Problem 21 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

PEDIATRICS There are 200 children in a certain school, and the weight of the children is a random variable \(X\) that is normally distributed with mean \(\mu=80\) pounds and standard deviation \(\sigma=7\) pounds. a. How many children weigh more than 90 pounds? b. How many children weigh less than 70 pounds? c. How many children weigh exactly 80 pounds?

Step-by-Step Solution

Answer

a. About 15 childrenb. About 15 childrenc. Zero children.

Step 1: Identify the given parameters

The mean weight \(\mu\) is 80 pounds, and the standard deviation \(\sigma\) is 7 pounds. There are 200 children in total.

Step 2: Convert the weights to z-scores

Use the z-score formula: \[ Z = \frac{X - \mu}{\sigma} \]For 90 pounds: \[ Z = \frac{90 - 80}{7} = \frac{10}{7} \approx 1.43 \]For 70 pounds: \[ Z = \frac{70 - 80}{7} = \frac{-10}{7} = -1.43 \]

Step 3: Find the corresponding probabilities

Using standard normal distribution tables or a calculator, find the probabilities.For \(Z\) > 1.43: \[ P(Z > 1.43) = 1 - P(Z \leq 1.43) \approx 1 - 0.9236 = 0.0764 \]For \(Z\) < -1.43: \[ P(Z < -1.43) \approx 0.0764 \]

Step 4: Calculate the number of children

To find the number of children, multiply the total number of children by the probabilities found.For more than 90 pounds: \[200 \times 0.0764 \approx 15.28 \approx 15\] (rounding down)For less than 70 pounds: \[200 \times 0.0764 \approx 15.28 \approx 15\] (rounding down)

Step 5: Evaluate the weight of exactly 80 pounds

For a continuous distribution, the probability of a single exact value (e.g., 80 pounds) is zero because there are infinitely many possible values. Hence, \( P(X = 80) = 0 \).

Key Concepts

z-scores

Understanding z-scores is crucial for solving normal distribution problems. A z-score is a measure that describes the position of a value within a normal distribution in terms of standard deviations from the mean.
Z-scores help you compare data points from different normal distributions or find probabilities under the standard normal curve.
To compute a z-score, you use the formula: \( Z = \frac{X - \mu}{\sigma} \).
In this formula,

• \(X\) is the value you are looking at,
• \(\mu\) is the mean of the distribution,
• and \(\sigma\) is the standard deviation.

For example, to find how many children weigh more than 90 pounds, you first need to convert 90 pounds to its corresponding z-score using the given mean (80 pounds) and standard deviation (7 pounds). The z-score calculation would be: \( Z = \frac{90 - 80}{7} \approx 1.43 \). This tells us that 90 pounds is 1.43 standard deviations above the mean.

standard normal distribution

The standard normal distribution is a special case of the normal distribution with a mean of 0 and a standard deviation of 1.
When you convert a value to a z-score, you are transforming it into the standard normal distribution.
This transformation allows you to use standard normal distribution tables or calculators to find probabilities.
For instance, after obtaining the z-score of 1.43 for the 90 pounds example, you would then refer to standard normal distribution tables to find the probability associated with this z-score. These tables give you the area under the curve to the left of the given z-score. For \( Z = 1.43 \), the area is approximately 0.9236. To find the probability of a child weighing more than 90 pounds, you calculate:
\( P(Z > 1.43) = 1 - P(Z \leq 1.43) \approx 1 - 0.9236 = 0.0764 \).
This probability, 0.0764, corresponds to about 7.64%.

probability calculations

Probability calculations in the context of normal distributions help you determine the likelihood of a specific outcome.
Once you have the z-scores, you can easily find these probabilities using the standard normal distribution table or a calculator. For the given exercise, you were asked to find the number of children weighing more or less than specific weights. For weights more than 90 pounds (\( Z = 1.43 \)), the probability is:
\( P(Z > 1.43) = 1 - P(Z \leq 1.43) \approx 0.0764 \). For weights less than 70 pounds (\( Z = -1.43 \)), the probability is approximately the same, because \( P(Z < -1.43) \approx P(Z > 1.43) = 0.0764 \). To find the number of children, multiply this probability by the total number of children (200):
\( 200 \times 0.0764 \approx 15.28 \). When rounded, this means approximately 15 children weigh more than 90 pounds and about 15 children weigh less than 70 pounds.

continuous random variables

Continuous random variables are variables that can take any value within a given range. In the context of the exercise, the weights of the children can be any value, which makes it a continuous random variable.
When dealing with continuous random variables, the probability of a single exact value (e.g., weighing exactly 80 pounds) is zero. This is because there are infinitely many possible values. For example, calculating the probability that a child weighs exactly 80 pounds gives:
\( P(X = 80) = 0 \). Instead, probabilities are calculated over intervals. For instance, to find the probability of a child's weight being between 70 and 90 pounds, you would integrate the probability density function over that range. However, standard normal distribution tables simplify this by providing cumulative probabilities for given z-scores, thus making these calculations manageable.