Problem 12 - Algebra 2 and Trigonometry

Question

In \(10-13,\) the mean and standard deviation of a normal distribution are given. Find each probability to the nearest hundredth. mean \(=50,\) standard deviation \(=2.5, P(44 \leq x \leq 50)\)

Step-by-Step Solution

Answer

The probability that \( 44 \leq x \leq 50 \) is approximately 0.49.

Step 1: Identify Given Data

We are given a normal distribution with a mean \( \mu = 50 \) and a standard deviation \( \sigma = 2.5 \). We need to find the probability that \( x \) is between 44 and 50.

Step 2: Convert to Z-Scores

To find the probability, we first convert the values 44 and 50 into Z-scores using the formula: \( Z = \frac{x - \mu}{\sigma} \). For \( x = 44 \), \( Z_{44} = \frac{44 - 50}{2.5} = -2.4 \). For \( x = 50 \), \( Z_{50} = \frac{50 - 50}{2.5} = 0 \).

Step 3: Use Standard Normal Distribution

Using the standard normal distribution table, find \( P(Z_{44} \leq Z \leq Z_{50}) \). From the table, \( P(Z \leq 0) = 0.5000 \) and \( P(Z \leq -2.4) = 0.0082 \).

Step 4: Calculate Probability

Calculate the probability of \( 44 \leq x \leq 50 \) by subtracting the smaller Z-area from the larger Z-area: \( 0.5000 - 0.0082 = 0.4918 \).

Step 5: Round the Probability

Round \( 0.4918 \) to the nearest hundredth to get \( 0.49 \).

Key Concepts

Mean

The mean, often represented by the symbol \( \mu \), is the average value in a set of data. In the case of a normal distribution, the mean is the central value around which data points are spread.
It acts as a balancing point for the data distribution.

• In our example, the mean is given as 50. This means the average of the dataset is 50.
• Any deviations from the mean reflect the tendency of data points to be higher or lower than this central value.

The mean is crucial, as it is the reference point used when calculating Z-scores, which in turn help find probabilities within the distribution.

Standard Deviation

Standard deviation, symbolized by \( \sigma \), measures the spread or dispersion of a set of values around the mean.
A smaller standard deviation indicates that data points tend to be close to the mean, while a larger standard deviation implies more spread out data points.

• In our example, the standard deviation is 2.5, meaning most data points fall within 2.5 units from the mean.
• Since the data follows a normal distribution, about 68% of data lies within one standard deviation of the mean in either direction.

Understanding standard deviation is necessary for calculating Z-scores, which indicate how many standard deviations a point is from the mean.

Z-score

A Z-score determines how many standard deviations a data point is from the mean. This is key for understanding its position in the distribution.
Z-scores are calculated using the formula:

• \( Z = \frac{x - \mu}{\sigma} \)

Here, \( x \) is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

• For example, 44 and 50 are converted to Z-scores:
  \( Z_{44} = \frac{44 - 50}{2.5} = -2.4 \)
  \( Z_{50} = \frac{50 - 50}{2.5} = 0 \)
• The Z-score tells us that 44 is 2.4 standard deviations below the mean, and 50 is at the mean.

Z-scores are vital for comparing values from different distributions.

Probability Calculation

Calculating probability in a normal distribution involves understanding areas under the curve of the standard normal distribution.
This relates to Z-scores, since they translate data points to the standard normal curve.

• In this exercise, the goal is to find \( P(44 \leq x \leq 50) \).
• Z-scores make this more manageable via standard normal distribution tables or software.

Continue by finding the area between \( Z_{44} = -2.4 \) and \( Z_{50} = 0 \):

• The probability for \( Z_{50} = 0 \) is 0.5000.
• The probability for \( Z_{44} = -2.4 \) is 0.0082.

Therefore,

• The probability of data falling between these Z-scores is:
  \( 0.5000 - 0.0082 = 0.4918 \)

Rounded to the nearest hundredth, the probability is 0.49. This calculation helps determine how likely a score falls within a specific range.