Finding Probabilities, Z Values, and X Values with the Normal Distribution-Excel: Videos & Practice Problems

Excel functions like norm.s.dist and norm.dist calculate cumulative probabilities for standard and general normal distributions, respectively, using inputs such as z-scores, x-values, mean, and standard deviation. The complement rule helps find right-tail probabilities by subtracting left-tail probabilities from one. Conversely, norm.s.inv and norm.inv convert cumulative probabilities into z-scores or x-values, essential for interpreting normal percentiles and applying the 68–95–99.7 Rule. These tools facilitate statistical analysis, hypothesis testing, and predictive modeling by linking probabilities, z-scores, and data values within normal distributions.

Finding Probabilities from Z-Scores and X Values-Excel

To find probabilities from z scores using Excel, the =NORM.S.DIST function is essential. This function applies to the standard normal distribution, which has a mean (μ) of 0 and a standard deviation (σ) of 1. When given a z score, =NORM.S.DIST(z, TRUE) returns the cumulative probability, representing the area under the curve to the left of that z score. For example, to find the probability that a z score is less than -1.5, you would use =NORM.S.DIST(-1.5, TRUE), which yields approximately 0.07, indicating a 7% chance of observing a z score below -1.5.

When dealing with a normal distribution that is not standardized—meaning the mean and standard deviation differ from 0 and 1—you use the =NORM.DIST function. This function requires four inputs: the x value, the population mean (μ), the population standard deviation (σ), and a logical value for cumulative probability (always TRUE to find the left tail probability). For instance, if the baking time for cookies is normally distributed with μ = 11 minutes and σ = 0.76 minutes, to find the probability that a batch takes 10 minutes or less, you would use =NORM.DIST(10, 11, 0.76, TRUE). This returns about 0.09, meaning there is a 9% chance the baking time is 10 minutes or less.

To find the probability of an event occurring above a certain value (a right tail probability), such as baking time exceeding 12.5 minutes, you can apply the complement rule. Since =NORM.DIST only calculates cumulative probabilities to the left, the probability that x is greater than 12.5 is calculated as:

\[P(X > 12.5) = 1 - P(X \leq 12.5) = 1 - \text{NORM.DIST}(12.5, \mu, \sigma, \text{TRUE})\]

Using the previous example, this becomes:

\[P(X > 12.5) = 1 - \text{NORM.DIST}(12.5, 11, 0.76, \text{TRUE}) \approx 0.02\]

This means there is about a 2% chance that the baking time exceeds 12.5 minutes.

Understanding how to use these Excel functions allows for efficient calculation of probabilities from both standardized z scores and general normal distributions. The key is recognizing when to use =NORM.S.DIST for standard normal cases and =NORM.DIST for other normal distributions, as well as applying the complement rule to find right tail probabilities. These tools are invaluable for statistical analysis and interpreting data within the framework of the normal distribution.

Find the following probabilities from z-scores.
=NORM.S.DIST(Z 'TRUE')
(A)
(B) $P(Z>2.4){=}_{---}$
(C) $P(-1.3\le Z\le 2.4){=}_{---}$

• A. (A) = $0.903199515$; (B) = $0.008197536$; (C) = $0.895001979$
• B. (A) = $0.096800485$; (B) = $0.008197536$; (C) = $0.895001979$
• C. (A) = $0.903199515$; (B) = $0.991802464$ (C) = $0.895001979$
• D. (A) = $0.096800485$; (B) = $0.008197536$; (C) = $1.088602949$

Answer: (B)

Finding Probabilities from Z-Scores and X Values-Excel Example 1

The weights of granola bars produced by a snack company follow a normal distribution with a mean (μ) of 35 grams and a standard deviation (σ) of 2.7 grams. To find probabilities related to these weights, the NORM.DIST function is used, which calculates the cumulative distribution function (CDF) for a normal distribution with any mean and standard deviation.

For example, to determine the probability that a randomly selected granola bar weighs less than 33 grams, we calculate the left-tail probability using the formula =NORM.DIST(33, 35, 2.7, TRUE). This returns approximately 0.23, indicating a 23% chance that a granola bar weighs under 33 grams.

