Non-Standard Normal Distribution: Videos & Practice Problems

The nonstandard normal distribution can be transformed into a standard normal distribution using the z-score formula: $(x-\mu )/\sigma )$. This allows for the calculation of probabilities and values associated with specific probabilities. For example, to find a value where only 5% of the population falls below it, first determine the corresponding z-score, then use $x=z\times \sigma +\mu$ to find the desired x value.

Finding Z-Scores for Non-Standard Normal Variables

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 can substitute our values into the z-score formula. Here, \( x = 10 \), \( \mu = 20 \), and \( \sigma = 5 \). Plugging in these values, we calculate:

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

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. The area to the left of a z-score of -2 corresponds to a probability of approximately 0.023. This means that there is a 2.3% chance that a randomly selected person will have a commute time of less than 10 minutes.

In summary, when dealing with nonstandard normal distributions, the process involves transforming the data into z-scores using the formula provided. This allows us to apply the techniques learned for standard normal distributions to solve problems effectively.

The arrival times of the bus Alex takes to work follow a normal distribution, with $\mu =12\min$ after the scheduled arrival rime & $\sigma =3\min$. If the bus is scheduled to arrive at Alex's work 10 min before opening, what is the probability that Alex arrives on time (i.e. the bus is less than 10 min late)?

• A. $74.86\%$
• B. $50.00\%$
• C. $25.14\%$
• D. $93.32\%$

Answer: (C)

Finding Z-Scores for Non-Standard Normal Variables Example 1

In this example, we explore the arrival times of a bus, which follow a normal distribution characterized by a mean (μ) of 12 minutes and a standard deviation (σ) of 3 minutes. We will calculate probabilities related to the bus arriving late.

First, we determine the probability that the bus arrives more than 15 minutes late. To do this, we need to find the area to the right of 15 minutes on the normal distribution curve. This can be expressed as the probability that the random variable X is greater than 15, or P(X > 15). To convert this to a standard normal distribution, we calculate the z-score using the formula:

\[ z = \frac{X - \mu}{\sigma} \]

Substituting the values, we find:

\[ z = \frac{15 - 12}{3} = 1 \]

Next, we look up the cumulative probability for z = 1 in the z-table, which gives us P(Z < 1) = 0.841. Since we need the area to the right, we calculate:

\[ P(Z > 1) = 1 - P(Z < 1) = 1 - 0.841 = 0.159 \]

This means there is a 15.9% probability that the bus will arrive more than 15 minutes late.

Next, we analyze the probability that the bus arrives between 15 and 20 minutes late, expressed as P(15 < X < 20). We need to find the z-scores for both 15 and 20 minutes. We already calculated the z-score for 15 minutes as 1. Now, we calculate the z-score for 20 minutes:

\[ z = \frac{20 - 12}{3} = \frac{8}{3} \approx 2.67 \]

Now, we need to find the probabilities for these z-scores. We look up P(Z < 2.67) in the z-table, which gives us approximately 0.9962. We already have P(Z < 1) = 0.841. To find the probability that the bus arrives between 15 and 20 minutes late, we subtract the two probabilities:

\[ P(15 < X < 20) = P(Z < 2.67) - P(Z < 1) = 0.9962 - 0.841 = 0.1552 \]

Converting this to a percentage, we find that there is a 15.52% probability that the bus will arrive between 15 and 20 minutes late.

In summary, we have learned how to calculate probabilities using the normal distribution, converting values to z-scores, and interpreting the results using the z-table. This method allows us to understand the likelihood of various outcomes based on the distribution of arrival times.

Finding Values of Non-Standard Normal Variables from Probabilities

Understanding the nonstandard normal distribution involves a process similar to that of the standard normal distribution, particularly in how we find values associated with given probabilities. To begin, we can transform any nonstandard normal variable into a standard one using the equation:

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

Here, \( z \) represents the z-score, \( x \) is the value we want to find, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. When given a probability, the first step is to determine the corresponding z-score. For instance, if we want to find an x value such that only 5% of the population has a value less than x, we first identify the z-score associated with a probability of 0.05, which is typically found using a z-table or calculator. In this case, the z-score is approximately -1.64.

