leq -1.36) = 0.5636 - 0.0869 = 0.4767 \]

Step 5 - Calculate the Number of Students

Multiply the probability by the total number of students to find how many students scored between 50 and 75. \[ 0.4767 \times 82 \approx 39.1 \] So, approximately 39 students scored between 50 and 75.

Key Concepts

mean

To start understanding problems involving normal distribution, it's essential to know about the mean.
In this exercise, the mean is given as \( \mu = 72.3 \).
This means that the average score of all students who took the calculus exam is 72.3.

The mean is the central value in a data set and is calculated by summing up all individual values and dividing by the number of values.
This central tendency helps us determine how scores are distributed around this central point.

Understanding the mean allows you to grasp how data points spread out around the center, setting the stage for exploring other critical statistics like the standard deviation.

standard deviation

Next, let's talk about the standard deviation.
This value is crucial as it measures the dispersion or spread of the data points within the dataset.

In our problem, the standard deviation \( \sigma = 16.4 \) indicates that the scores typically vary by 16.4 points from the mean.

A small standard deviation means that the data points are closely clustered around the mean, while a large standard deviation suggests a wide spread of scores.
This helps us understand the variability or consistency of the exam scores.

Knowing the standard deviation allows us to asses how much individual scores deviate from the average score.

Z-score

Understanding the Z-score is critical when dealing with normal distributions.
The Z-score helps us determine how many standard deviations an individual data point is from the mean.

The formula for calculating the Z-score is: \[ Z = \frac{X - \mu}{\sigma} \] where \( X \) is the individual score, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

In our example, to find the Z-scores for scores of 50 and 75, we use the mean (72.3) and the standard deviation (16.4).
The Z-scores are calculated as: \[ Z_1 = \frac{50 - 72.3}{16.4} = -1.36 \] and \[ Z_2 = \frac{75 - 72.3}{16.4} = 0.16 \]
These Z-scores tell us how far these scores are from the mean in terms of standard deviations.

probability

Finally, let's discuss probability. In this context, probability tells us how likely it is for a randomly chosen student's score to fall within a certain range.

Using the Z-scores we calculated, we can use the standard normal distribution table to find corresponding probabilities.
For our Z-scores: \[ P(Z \leq -1.36) = 0.0869 \] and \[ P(Z \leq 0.16) = 0.5636 \]

To find the probability that a student's score is between 50 and 75, we subtract the smaller probability from the larger probability: \[ P(50 < X < 75) = 0.5636 - 0.0869 = 0.4767 \]
This indicates that there's a 47.67% chance that a student's score will fall within this range.
To find the number of students with scores between 50 and 75, we simply multiply this probability by the total number of students: \[ 0.4767 \times 82 \approx 39.1 \]
This means approximately 39 students will have scores between 50 and 75.