When finding the probability that a granola bar weighs more than 38 grams, we need the right-tail probability. Since NORM.DIST provides the left-tail probability, we subtract this value from 1. The calculation is =1 - NORM.DIST(38, 35, 2.7, TRUE), yielding about 0.13 or a 13% chance that a granola bar weighs over 38 grams.

To find the probability that a granola bar weighs between 33 and 38 grams, we consider the area under the normal curve between these two values. This is the total area (1) minus the sum of the left-tail probability (weight less than 33 grams) and the right-tail probability (weight greater than 38 grams). Mathematically, this is expressed as:

\[P(33 < X < 38) = 1 - P(X < 33) - P(X > 38)\]

Using the previously calculated probabilities, this results in approximately 0.64, meaning there is a 64% chance that a granola bar weighs between 33 and 38 grams.

Understanding how to manipulate the normal distribution and use cumulative distribution functions like NORM.DIST is essential for calculating probabilities involving any normally distributed variable with known mean and standard deviation. This approach allows for precise probability assessments without converting to the standard normal distribution, streamlining calculations in practical scenarios.

Finding Z-Scores and X Values from Probabilities-Excel

To convert probabilities into z scores and x values using Excel, you can use specific functions designed for the normal distribution. When working with the standard normal distribution, where the mean is zero and the standard deviation is one, the function =NORM.S.INV(probability) is used. This function requires the input probability to be a left tail probability, meaning the probability that a z score is less than a certain value. The output is the corresponding z score on the standard normal curve. If the given probability is not a left tail probability, it must be converted into one before using this function.

For example, if there is a 0.58 probability that a z score is below a certain value, entering =NORM.S.INV(0.58) in Excel returns approximately 0.202. This means the z score corresponding to the 58th percentile of the standard normal distribution is 0.202.

When dealing with a normal distribution that is not standard (i.e., with a mean μ not equal to zero and a standard deviation σ not equal to one), Excel’s =NORM.INV(probability, mean, standard_dev) function is used. This function also requires a left tail probability as the first input, followed by the population mean and standard deviation. It directly returns the x value corresponding to the given cumulative probability.

For instance, if song lengths are normally distributed with a mean of 3.56 minutes and a standard deviation of 0.32 minutes, and you want to find the length below which 33% of songs fall, you would use =NORM.INV(0.33, 3.56, 0.32). This yields approximately 3.419 minutes, indicating a 33% chance that a song is shorter than this length.

In cases where the probability given is a right tail probability (the probability that a value is greater than a certain number), it must be converted to a left tail probability using the complement rule: the left tail probability equals 1 minus the right tail probability. For example, if there is a 0.2 chance that a song is longer than a certain length, the left tail probability is 1 − 0.2 = 0.8. Using =NORM.INV(0.8, 3.56, 0.32) returns approximately 3.829 minutes, meaning there is a 20% chance that a song is longer than this length.

These Excel functions streamline the process of finding z scores and x values from probabilities, eliminating the need for manual calculations involving z score tables or standardization formulas. Understanding how to manipulate probabilities and apply these functions effectively allows for efficient analysis of normally distributed data in various contexts.

Find the following z-scores.
=NORM.S.INV(P(Z<z))
(A) There is $29\%$ chance of a z-score being below what value?
$z{=}_{---}$
(B) There is a $67\%$ chance that a z-score is larger than what value?
$z{=}_{---}$

• A. (A) $z=-0.55$; (B) $z=0.44$
• B. (A) $z=0.55$; (B) $z=-0.44$
• C. (A) $z=-0.55$ ; (B) $z=-0.44$
• D. (A) $z=1.9$; (B) $z=-0.44$

Answer: (C)

Finding Z-Scores and X Values from Probabilities-Excel Example 2

The running time of a fitness trainer for a mile follows a normal distribution with a mean (μ) of 7.65 minutes and a standard deviation (σ) of 0.39 minutes. To analyze probabilities related to this distribution, it is essential to understand how to find specific time values corresponding to given probabilities using the inverse normal distribution function.

For example, if there is a 67% chance that a run is shorter than a certain time, this represents a left-tail probability of 0.67. To find the corresponding time value (x), the inverse normal function can be applied with inputs: the cumulative probability (0.67), the mean (7.65), and the standard deviation (0.39). This calculation yields \(x = 7.822\) minutes, meaning there is a 67% probability that the running time is less than 7.822 minutes.

