Problem 36 - Calculus for Biology and Medicine

Question

The mean weight of female students at a small college is \(123 \mathrm{lb}\), and the standard deviation is \(9 \mathrm{lb}\). If the weights are normally distributed, determine what percentage of female students weigh (a) between 110 and \(130 \mathrm{lb}\), (b) less than \(100 \mathrm{lb}\), and (c) more than \(150 \mathrm{lb}\).

Step-by-Step Solution

Answer

69.15% weigh between 110 and 130 lb, 0.52% weigh less than 100 lb, and 0.13% weigh more than 150 lb.

Step 1: Determine Z-scores

To find the percentage of students within certain weight ranges, we first compute the Z-scores for the given weights using the formula \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the weight, \( \mu = 123 \mathrm{lb} \) is the mean, and \( \sigma = 9 \mathrm{lb} \) is the standard deviation. For part (a), calculate \( Z \) for \( 110 \mathrm{lb} \) and \( 130 \mathrm{lb} \). For part (b), calculate \( Z \) for \( 100 \mathrm{lb} \). For part (c), calculate \( Z \) for \( 150 \mathrm{lb} \).

Step 2: Calculate Z-scores for Part (a)

For \( 110 \mathrm{lb} \), \( Z = \frac{110 - 123}{9} = -1.44 \).For \( 130 \mathrm{lb} \), \( Z = \frac{130 - 123}{9} = 0.78 \).

Step 3: Calculate Z-score for Part (b)

For \( 100 \mathrm{lb} \), \( Z = \frac{100 - 123}{9} = -2.56 \).

Step 4: Calculate Z-score for Part (c)

For \( 150 \mathrm{lb} \), \( Z = \frac{150 - 123}{9} = 3.00 \).

Step 5: Use Z-table to Find Probabilities

Using a Z-table, find the probability for each calculated Z-score:- For part (a), find probabilities for \( Z = -1.44 \) and \( Z = 0.78 \), then calculate the difference to get the percentage of students between the weights.- For part (b), find the probability for \( Z = -2.56 \).- For part (c), find the probability for \( Z = 3.00 \).

Step 6: Interpret Z-table Values for Part (a)

\( P(Z < -1.44) = 0.0749 \), and \( P(Z < 0.78) = 0.7823 \). The percentage of students weighing between 110 and 130 lb is \( P(110 < X < 130) = 0.7823 - 0.0749 = 0.7074 \) or 70.74%.

Step 7: Interpret Z-table Value for Part (b)

\( P(Z < -2.56) = 0.0052 \). Therefore, 0.52% of female students weigh less than 100 lb.

Step 8: Interpret Z-table Value for Part (c)

Since \( Z = 3.00 \) is off the standard Z-table, approximate the percentage as very small, around \( 0.0013 \) or 0.13%. Thus, 0.13% of students weigh more than 150 lb.

Key Concepts

Z-Score Calculation

In statistics, a Z-score is a numerical measurement that describes a value's relationship to the mean of a group of values. The Z-score calculation helps us understand how far away a particular value is from the mean in terms of standard deviations. This is significant because it allows us to standardize different sets of data and compare them on a common scale.

The formula for calculating a Z-score is given by:\[ Z = \frac{X - \mu}{\sigma} \]where:- \( X \) is the value of interest (such as a specific weight).- \( \mu \) represents the mean of the dataset (in this case, the mean weight of the female students).- \( \sigma \) is the standard deviation of the dataset.

By inputting our values into this formula, we can compute Z-scores for different weight measurements to understand their position relative to the average weight.

Standard Deviation

Standard deviation is a key statistical measure that quantifies the amount of variation or spread in a set of data values. A low standard deviation indicates that the data points are generally close to the mean, while a high standard deviation implies that the data points are spread out over a wider range.

In the context of the given problem, the standard deviation is 9 lbs. This tells us how much the weights of female students typically deviate from the average weight of 123 lbs. By understanding this, we can expect most weights to lie within a certain range from the mean, as defined by the standard deviation.

Standard deviation is crucial when conducting Z-score calculations, as it provides the scale by which we measure deviation from the mean. This makes it possible to apply the Z-score formula and interpret the normal distribution effectively.

Probability Interpretation

Interpreting probabilities derived from Z-scores helps in making conclusions about the data. In a normal distribution, probabilities indicate the likelihood of a value or range of values occurring. For instance, if you have a Z-score of 0, it means the value is exactly at the mean and corresponds to the 50th percentile.

In our problem, we calculate the probability of weights falling within certain ranges:

• For weights between 110 and 130 lbs, the combined probability from the Z-table tells us the percentage of data points within this range.
• Lower probabilities, such as for weights below 100 lbs, suggest the event is quite rare in this dataset.
• Extremely high Z-scores, like for weights above 150 lbs, show the probability is very low, indicating such values are unlikely.

Probability interpretations make the Z-scores practical, thus helping us understand the distribution of data across the range.

Z-Table Lookup

A Z-table, sometimes known as a standard normal table, is a mathematical chart used to find the probability of a standard normal distribution up to a given Z-score. By looking up the Z-score in the table, you can determine the cumulative probability for values up to that Z-score.

The process generally involves:

• Finding the Z-score from the first column, which represents the whole number and tenths decimal of the Z-score.
• Matching it with the top row that gives final decimals, helping use the exact calculated Z-score value.
• Reading the corresponding probability from the table, which gives the cumulative probability from the Z-score to the left of the curve.

Z-tables are essential for quickly determining how typical or atypical a particular observation is within a normally distributed data set. They help transform complex statistical analysis into manageable data interpretation, providing insight into the likelihood of various outcomes.