Next, we can rearrange the z-score formula to solve for \( x \):

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

Substituting the known values, where \( \mu = 20 \) (the mean) and \( \sigma = 5 \) (the standard deviation), we can calculate:

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

Calculating this gives:

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

This means that approximately 5% of the population has a commute time of less than 11.8 minutes. This process illustrates how to transition from a known probability to finding a specific value in a nonstandard normal distribution, reinforcing the concept that the same principles apply as with the standard normal distribution.

The heights of adult women are approximately normally distributed with a mean of $160\mathrm{cm}$ and a standard deviation of $7\mathrm{cm}$. Find the height $x$ such that $5\%$ of women are shorter than $x$.

• A. $145.2\mathrm{cm}$
• B. $148.5\mathrm{cm}$
• C. $151.3\mathrm{cm}$
• D. $160.0\mathrm{cm}$

Answer: (B)

Here’s what students ask on this topic:

What is the difference between a standard normal distribution and a nonstandard normal distribution?

The standard normal distribution is a specific type of normal distribution with a mean (μ) of 0 and a standard deviation (σ) of 1. It is often used as a reference for calculating probabilities and z-scores. A nonstandard normal distribution, on the other hand, has a mean and standard deviation that differ from 0 and 1, respectively. For example, a distribution with μ = 20 and σ = 5 is nonstandard. To work with nonstandard distributions, we convert them to the standard normal form using the z-score formula: $z=\frac{x-\mu }{\sigma }$. This transformation allows us to use standard z-tables or calculators to find probabilities or critical values.

How do you calculate the z-score for a nonstandard normal distribution?

To calculate the z-score for a nonstandard normal distribution, use the formula: $z=\frac{x-\mu }{\sigma }$, where $x$ is the data point, $\mu$ is the mean, and $\sigma$ is the standard deviation. For example, if the mean commute time is 20 minutes ($\mu =20$) and the standard deviation is 5 minutes ($\sigma =5$), the z-score for a commute time of 10 minutes ($x=10$) is: $z=\frac{10-20}{5}=-2$. This means the value is 2 standard deviations below the mean.

How do you find the probability for a value in a nonstandard normal distribution?

To find the probability for a value in a nonstandard normal distribution, follow these steps:
1. Calculate the z-score using the formula: $z=\frac{x-\mu }{\sigma }$, where $x$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation.
2. Use a z-table or calculator to find the cumulative probability associated with the z-score.
For example, if $\mu =20$, $\sigma =5$, and $x=10$, the z-score is $-2$. Using a z-table, the probability to the left of $-2$ is approximately 0.023, or 2.3%.

How do you find a value (x) for a given probability in a nonstandard normal distribution?

To find a value ($x$) for a given probability in a nonstandard normal distribution:
1. Use a z-table or calculator to find the z-score corresponding to the given cumulative probability.
2. Apply the formula: $x=z\times \sigma +\mu$, where $z$ is the z-score, $\sigma$ is the standard deviation, and $\mu$ is the mean.
For example, if $\mu =20$, $\sigma =5$, and the probability is 5% ($0.05$), the z-score is approximately $-1.64$. Substituting, $x=-1.64\times 5+20=11.8$. Thus, 5% of the data lies below 11.8.

Why is the z-score formula important for solving nonstandard normal distribution problems?

The z-score formula, $z=\frac{x-\mu }{\sigma }$, is crucial for solving nonstandard normal distribution problems because it standardizes data points. By converting a value ($x$) from a distribution with any mean ($\mu$) and standard deviation ($\sigma$) into a z-score, we can use standard normal distribution tools like z-tables or calculators. This simplifies finding probabilities or critical values. For example, if $\mu =20$ and $\sigma =5$, the z-score for $x=10$ is $-2$. This allows us to find probabilities using standard tools, making the formula indispensable for solving such problems.