Similarly, if there is a 45% chance that a run is longer than a certain time, this corresponds to a right-tail probability of 0.45. To use the inverse normal function, convert this to a left-tail probability by subtracting from 1, resulting in 0.55. Using the mean and standard deviation as before, the inverse normal function gives \(x = 7.699\) minutes. Thus, there is a 45% chance that the running time exceeds 7.699 minutes.

These calculations rely on the properties of the normal distribution and the use of the inverse cumulative distribution function, often denoted as \(x = \text{norm.inv}(p, \mu, \sigma)\), where \(p\) is the cumulative probability, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. This approach allows for determining specific values of running time associated with given probabilities, which is crucial for interpreting and predicting outcomes in normally distributed data.

Here’s what students ask on this topic:

How do you find the probability of a z score using Excel's normal distribution functions?

To find the probability of a z score in Excel, use the function =NORM.S.DIST(z, TRUE), where z is your z score. This function calculates the cumulative probability from the left tail up to that z score in the standard normal distribution (mean 0, standard deviation 1). For example, if you want the probability of a z score less than -1.5, enter =NORM.S.DIST(-1.5, TRUE), and Excel will return approximately 0.07, meaning a 7% chance. This method is useful when you have standardized data and want to find the likelihood of observing a value below a certain z score.

How can you calculate the probability of an x value in a normal distribution with any mean and standard deviation using Excel?

When dealing with a normal distribution with mean $\mu$ and standard deviation $\sigma$, use Excel's =NORM.DIST(x, mean, standard_dev, TRUE) function. Here, x is the value you're interested in. This function returns the cumulative probability that a randomly selected value is less than or equal to x. For example, if baking times are normally distributed with $\mu =11$ minutes and $\sigma =0.76$ minutes, to find the probability of baking time less than 10 minutes, use =NORM.DIST(10, 11, 0.76, TRUE). Excel will give the left-tail probability, which is about 0.09 or 9% chance.

How do you find the probability of a value being greater than a certain x in Excel using the normal distribution?

Excel's NORM.DIST function calculates cumulative probabilities for values less than or equal to x, so to find the probability of a value greater than x, use the complement rule: $P(X>x)=1-P(X\le x)$. For example, if you want the probability that baking time is longer than 12.5 minutes with mean 11 and standard deviation 0.76, calculate =1 - NORM.DIST(12.5, 11, 0.76, TRUE). This returns the right-tail probability, about 0.02 or 2% chance.

How can you find a z score from a given cumulative probability using Excel?

To find a z score corresponding to a cumulative probability in the standard normal distribution, use Excel's =NORM.S.INV(probability) function. The input probability must be a left-tail probability (the area under the curve to the left of the z score). For example, if you want the z score where 58% of values lie below it, enter =NORM.S.INV(0.58), and Excel returns approximately 0.202. This means there's a 58% chance a z score is less than 0.202. If you have a right-tail probability, convert it to left-tail by subtracting from 1 before using this function.

How do you find an x value from a given cumulative probability in a normal distribution with any mean and standard deviation using Excel?

Use Excel's =NORM.INV(probability, mean, standard_dev) function to find the x value corresponding to a cumulative left-tail probability in a normal distribution with mean $\mu$ and standard deviation $\sigma$. For example, if song lengths are normally distributed with $\mu =3.56$ minutes and $\sigma =0.32$ minutes, and you want the length below which 33% of songs fall, enter =NORM.INV(0.33, 3.56, 0.32). Excel returns approximately 3.419 minutes. For right-tail probabilities, convert to left-tail by subtracting from 1 before using this function.

What is the complement rule and how is it used with Excel's normal distribution functions?

The complement rule states that the probability of an event is one minus the probability of its complement: $P\left(A\right)=1-P(\neg A)$. In Excel, since NORM.DIST and NORM.S.DIST calculate left-tail probabilities, to find the probability of a value being greater than x, use =1 - NORM.DIST(x, mean, standard_dev, TRUE). This approach helps find right-tail probabilities, which Excel functions do not directly provide. For example, to find the chance a batch takes longer than 12.5 minutes, calculate =1 - NORM.DIST(12.5, 11, 0.76, TRUE).