Non-Standard Normal Distribution: Videos & Practice Problems

The standard normal distribution is characterized by a mean (μ) of zero and a standard deviation (σ) of one. However, in real-world scenarios, data often does not conform to this standard form. For instance, consider a situation where the mean commute time for a group of people is 20 minutes, with a standard deviation of 5 minutes. To analyze such nonstandard distributions, we can transform the data into a standard normal distribution using a specific formula.

The key to this transformation lies in the concept of the z-score, which quantifies how many standard deviations a data point is from the mean. The formula for calculating the z-score is given by:

\( z = \frac{x - \mu}{\sigma} \)

In this formula, \( x \) represents the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. For example, if we want to find the probability that a randomly selected person commutes for less than 10 minutes, we first calculate the z-score using the provided mean and standard deviation:

\( z = \frac{10 - 20}{5} = -2 \)

This z-score of -2 indicates that 10 minutes is two standard deviations below the mean of 20 minutes. To find the probability associated with this z-score, we can refer to a z-table or use a calculator. In this case, the probability that a randomly selected person has a commute time of less than 10 minutes corresponds to the area to the left of \( z = -2 \), which is approximately 0.023.

Thus, when dealing with nonstandard normal distributions, the process involves transforming the data into z-scores, allowing us to apply the same principles and techniques used for standard normal distributions. This method simplifies the analysis and enables us to find probabilities effectively.

In this example, we explore the manufacturing of metal rods for automotive parts, focusing on their lengths, which are normally distributed with a mean (\(\mu\)) of 100 cm and a standard deviation (\(\sigma\)) of 0.8 cm. The first part of the problem involves determining the probability that a randomly selected rod is labeled defective, defined as being longer than 101.3 cm. To find this probability, we convert the length into a z-score using the formula:

\( z = \frac{x - \mu}{\sigma} \)

Substituting the values, we have:

\( z = \frac{101.3 - 100}{0.8} = 1.625 \)

This z-score indicates how many standard deviations the value is from the mean. To find the probability that a rod is longer than 101.3 cm, we look up the z-score in a z-table, which provides the area to the left of the z-score. The area to the left of \( z = 1.625 \) is approximately 0.9479. Therefore, the probability of selecting a rod longer than 101.3 cm is:

\( P(X > 101.3) = 1 - P(Z < 1.625) = 1 - 0.9479 = 0.0521 \)

This translates to a 5.21% chance that a rod is defective. Since the manufacturing plant aims for fewer than 5% of rods to be defective, the current tolerance should indeed be adjusted.

The second part of the problem examines a new tolerance range for non-defective rods, specifically between 98.7 cm and 101.3 cm. To find the probability that a rod falls within this range, we need to calculate the z-scores for both endpoints:

For \( x_1 = 98.7 \):

\( z_1 = \frac{98.7 - 100}{0.8} = -1.625 \)

We already calculated \( z_2 = 1.625 \) for \( x_2 = 101.3 \). Now, we find the probabilities associated with these z-scores:

\( P(Z < 1.625) = 0.9479 \)

\( P(Z < -1.625) = 1 - P(Z < 1.625) = 0.0521 \)

To find the probability that a rod length is between 98.7 cm and 101.3 cm, we subtract the area to the left of \( z_1 \) from the area to the left of \( z_2 \):

\( P(98.7 < X < 101.3) = P(Z < 1.625) - P(Z < -1.625) = 0.9479 - 0.0521 = 0.8958 \)

This result indicates that approximately 89.58% of rods fall within the specified range. Since the quality control manager desires at least 90% of rods to meet this tolerance, the company should consider investing in new equipment to improve quality control.

In the study of nonstandard normal distributions, a key process involves finding specific values (x) associated with given probabilities. This mirrors the approach used in standard normal distributions, where we first determine probabilities from z-scores and then reverse the process to find z-scores from probabilities. The foundational concept is transforming a nonstandard normal variable into a standard one using the z-score formula.

The z-score is calculated using the formula:

\( z = \frac{x - \mu}{\sigma} \)

where \( \mu \) is the mean and \( \sigma \) is the standard deviation. For a nonstandard normal distribution, we can rearrange this formula to find the x value associated with a specific probability. The process begins by identifying the z-score that corresponds to the given probability using a z-table or calculator.

For example, consider a scenario where the distribution of commute times for 1,000 people is approximately normal, with a mean (\( \mu \)) of 20 minutes and a standard deviation (\( \sigma \)) of 5 minutes. If we want to find the commute time (x) such that only 5% of people have a commute time less than x, we first determine the z-score associated with a probability of 0.05, which is approximately -1.64.

Next, we apply the rearranged z-score formula to find x:

\( x = z \cdot \sigma + \mu \)

Substituting the known values, we have:

\( x = (-1.64) \cdot 5 + 20 \)

Calculating this gives:

\( x = -8.2 + 20 = 11.8 \)

In this example, we explore how to determine the typical weekly sales range for a company launching a new line of smart thermostats, assuming that the weekly sales follow a normal distribution. The mean (μ) of the sales is 4,000 units, and the standard deviation (σ) is 500 units. The goal is to find the range that captures the central 80% of the weekly sales.

To visualize the distribution, we can sketch a bell curve centered at μ = 4,000. The central 80% of the distribution means we need to find two values, x₁ and x₂, that correspond to the lower and upper bounds of this range. The area in the middle of the curve represents 80%, which leaves 10% in each tail of the distribution. Therefore, the area to the left of x₁ is 0.1, and the area to the left of x₂ is 0.9.

To find the z-scores corresponding to these areas, we can refer to a z-table or use a calculator. For the area of 0.1, the z-score (z₁) is approximately -1.28. For the area of 0.9, the z-score (z₂) is +1.28, due to the symmetry of the normal distribution.

Using the z-score formula, we can calculate x₁ and x₂:

x₁ = z₁ × σ + μ = (-1.28) × 500 + 4,000 = 3,360

x₂ = z₂ × σ + μ = (1.28) × 500 + 4,000 = 4,640

Thus, the range of typical weekly sales volumes that captures the central 80% of demand is from 3,360 to 4,640 units. This method illustrates how to apply the properties of the normal distribution to forecast sales effectively.