Problem 13 - Algebra 2

Question

HEALTH For Exercises 12 and \(13,\) use the following information. The cholesterol level for adult males of a specific racial group is normally distributed with a mean of 4.8 and a standard deviation of \(0.6 .\) About how many of the 900 men in a study have cholesterol between 4.2 and 6.0\(?\)

Step-by-Step Solution

Answer

Approximately 737 men have cholesterol levels between 4.2 and 6.0.

Step 1: Understand the Problem Parameters

We have a normal distribution with a mean \( \mu = 4.8 \) and a standard deviation \( \sigma = 0.6 \). We are to find out how many among 900 men have cholesterol levels between 4.2 and 6.0.

Step 2: Convert to Z-scores

To find the probability, convert the cholesterol values to Z-scores using the formula: \[Z = \frac{X - \mu}{\sigma}\]Calculate the Z-score for 4.2: \[Z_{4.2} = \frac{4.2 - 4.8}{0.6} = -1.0\]Calculate the Z-score for 6.0: \[Z_{6.0} = \frac{6.0 - 4.8}{0.6} = 2.0\]

Step 3: Find Probabilities from Z-table

Using a standard normal distribution table (Z-table), locate the probability for these Z-scores.- For \(Z = -1.0\), the probability is approximately 0.1587.- For \(Z = 2.0\), the probability is approximately 0.9772.

Step 4: Calculate Probability Between Z-scores

The probability that a cholesterol level is between 4.2 and 6.0 is found by subtracting the two probabilities: \[P(4.2 < X < 6.0) = P(Z < 2.0) - P(Z < -1.0)\]\[P(4.2 < X < 6.0) = 0.9772 - 0.1587 = 0.8185\]

Step 5: Calculate Number of Men in the Interval

Finally, to find how many men have a cholesterol level in this interval, multiply the total number of men by the probability: \[0.8185 \times 900 = 736.65\]Round to the nearest whole number, 737.

Key Concepts

Z-scores

Z-scores help us understand how far a specific value is from the mean in a normal distribution. They are crucial when dealing with normally distributed data because they standardize different data points. This allows us to compare them easily. To calculate a Z-score, we use the formula:

• \(Z = \frac{X - \mu}{\sigma}\)

The Z-score measures the number of standard deviations a data point is from the mean. A Z-score of zero indicates that the value is exactly at the mean, whereas a positive or negative score shows how many standard deviations the value is above or below the mean respectively. By transforming the cholesterol values of 4.2 and 6.0 into Z-scores in the provided exercise, we compare these values against a standard normal distribution. This helps to find out what portion of men fall between these two cholesterol levels.

Mean and Standard Deviation

The mean and standard deviation are key concepts in understanding normal distribution. The mean \((\mu)\) is the average of all values in the distribution. It is the central point that data points pivot around in a normal distribution. In our exercise, the mean cholesterol level is 4.8.
On the other hand, the standard deviation \((\sigma)\) measures the spread or dispersion of the data points around the mean. A smaller standard deviation indicates that data points are closer to the mean, whereas a larger one suggests a wider range of values. In this particular example, the standard deviation is 0.6, meaning most cholesterol values are within 0.6 units of the mean.

• Mean = 4.8 (central average)
• Standard Deviation = 0.6 (typical deviation from the mean)

These two statistics form the foundation for calculating Z-scores, which then assists in determining probabilities and making predictions based on the normal distribution.

Probability Calculation

Probability calculation in the context of a normal distribution involves determining how likely it is for a data point to fall within a certain range. Once we have the Z-scores, we can use a Z-table to find probabilities associated with these scores.
In the exercise, after calculating the Z-scores for 4.2 and 6.0, we used a Z-table to obtain the probability that a cholesterol level is below each of these Z-scores. We found:

• For \(Z = -1.0\), the probability is approximately 0.1587.
• For \(Z = 2.0\), the probability is approximately 0.9772.

The next step is finding the probability that a value falls between these scores, calculated by subtracting the smaller probability from the larger one:

• Probability between 4.2 and 6.0 = 0.9772 - 0.1587 = 0.8185.

Finally, by multiplying this probability (0.8185) by the total number of men (900), we determine the number of men with cholesterol levels in this range, resulting in approximately 737 men.